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		<id>http://www.physicsbook.gatech.edu/index.php?title=Faraday%27s_Law&amp;diff=39299</id>
		<title>Faraday&#039;s Law</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Faraday%27s_Law&amp;diff=39299"/>
		<updated>2021-12-01T09:40:50Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Claimed by Ananya Ghose Fall 2021&lt;br /&gt;
&lt;br /&gt;
Note to editors: need a computational model&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s Law &lt;br /&gt;
 focuses on how a time-varying magnetic field produces a &amp;quot;curly&amp;quot; non-Coulomb electric field, thereby inducing an emf. &lt;br /&gt;
&lt;br /&gt;
==Faraday&#039;s Law==&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s Law summarizes the ways voltage can be generated as a result of a time-varying magnetic flux. And it gives a way to connect the magnetic and electric fields in a quantifiable way (will elaborate later). Faraday&#039;s law is one of four laws in Maxwell&#039;s equations. It tells us that in the presence of a time-varying magnetic field or current (which induces a time-varying magnetic field), there is an emf with a magnitude equal to the change in magnetic flux. It serves as a succinct summary of the ways a voltage (or emf) may be generated by a changing magnetic environment. The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with the magnetic field.&lt;br /&gt;
&lt;br /&gt;
==Curly Electric Field==&lt;br /&gt;
&lt;br /&gt;
[[File:curly.jpg]]  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Faraday&#039;s Law&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
emf = &amp;lt;math&amp;gt;{\frac{-d{{&amp;amp;Phi;}}_{mag}}{dt}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where emf = &amp;lt;math&amp;gt;\oint\vec{E}_{NC}\bullet d\vec{l}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{{&amp;amp;Phi;}}_{mag}\equiv\int\vec{B}\bullet\hat{n}dA&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In other words: The emf along a round-trip is equal to the rate of change of the magnetic flux on the area encircled by the path. &lt;br /&gt;
&lt;br /&gt;
Direction: With the thumb of your right hand pointing in the direction of the &#039;&#039;-dB/dt&#039;&#039;, your fingers curl around in the direction of Enc.&lt;br /&gt;
&lt;br /&gt;
The meaning of the minus sign: If the thumb of your right hand points in the direction of &#039;&#039;-dB/dt&#039;&#039; (that is, the opposite of the direction in which the magnetic field is increasing), your fingers curl around in the direction along which the path integral of electric field is positive. Similarly, the direction of the induced current can be explained using Lenz&#039;s Law. Lenz&#039;s law states that the induced current from the non-Coulombic electric field is induced in such a way that it produces a magnetic field that opposes the first magnetic field to keep the magnetic flux constant.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Formal Version of Faraday&#039;s Law&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\oint\vec{E}_{NC}\bullet d\vec{l} = {\frac{-d}{dt}}\int\vec{B}\bullet\hat{n}dA&amp;lt;/math&amp;gt;    (sign given by right-hand rule)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Fiding the direction of the induced conventional current===&lt;br /&gt;
To find the direction of the induced conventional current by the change in the magnetic flux one must find the direction of the Non-Coulomb electric filed generated by the change in flux as the conventional current is the direction of the Non-Coulomb electric field. &lt;br /&gt;
To find the direction of the the Non-Coulomb Electic field, one must find the direction of &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;. Do this using the change in magnetic field as the basis of finding the &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
As stated previously the negative sign in front of the change in magnetic flux in the Law is a representative of Lenz&#039;s law or in other words, it&#039;s there to remind us to apply Lenz&#039;s law. Lenz&#039;s law is basically there to make us abide by the law of conservation of energy. That said, thinking in terms of conservation of energy provides the simplest way to figure out the direction of the Non-Coulomb electric field. &lt;br /&gt;
The external magnetic field induces the Non-Coulomb electric field which drives the current which in turn creates a new magnetic field which we will call the induced magnetic field. This is the magnetic field whose direction we can deduce which in turn will help us find the direction of the current. &lt;br /&gt;
The easiest way to do this is to imagine a loop of wire with and an external magnetic field perpendicular to the surface of the plane of the loop. There is a change in magnetic flux generated by the change in the magnitude of the magnetic field. vector for the initial external magnetic field and a vector for the final magnetic field. Then, draw the change in magnetic field vector, &amp;lt;math&amp;gt; \Delta \mathbf{B} &amp;lt;/math&amp;gt;, and then the negative vector of that change in magnetic field gives &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
[[File:neg_change_B_dt.jpg]]&lt;br /&gt;
&lt;br /&gt;
Pointing the thumb of your right hand in the direction of &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt; allows you to curl your fingers in the direction of &amp;lt;math&amp;gt; \mathbf{E_{NC}} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In this chapter we have seen that a changing magnetic flux induces an emf: &lt;br /&gt;
 &lt;br /&gt;
[[File:tips5.png]]&lt;br /&gt;
 &lt;br /&gt;
according to Faraday’s law of induction. For a conductor which forms a closed loop, the &lt;br /&gt;
emf sets up an induced current &#039;&#039;I =|ε|/R&#039;&#039; , where &#039;&#039;R&#039;&#039; is the resistance of the loop. To &lt;br /&gt;
compute the induced current and its direction, we follow the procedure below: &lt;br /&gt;
 &lt;br /&gt;
1. For the closed loop of area on a plane, define an area vector A and let it point in &lt;br /&gt;
the direction of your thumb, for the convenience of applying the right-hand rule later. &lt;br /&gt;
Compute the magnetic flux through the loop using&lt;br /&gt;
&lt;br /&gt;
[[File:tips4.png]]&lt;br /&gt;
&lt;br /&gt;
Determine the sign of the magnetic flux [[File:tips3.png]]&lt;br /&gt;
 &lt;br /&gt;
2. Evaluate the rate of change of magnetic flux [[File:tips2.png]] . Keep in mind that the change &lt;br /&gt;
could be caused by &lt;br /&gt;
&lt;br /&gt;
[[File:tips.png]]&lt;br /&gt;
&lt;br /&gt;
Determine the sign of [[File:tips2.png]]&lt;br /&gt;
&lt;br /&gt;
3. The sign of the induced emf is the opposite of that of [[File:tips2.png]]. The direction of the&lt;br /&gt;
induced current can be found by using Lenz’s law or right-hand rule (discussed previously).&lt;br /&gt;
&lt;br /&gt;
==Computational Model==&lt;br /&gt;
The following simulations demonstrate Faraday&#039;s Law in action. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==More on Faraday&#039;s Law==&lt;br /&gt;
&lt;br /&gt;
Moving a magnet near a coil is not the only way to induce an emf in the coil. Another way to induce emf in a coil is to bring another coil with a steady current near the first coil, thereby changing the magnetic field (and flux) surrounding the first coil, inducing an emf and a current. Also, rotating a bar magnet (or coil) near a coil produces a time-varying magnetic field in the coil since rotating the magnet changes the magnetic field in the coil. The key to inducing the emf in the second coil is to change the magnetic field around it somehow, either by bringing an object that has its own magnetic field around that coil, or changing the current in that object, changing its magnetic field.&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s law can be used to calculate motional emf as well. A bar on two current-carrying rails connected by a resistor moves along the rails, using a magnetic force to induce a current in the wire. There is a magnetic field going into the page. One way to calculate the motional emf is to use the [http://www.physicsbook.gatech.edu/Motional_Emf magnetic force], but an easier way is to use Faraday&#039;s law. &lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s law, using the change in magnetic flux, can be used to find the motional emf, where the changing factor in the magnetic flux is the area of the circuit as the bar moves, while the magnetic field is kept constant.&lt;br /&gt;
&lt;br /&gt;
[[File:motionalemf.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
[[File:solenoid.ring.jpg|center|alt=Diagram for simple example]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Adapted from the&#039;&#039; Matter &amp;amp; Interactions &#039;&#039;textbook, variation of P12 (4th ed)&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
The solenoid radius is 4 cm and the ring radius is 20 cm. B = 0.8 T inside the solenoid and approximately 0 outside the solenoid. What is the magnetic flux through the outer ring?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Solution:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because the magnetic field outside the solenoid is 0, there is no flux between the ring and solenoid. So the flux in the ring is due to the area of the solenoid, so we use the area of the solenoid to find the flux through the outer ring rather than the area of the ring itself:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \phi = BAcos(\theta)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;= (0.8 T)(\pi)(0.04 m)^2cos(0) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;= 4.02 x 10^{-3} T*m^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middle===&lt;br /&gt;
&lt;br /&gt;
[[File:rectanglecoilsolenoid.jpg|center|alt=Diagram for simple example]]&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;Adapted from the&#039;&#039; Matter &amp;amp; Interactions &#039;&#039;textbook, variation of P27 (4th ed)&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
A very long, tightly wound solenoid has a circular cross-section of radius 2 cm (only a portion of the very long solenoid is shown). The magnetic field outside the solenoid is negligible. Throughout the inside of the solenoid the magnetic field &#039;&#039;B&#039;&#039; is uniform, to the left as shown, but varying with time &#039;&#039;t: B&#039;&#039; = (.06+.02&amp;lt;math&amp;gt;t^2&amp;lt;/math&amp;gt;)T. Surrounding the circular solenoid is a loop of 7 turns of wire in the shape of a rectangle 6 cm by 12 cm. The total resistance of the 7-turn loop is 0.2 ohms.&lt;br /&gt;
&lt;br /&gt;
(a) At &#039;&#039;t&#039;&#039; = 2 s, what is the direction of the current in the 7-turn loop? Explain briefly.&lt;br /&gt;
&lt;br /&gt;
(b) At &#039;&#039;t&#039;&#039; = 2 s, what is the magnitude of the current in the 7-turn loop? Explain briefly.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Solution&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(a)&#039;&#039;&#039; The direction of the current in the loop is clockwise.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(b)&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
B(t) = (.06+.02&amp;lt;math&amp;gt;t^2&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
A = (π)(0.02 m)^2 = .00126 &amp;lt;math&amp;gt;m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}| = AN\frac{dB(t)}{dt}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}|&amp;lt;/math&amp;gt; = (.00126 &amp;lt;math&amp;gt;m^2&amp;lt;/math&amp;gt;)(7)&amp;lt;math&amp;gt;\frac{d(.06+.02t^2)}{dt}&amp;lt;/math&amp;gt; = (.00882)(.02)(2t) = .0003528t&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;At &#039;&#039;t&#039;&#039; = 2 s:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}|&amp;lt;/math&amp;gt; = .0003528(2) = .0007056 V&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;i = \frac{{&amp;amp;epsilon;}}{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;i = \frac{{.0007056 V}}{0.2 ohms}&amp;lt;/math&amp;gt; = &#039;&#039;&#039;.00353 A&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:difficultfaraday.png]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A square loop (dimensions L⇥L, total resistance R) is located halfway inside a region with uniform magnetic field B0. The magnitude of the magnetic field suddenly begins to increase linearly in time, eventually quadrupling in a time T.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(a) What current (magnitude and direction), if any, is induced in the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |emf| = \frac{-{&amp;amp;Phi;}_{B}}{&amp;amp;Delta;t} = \frac{A(B_f - B_i)}{T} = \frac{L^2(4B_o - B_o)}{T} = \frac{3B_oL^2}{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
emf = IR = &amp;lt;math&amp;gt;\frac{3B_oL^2}{TR}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(b) What net force (magnitude and direction), if any, is induced on the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; F_{top} &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; F_{bottom} &amp;lt;/math&amp;gt; cancel out.&lt;br /&gt;
&amp;lt;math&amp;gt; F_{left} &amp;lt;/math&amp;gt; = 0 because the left side is out of &amp;lt;math&amp;gt; \vec{B} &amp;lt;/math&amp;gt; region.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{F}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \vec{F}_{right} &amp;lt;/math&amp;gt; = I &amp;lt;math&amp;gt; \vec{L} \times \vec{B} = (ILB)[(\hat{y} \times - \hat{z} )] = \frac{3B_oL^2}{TR}(4B_o L)(- \hat{x}) = \frac{3{B_o}^2 L^3}{TR}(- \hat{x})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(c) What net torque (magnitude and direction), if any, is induced on the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{&amp;amp;tau;} = \vec{&amp;amp;mu;} \times \vec{B} = 0 &amp;lt;/math&amp;gt; because &amp;lt;math&amp;gt;\vec{&amp;amp;mu;}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\vec{B}&amp;lt;/math&amp;gt; are anti-parallel.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s Law is one of Maxwell&#039;s equations which describe the essence of electric and magnetic fields. Maxwell&#039;s equations effectively summarize and connect all that we have learned throughout the course of Physics 2.&lt;br /&gt;
&lt;br /&gt;
As an electrical engineer, Faraday&#039;s Law is relevant to my major.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Faraday’s Law Applications ==&lt;br /&gt;
     &lt;br /&gt;
Physics 2 content has a lot of important concepts that we as engineers can use to make our jobs easier. Whether it be a direct application of a rule or some derivation of a rule. I know I personally struggle with a concept until I get a concrete real life application that I can see the material applied in. This section of the page will discuss how Faraday’s law is applied to concepts that you as students maybe more familiar with your day to day life.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Hydroelectric Generators ==&lt;br /&gt;
     Generators create energy by transforming mechanical motion into electrical energy, but hydroelectric generators use the power of falling water to turn a large turbine which is connected to a large magnet. Around this magnet is a large coil of tightly wound wire. The conceptual creation of electricity is the same as Faraday’s Law except alternating current is being produced, but the idea that a changing magnetic field in a coil of wire induces an electromotive force is still the same. The difference is the magnetic field changes sign and flips resulting in the same thing to occur in the induced EMF. Although the calculations here are slightly more difficult the concepts are the same.&lt;br /&gt;
&lt;br /&gt;
== Transformers ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
	Transformers use a similar concept for Faraday’s Law but it’s slightly different. The job of a transformer is to either step up or step down the voltage on the power line. Transformers have a constant magnetic field associated with it due to an iron core. The power supply voltage is adjusted by altering the number of turns of wire around the iron core which in turn alters the EMF of the electricity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Cartoon of Hydroelectric Plant&lt;br /&gt;
https://etrical.files.wordpress.com/2009/12/hydrohow.jpg &lt;br /&gt;
Turbine Picture &lt;br /&gt;
http://theprepperpodcast.com/wp-content/uploads/2016/02/108-All-About-Hydro-Power-Generators-1054x500.jpg  &lt;br /&gt;
Transformer Diagram https://en.wikipedia.org/wiki/Transformer#/media/File:Transformer3d_col3.svg&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
In 1831, eletromagnetic induction was discovered by Michael Faraday.&lt;br /&gt;
&lt;br /&gt;
===Faraday&#039;s Law Experiment ===&lt;br /&gt;
&lt;br /&gt;
[[File:experiment.png]]&lt;br /&gt;
&lt;br /&gt;
Faraday showed that no current is registered in the galvanometer when bar magnet is &lt;br /&gt;
stationary with respect to the loop. However, a current is induced in the loop when a &lt;br /&gt;
relative motion exists between the bar magnet and the loop. In particular, the &lt;br /&gt;
galvanometer deflects in one direction as the magnet approaches the loop, and the &lt;br /&gt;
opposite direction as it moves away. &lt;br /&gt;
 &lt;br /&gt;
Faraday’s experiment demonstrates that an electric current is induced in the loop by &lt;br /&gt;
changing the magnetic field. The coil behaves as if it were connected to an emf source. &lt;br /&gt;
Experimentally it is found that the induced emf depends on the rate of change of &lt;br /&gt;
magnetic flux through the coil.&lt;br /&gt;
&lt;br /&gt;
Test it out yourself [https://phet.colorado.edu/en/simulation/faradays-law here]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further Readings===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;The Electric Life of Michael Faraday&#039;&#039; (2009) by Alan Hirshfield&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Electromagnetic Induction Phenomena&#039;&#039; (2012) by D. Schieber&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=KGTZPTnZBFE&lt;br /&gt;
&lt;br /&gt;
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/electroinduction.htm&lt;br /&gt;
&lt;br /&gt;
http://www.famousscientists.org/michael-faraday/&lt;br /&gt;
&lt;br /&gt;
http://www.bbc.co.uk/history/historic_figures/faraday_michael.shtml&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html&lt;br /&gt;
&lt;br /&gt;
https://files.t-square.gatech.edu/access/content/group/gtc-970b-7c13-52a7-9627-cdc3154438c6/Test%20Preparation/Old%20Test/2212_Test4_Key-1.pdf&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Faraday%27s_Law&amp;diff=39298</id>
		<title>Faraday&#039;s Law</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Faraday%27s_Law&amp;diff=39298"/>
		<updated>2021-12-01T09:40:30Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Claimed by Ananya Ghose Fall 2021&lt;br /&gt;
&lt;br /&gt;
Note to editors: need a computational model&lt;br /&gt;
Faraday&#039;s Law &lt;br /&gt;
 focuses on how a time-varying magnetic field produces a &amp;quot;curly&amp;quot; non-Coulomb electric field, thereby inducing an emf. &lt;br /&gt;
&lt;br /&gt;
==Faraday&#039;s Law==&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s Law summarizes the ways voltage can be generated as a result of a time-varying magnetic flux. And it gives a way to connect the magnetic and electric fields in a quantifiable way (will elaborate later). Faraday&#039;s law is one of four laws in Maxwell&#039;s equations. It tells us that in the presence of a time-varying magnetic field or current (which induces a time-varying magnetic field), there is an emf with a magnitude equal to the change in magnetic flux. It serves as a succinct summary of the ways a voltage (or emf) may be generated by a changing magnetic environment. The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with the magnetic field.&lt;br /&gt;
&lt;br /&gt;
==Curly Electric Field==&lt;br /&gt;
&lt;br /&gt;
[[File:curly.jpg]]  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Faraday&#039;s Law&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
emf = &amp;lt;math&amp;gt;{\frac{-d{{&amp;amp;Phi;}}_{mag}}{dt}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where emf = &amp;lt;math&amp;gt;\oint\vec{E}_{NC}\bullet d\vec{l}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{{&amp;amp;Phi;}}_{mag}\equiv\int\vec{B}\bullet\hat{n}dA&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In other words: The emf along a round-trip is equal to the rate of change of the magnetic flux on the area encircled by the path. &lt;br /&gt;
&lt;br /&gt;
Direction: With the thumb of your right hand pointing in the direction of the &#039;&#039;-dB/dt&#039;&#039;, your fingers curl around in the direction of Enc.&lt;br /&gt;
&lt;br /&gt;
The meaning of the minus sign: If the thumb of your right hand points in the direction of &#039;&#039;-dB/dt&#039;&#039; (that is, the opposite of the direction in which the magnetic field is increasing), your fingers curl around in the direction along which the path integral of electric field is positive. Similarly, the direction of the induced current can be explained using Lenz&#039;s Law. Lenz&#039;s law states that the induced current from the non-Coulombic electric field is induced in such a way that it produces a magnetic field that opposes the first magnetic field to keep the magnetic flux constant.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Formal Version of Faraday&#039;s Law&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\oint\vec{E}_{NC}\bullet d\vec{l} = {\frac{-d}{dt}}\int\vec{B}\bullet\hat{n}dA&amp;lt;/math&amp;gt;    (sign given by right-hand rule)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Fiding the direction of the induced conventional current===&lt;br /&gt;
To find the direction of the induced conventional current by the change in the magnetic flux one must find the direction of the Non-Coulomb electric filed generated by the change in flux as the conventional current is the direction of the Non-Coulomb electric field. &lt;br /&gt;
To find the direction of the the Non-Coulomb Electic field, one must find the direction of &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;. Do this using the change in magnetic field as the basis of finding the &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
As stated previously the negative sign in front of the change in magnetic flux in the Law is a representative of Lenz&#039;s law or in other words, it&#039;s there to remind us to apply Lenz&#039;s law. Lenz&#039;s law is basically there to make us abide by the law of conservation of energy. That said, thinking in terms of conservation of energy provides the simplest way to figure out the direction of the Non-Coulomb electric field. &lt;br /&gt;
The external magnetic field induces the Non-Coulomb electric field which drives the current which in turn creates a new magnetic field which we will call the induced magnetic field. This is the magnetic field whose direction we can deduce which in turn will help us find the direction of the current. &lt;br /&gt;
The easiest way to do this is to imagine a loop of wire with and an external magnetic field perpendicular to the surface of the plane of the loop. There is a change in magnetic flux generated by the change in the magnitude of the magnetic field. vector for the initial external magnetic field and a vector for the final magnetic field. Then, draw the change in magnetic field vector, &amp;lt;math&amp;gt; \Delta \mathbf{B} &amp;lt;/math&amp;gt;, and then the negative vector of that change in magnetic field gives &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
[[File:neg_change_B_dt.jpg]]&lt;br /&gt;
&lt;br /&gt;
Pointing the thumb of your right hand in the direction of &amp;lt;math&amp;gt; \frac{-dB}{dt} &amp;lt;/math&amp;gt; allows you to curl your fingers in the direction of &amp;lt;math&amp;gt; \mathbf{E_{NC}} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In this chapter we have seen that a changing magnetic flux induces an emf: &lt;br /&gt;
 &lt;br /&gt;
[[File:tips5.png]]&lt;br /&gt;
 &lt;br /&gt;
according to Faraday’s law of induction. For a conductor which forms a closed loop, the &lt;br /&gt;
emf sets up an induced current &#039;&#039;I =|ε|/R&#039;&#039; , where &#039;&#039;R&#039;&#039; is the resistance of the loop. To &lt;br /&gt;
compute the induced current and its direction, we follow the procedure below: &lt;br /&gt;
 &lt;br /&gt;
1. For the closed loop of area on a plane, define an area vector A and let it point in &lt;br /&gt;
the direction of your thumb, for the convenience of applying the right-hand rule later. &lt;br /&gt;
Compute the magnetic flux through the loop using&lt;br /&gt;
&lt;br /&gt;
[[File:tips4.png]]&lt;br /&gt;
&lt;br /&gt;
Determine the sign of the magnetic flux [[File:tips3.png]]&lt;br /&gt;
 &lt;br /&gt;
2. Evaluate the rate of change of magnetic flux [[File:tips2.png]] . Keep in mind that the change &lt;br /&gt;
could be caused by &lt;br /&gt;
&lt;br /&gt;
[[File:tips.png]]&lt;br /&gt;
&lt;br /&gt;
Determine the sign of [[File:tips2.png]]&lt;br /&gt;
&lt;br /&gt;
3. The sign of the induced emf is the opposite of that of [[File:tips2.png]]. The direction of the&lt;br /&gt;
induced current can be found by using Lenz’s law or right-hand rule (discussed previously).&lt;br /&gt;
&lt;br /&gt;
==Computational Model==&lt;br /&gt;
The following simulations demonstrate Faraday&#039;s Law in action. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==More on Faraday&#039;s Law==&lt;br /&gt;
&lt;br /&gt;
Moving a magnet near a coil is not the only way to induce an emf in the coil. Another way to induce emf in a coil is to bring another coil with a steady current near the first coil, thereby changing the magnetic field (and flux) surrounding the first coil, inducing an emf and a current. Also, rotating a bar magnet (or coil) near a coil produces a time-varying magnetic field in the coil since rotating the magnet changes the magnetic field in the coil. The key to inducing the emf in the second coil is to change the magnetic field around it somehow, either by bringing an object that has its own magnetic field around that coil, or changing the current in that object, changing its magnetic field.&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s law can be used to calculate motional emf as well. A bar on two current-carrying rails connected by a resistor moves along the rails, using a magnetic force to induce a current in the wire. There is a magnetic field going into the page. One way to calculate the motional emf is to use the [http://www.physicsbook.gatech.edu/Motional_Emf magnetic force], but an easier way is to use Faraday&#039;s law. &lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s law, using the change in magnetic flux, can be used to find the motional emf, where the changing factor in the magnetic flux is the area of the circuit as the bar moves, while the magnetic field is kept constant.&lt;br /&gt;
&lt;br /&gt;
[[File:motionalemf.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
[[File:solenoid.ring.jpg|center|alt=Diagram for simple example]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Adapted from the&#039;&#039; Matter &amp;amp; Interactions &#039;&#039;textbook, variation of P12 (4th ed)&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
The solenoid radius is 4 cm and the ring radius is 20 cm. B = 0.8 T inside the solenoid and approximately 0 outside the solenoid. What is the magnetic flux through the outer ring?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Solution:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Because the magnetic field outside the solenoid is 0, there is no flux between the ring and solenoid. So the flux in the ring is due to the area of the solenoid, so we use the area of the solenoid to find the flux through the outer ring rather than the area of the ring itself:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \phi = BAcos(\theta)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;= (0.8 T)(\pi)(0.04 m)^2cos(0) &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;= 4.02 x 10^{-3} T*m^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middle===&lt;br /&gt;
&lt;br /&gt;
[[File:rectanglecoilsolenoid.jpg|center|alt=Diagram for simple example]]&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;Adapted from the&#039;&#039; Matter &amp;amp; Interactions &#039;&#039;textbook, variation of P27 (4th ed)&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
A very long, tightly wound solenoid has a circular cross-section of radius 2 cm (only a portion of the very long solenoid is shown). The magnetic field outside the solenoid is negligible. Throughout the inside of the solenoid the magnetic field &#039;&#039;B&#039;&#039; is uniform, to the left as shown, but varying with time &#039;&#039;t: B&#039;&#039; = (.06+.02&amp;lt;math&amp;gt;t^2&amp;lt;/math&amp;gt;)T. Surrounding the circular solenoid is a loop of 7 turns of wire in the shape of a rectangle 6 cm by 12 cm. The total resistance of the 7-turn loop is 0.2 ohms.&lt;br /&gt;
&lt;br /&gt;
(a) At &#039;&#039;t&#039;&#039; = 2 s, what is the direction of the current in the 7-turn loop? Explain briefly.&lt;br /&gt;
&lt;br /&gt;
(b) At &#039;&#039;t&#039;&#039; = 2 s, what is the magnitude of the current in the 7-turn loop? Explain briefly.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Solution&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(a)&#039;&#039;&#039; The direction of the current in the loop is clockwise.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(b)&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
B(t) = (.06+.02&amp;lt;math&amp;gt;t^2&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
A = (π)(0.02 m)^2 = .00126 &amp;lt;math&amp;gt;m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}| = AN\frac{dB(t)}{dt}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}|&amp;lt;/math&amp;gt; = (.00126 &amp;lt;math&amp;gt;m^2&amp;lt;/math&amp;gt;)(7)&amp;lt;math&amp;gt;\frac{d(.06+.02t^2)}{dt}&amp;lt;/math&amp;gt; = (.00882)(.02)(2t) = .0003528t&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;At &#039;&#039;t&#039;&#039; = 2 s:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|{&amp;amp;epsilon;}|&amp;lt;/math&amp;gt; = .0003528(2) = .0007056 V&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;i = \frac{{&amp;amp;epsilon;}}{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;i = \frac{{.0007056 V}}{0.2 ohms}&amp;lt;/math&amp;gt; = &#039;&#039;&#039;.00353 A&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:difficultfaraday.png]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A square loop (dimensions L⇥L, total resistance R) is located halfway inside a region with uniform magnetic field B0. The magnitude of the magnetic field suddenly begins to increase linearly in time, eventually quadrupling in a time T.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(a) What current (magnitude and direction), if any, is induced in the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; |emf| = \frac{-{&amp;amp;Phi;}_{B}}{&amp;amp;Delta;t} = \frac{A(B_f - B_i)}{T} = \frac{L^2(4B_o - B_o)}{T} = \frac{3B_oL^2}{T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
emf = IR = &amp;lt;math&amp;gt;\frac{3B_oL^2}{TR}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(b) What net force (magnitude and direction), if any, is induced on the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; F_{top} &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; F_{bottom} &amp;lt;/math&amp;gt; cancel out.&lt;br /&gt;
&amp;lt;math&amp;gt; F_{left} &amp;lt;/math&amp;gt; = 0 because the left side is out of &amp;lt;math&amp;gt; \vec{B} &amp;lt;/math&amp;gt; region.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{F}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \vec{F}_{right} &amp;lt;/math&amp;gt; = I &amp;lt;math&amp;gt; \vec{L} \times \vec{B} = (ILB)[(\hat{y} \times - \hat{z} )] = \frac{3B_oL^2}{TR}(4B_o L)(- \hat{x}) = \frac{3{B_o}^2 L^3}{TR}(- \hat{x})&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(c) What net torque (magnitude and direction), if any, is induced on the loop at time T?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \vec{&amp;amp;tau;} = \vec{&amp;amp;mu;} \times \vec{B} = 0 &amp;lt;/math&amp;gt; because &amp;lt;math&amp;gt;\vec{&amp;amp;mu;}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\vec{B}&amp;lt;/math&amp;gt; are anti-parallel.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
Faraday&#039;s Law is one of Maxwell&#039;s equations which describe the essence of electric and magnetic fields. Maxwell&#039;s equations effectively summarize and connect all that we have learned throughout the course of Physics 2.&lt;br /&gt;
&lt;br /&gt;
As an electrical engineer, Faraday&#039;s Law is relevant to my major.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Faraday’s Law Applications ==&lt;br /&gt;
     &lt;br /&gt;
Physics 2 content has a lot of important concepts that we as engineers can use to make our jobs easier. Whether it be a direct application of a rule or some derivation of a rule. I know I personally struggle with a concept until I get a concrete real life application that I can see the material applied in. This section of the page will discuss how Faraday’s law is applied to concepts that you as students maybe more familiar with your day to day life.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Hydroelectric Generators ==&lt;br /&gt;
     Generators create energy by transforming mechanical motion into electrical energy, but hydroelectric generators use the power of falling water to turn a large turbine which is connected to a large magnet. Around this magnet is a large coil of tightly wound wire. The conceptual creation of electricity is the same as Faraday’s Law except alternating current is being produced, but the idea that a changing magnetic field in a coil of wire induces an electromotive force is still the same. The difference is the magnetic field changes sign and flips resulting in the same thing to occur in the induced EMF. Although the calculations here are slightly more difficult the concepts are the same.&lt;br /&gt;
&lt;br /&gt;
== Transformers ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
	Transformers use a similar concept for Faraday’s Law but it’s slightly different. The job of a transformer is to either step up or step down the voltage on the power line. Transformers have a constant magnetic field associated with it due to an iron core. The power supply voltage is adjusted by altering the number of turns of wire around the iron core which in turn alters the EMF of the electricity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Cartoon of Hydroelectric Plant&lt;br /&gt;
https://etrical.files.wordpress.com/2009/12/hydrohow.jpg &lt;br /&gt;
Turbine Picture &lt;br /&gt;
http://theprepperpodcast.com/wp-content/uploads/2016/02/108-All-About-Hydro-Power-Generators-1054x500.jpg  &lt;br /&gt;
Transformer Diagram https://en.wikipedia.org/wiki/Transformer#/media/File:Transformer3d_col3.svg&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
In 1831, eletromagnetic induction was discovered by Michael Faraday.&lt;br /&gt;
&lt;br /&gt;
===Faraday&#039;s Law Experiment ===&lt;br /&gt;
&lt;br /&gt;
[[File:experiment.png]]&lt;br /&gt;
&lt;br /&gt;
Faraday showed that no current is registered in the galvanometer when bar magnet is &lt;br /&gt;
stationary with respect to the loop. However, a current is induced in the loop when a &lt;br /&gt;
relative motion exists between the bar magnet and the loop. In particular, the &lt;br /&gt;
galvanometer deflects in one direction as the magnet approaches the loop, and the &lt;br /&gt;
opposite direction as it moves away. &lt;br /&gt;
 &lt;br /&gt;
Faraday’s experiment demonstrates that an electric current is induced in the loop by &lt;br /&gt;
changing the magnetic field. The coil behaves as if it were connected to an emf source. &lt;br /&gt;
Experimentally it is found that the induced emf depends on the rate of change of &lt;br /&gt;
magnetic flux through the coil.&lt;br /&gt;
&lt;br /&gt;
Test it out yourself [https://phet.colorado.edu/en/simulation/faradays-law here]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further Readings===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;The Electric Life of Michael Faraday&#039;&#039; (2009) by Alan Hirshfield&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Electromagnetic Induction Phenomena&#039;&#039; (2012) by D. Schieber&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
https://www.youtube.com/watch?v=KGTZPTnZBFE&lt;br /&gt;
&lt;br /&gt;
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/electroinduction.htm&lt;br /&gt;
&lt;br /&gt;
http://www.famousscientists.org/michael-faraday/&lt;br /&gt;
&lt;br /&gt;
http://www.bbc.co.uk/history/historic_figures/faraday_michael.shtml&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Matter and Interactions, 4th Edition&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html&lt;br /&gt;
&lt;br /&gt;
https://files.t-square.gatech.edu/access/content/group/gtc-970b-7c13-52a7-9627-cdc3154438c6/Test%20Preparation/Old%20Test/2212_Test4_Key-1.pdf&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39191</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39191"/>
		<updated>2021-11-29T01:05:25Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
1. Arago&#039;s Rotations utilize the Lorentz force created by a spinning conductive disk. Induction motors are based on Arago&#039;s Rotations and are cheaper to produce than &lt;br /&gt;
synchronous motors because they do not require permanent magnets. &lt;br /&gt;
&lt;br /&gt;
2. The calculation of the magnetic field produced by a spinning disk requires the use of Calculus, which is a is a large branch of Mathematics. &lt;br /&gt;
&lt;br /&gt;
3. Neodymium disk magnets are used in products that require strong permanent magnets like wind turbines, sensors, speakers, alarms, and more. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The magnetic field produced by a spinning disk was discovered by Francois Arago in 1824. Arago rapidly rotated disks made of conductive material beneath a magnetized needle, and the needle reacted to the magnetic field being produced by the disk. At the time, the phenomenon was difficult to explain. Several scientists investigated Arago&#039;s rotations including Joseph Louis Gay-Lussac, Peter Barlow, Charles Babbage, and John Frederick William Herschel. Babbage and Herschel worked together to conduct experiments that tested the phenomenon on different materials, both conductive and non-conductive, and with different sizes of disks and needles. They found that Arago&#039;s rotations only worked with a big enough needle and a conductive disk. Later on, Michael Faraday wrote the theory of electromagnetic induction, which explained that Arago&#039;s rotations were the product of the disk be an inducted magnet that produced a magnetic field that influenced the magnetic needle above it.&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://royalsocietypublishing.org/doi/10.1098/rstl.1825.0023 The Royal Society Volume 115]&lt;br /&gt;
*[https://www.sciencedirect.com/topics/engineering/rotating-magnetic-field ScienceDirect]&lt;br /&gt;
*[https://aapt.scitation.org/doi/full/10.1119/1.4901191?casa_token=Bg-09Hghc4EAAAAA%3A-I0qc7-QzU2vao3mc8zzY8qaKEDhNG4Hrl5V6S55Mep_oTZjOcqt3sCOL8hPr537dIID-R5cSQ American Journal of Physics, Volume 83, Issue 3]&lt;br /&gt;
*[https://www.stanfordmagnets.com/neodymium-magnets.html Stanford Magnets]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39189</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39189"/>
		<updated>2021-11-29T01:00:15Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
1. Arago&#039;s Rotations utilize the Lorentz force created by a spinning conductive disk. Induction motors are based on Arago&#039;s Rotations and are cheaper to produce than &lt;br /&gt;
synchronous motors because they do not require permanent magnets. &lt;br /&gt;
&lt;br /&gt;
2. The calculation of the magnetic field produced by a spinning disk requires the use of Calculus, which is a is a large branch of Mathematics. &lt;br /&gt;
&lt;br /&gt;
3. Neodymium disk magnets are used in products that require strong permanent magnets like wind turbines, sensors, speakers, alarms, and more. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The magnetic field produced by a spinning disk was discovered by Francois Arago in 1824. Arago rapidly rotated disks made of conductive material beneath a magnetized needle, and the needle reacted to the magnetic field being produced by the disk. At the time, the phenomenon was difficult to explain. Several scientists investigated Arago&#039;s rotations including Joseph Louis Gay-Lussac, Peter Barlow, Charles Babbage, and John Frederick William Herschel. Babbage and Herschel worked together to conduct experiments that tested the phenomenon on different materials, both conductive and non-conductive, and with different sizes of disks and needles. They found that Arago&#039;s rotations only worked with a big enough needle and a conductive disk. Later on, Michael Faraday wrote the theory of electromagnetic induction, which explained that Arago&#039;s rotations were the product of the disk be an inducted magnet that produced a magnetic field that influenced the magnetic needle above it.&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://royalsocietypublishing.org/doi/10.1098/rstl.1825.0023 The Royal Society Volume 115]&lt;br /&gt;
*[https://www.sciencedirect.com/topics/engineering/rotating-magnetic-field ScienceDirect]&lt;br /&gt;
*[https://aapt.scitation.org/doi/full/10.1119/1.4901191?casa_token=Bg-09Hghc4EAAAAA%3A-I0qc7-QzU2vao3mc8zzY8qaKEDhNG4Hrl5V6S55Mep_oTZjOcqt3sCOL8hPr537dIID-R5cSQ American Journal of Physics, Volume 83, Issue 3]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39188</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39188"/>
		<updated>2021-11-29T00:52:18Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
1. Arago&#039;s Rotations utilize the Lorentz force created by a spinning conductive disk. Induction motors are based on Arago&#039;s Rotations and are cheaper to produce than &lt;br /&gt;
synchronous motors because they do not require permanent magnets. &lt;br /&gt;
&lt;br /&gt;
2. The calculation of the magnetic field produced by a spinning disk requires the use of Calculus, which is a is a large branch of Mathematics. &lt;br /&gt;
&lt;br /&gt;
3. Neodymium disk magnets are used in products that require strong permanent magnets like wind turbines, sensors, speakers, alarms, and more. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The magnetic field produced by a spinning disk was discovered by Francois Arago in 1824. Arago rapidly rotated disks made of conductive material beneath a magnetized needle, and the needle reacted to the magnetic field being produced by the disk. At the time, the phenomenon was difficult to explain. Several scientists investigated Arago&#039;s rotations including Joseph Louis Gay-Lussac, Peter Barlow, Charles Babbage, and John Frederick William Herschel. Babbage and Herschel worked together to conduct experiments that tested the phenomenon on different materials, both conductive and non-conductive, and with different sizes of disks and needles. They found that Arago&#039;s rotations only worked with a big enough needle and a conductive disk. Later on, Michael Faraday wrote the theory of electromagnetic induction, which explained that Arago&#039;s rotations were the product of the disk be an inducted magnet that produced a magnetic field that influenced the magnetic needle above it.&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39187</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39187"/>
		<updated>2021-11-29T00:36:27Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: /* History */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
1. Arago&#039;s Rotations utilize the Lorentz force created by a spinning conductive disk. Induction motors are based on Arago&#039;s Rotations and are cheaper to produce than &lt;br /&gt;
synchronous motors because they do not require permanent magnets. &lt;br /&gt;
&lt;br /&gt;
2. The calculation of the magnetic field produced by a spinning disk requires the use of Calculus, which is a is a large branch of Mathematics. &lt;br /&gt;
&lt;br /&gt;
3. Neodymium disk magnets are used in products that require strong permanent magnets like wind turbines, sensors, speakers, alarms, and more. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The magnetic field produced by a spinning disk was discovered by Francois Arago in 1824. Arago rapidly rotated disks made of conductive material beneath a magnetized needle, and the needle reacted to the magnetic field being produced by the disk. At the time, the phenomenon was difficult to explain. Later on, Michael Faraday wrote the theory of electromagnetic induction, which explained how Arago&#039;s rotations worked.&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39177</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39177"/>
		<updated>2021-11-28T13:26:38Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
1. Arago&#039;s Rotations utilize the Lorentz force created by a spinning conductive disk. Induction motors are based on Arago&#039;s Rotations and are cheaper to produce than &lt;br /&gt;
synchronous motors because they do not require permanent magnets. &lt;br /&gt;
&lt;br /&gt;
2. The calculation of the magnetic field produced by a spinning disk requires the use of Calculus, which is a is a large branch of Mathematics. &lt;br /&gt;
&lt;br /&gt;
3. Neodymium disk magnets are used in products that require strong permanent magnets like wind turbines, sensors, speakers, alarms, and more. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
::&#039;&#039;&#039;Insert History Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39174</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39174"/>
		<updated>2021-11-27T04:45:07Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
[[File:Disk_model_preview.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
::&#039;&#039;&#039;Insert Connectedness Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
::&#039;&#039;&#039;Insert History Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Disk_model_preview.JPG&amp;diff=39173</id>
		<title>File:Disk model preview.JPG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Disk_model_preview.JPG&amp;diff=39173"/>
		<updated>2021-11-27T04:43:14Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39172</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39172"/>
		<updated>2021-11-27T04:31:42Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Model made by MaKenna Kelly (undergraduate student and an editor of this page)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
::&#039;&#039;&#039;Insert Connectedness Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
::&#039;&#039;&#039;Insert History Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39171</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39171"/>
		<updated>2021-11-27T04:28:18Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
Click the &amp;quot;1&amp;quot; on the right to view and/or edit a computational model of a disk&#039;s magnetic field. [https://www.glowscript.org/#/user/makennak/folder/MyPrograms/program/PHYS2Wiki]&lt;br /&gt;
&lt;br /&gt;
Notes about the model:&lt;br /&gt;
&lt;br /&gt;
- the disk is represented by the gray object in the center of the screen&lt;br /&gt;
&lt;br /&gt;
- the arrows represent the magnetic field created by the disk at different observation locations &lt;br /&gt;
&lt;br /&gt;
- the orange arrow represents B1 (magnetic field to the right of the disk)&lt;br /&gt;
&lt;br /&gt;
- the red arrow represents B2 (magnetic field above the disk)&lt;br /&gt;
&lt;br /&gt;
- the green arrow represents B3 (magnetic field to the left of the disk)&lt;br /&gt;
&lt;br /&gt;
- the blue arrow represents B4 (magnetic field below the disk)&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
::&#039;&#039;&#039;Insert Connectedness Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
::&#039;&#039;&#039;Insert History Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39170</id>
		<title>Magnetic Field of a Disk</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Field_of_a_Disk&amp;diff=39170"/>
		<updated>2021-11-27T02:55:10Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Claimed by MaKenna Kelly Fall 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Main Idea==&lt;br /&gt;
On this page, we will explore how to solve for the magnetic field produced by a rotating, charged, circular disk. &lt;br /&gt;
&lt;br /&gt;
First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object can produce unique electric fields. &lt;br /&gt;
&lt;br /&gt;
In order to figure out the magnetic field of a disk, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by a disk.&lt;br /&gt;
&lt;br /&gt;
===Mathematical Model===&lt;br /&gt;
Say we have a &amp;quot;flat&amp;quot; circular disk with radius &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt; and carrying a uniform surface charge density &amp;lt;math&amp;gt;\sigma&amp;lt;/math&amp;gt;. It rotates with an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt; about the z-axis. &lt;br /&gt;
&lt;br /&gt;
[[File:magfielddisk1.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
To find the magnetic field &amp;lt;math&amp;gt;B&amp;lt;/math&amp;gt; at any point &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width &amp;lt;math&amp;gt;dr&amp;lt;/math&amp;gt; and with radius &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt;, we will have an infinitesimal area for loop :&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dA = 2\pi r   dr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;A = \pi r^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And therefore have an infinitesimal charge on each loop of: &lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dQ = 2\pi \sigma rdr&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;Q = \sigma A&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This loop is rotating at an angular velocity &amp;lt;math&amp;gt;\omega&amp;lt;/math&amp;gt;, which means the charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; on our loop makes a full rotation around the axis every period,&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; T = \frac{2\pi}{\omega} &amp;lt;/math&amp;gt; seconds.&lt;br /&gt;
&lt;br /&gt;
Since a charge &amp;lt;math&amp;gt;dQ&amp;lt;/math&amp;gt; makes a full rotation every &amp;lt;math&amp;gt;T&amp;lt;/math&amp;gt; seconds, our infinitesimally thin ring is a current loop with an infinitesimal current:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt;dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{\frac{2\pi}{\omega}} = \sigma \omega rdr &amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;I = \frac{Q}{t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For this single loop, equivalent to a current &amp;lt;math&amp;gt;dI&amp;lt;/math&amp;gt;, we can calculate the field a distance &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; above the axis. Using this proof from the [[Magnetic Field of a Loop]] page, we can get:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now to find the total magnetic field for all of the concentric rings, we must integrate from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The absolute value around the &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is important because we want the field to be the same on either side of the disk, whether we are in the positive z or negative z direction.&lt;br /&gt;
&lt;br /&gt;
===Computational Model===&lt;br /&gt;
::&#039;&#039;&#039;Insert Computational Model here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
===Simple===&lt;br /&gt;
An infinitely thin disk with radius &amp;lt;math&amp;gt;R = 2 \text{ m}&amp;lt;/math&amp;gt; is spinning counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega = 15\pi \ \frac{rads}{s}&amp;lt;/math&amp;gt;. The disk has a uniform charge density &amp;lt;math&amp;gt;\rho = 2 \ \frac{C}{A}&amp;lt;/math&amp;gt;. Imagine the disk is lying flat in the xy-plane.&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude and direction of the magnetic field &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; above the axis of rotation. What about &amp;lt;math&amp;gt;5 \text{ m}&amp;lt;/math&amp;gt; below the axis of rotation?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using a convenient right-hand rule, we curl our fingers in the direction of rotation (AKA the direction of the current). the way our thumb points is the direction of the magnetic field: the (+z) direction. &lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters above the disk, we will call upon our [[Magnetic Field of a Disk# Mathematical Model| Mathematical Model]]:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = 5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(5)^2 + (2)^2}{\sqrt{(5)^2 + (2)^2}} - 2|5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field above the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::To find the direction of the magnetic field below the disk, we once again employ the right-hand rule: our fingers curl in the direction of rotation and we find the magnetic field points in the (+z) direction again.&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field &amp;lt;math&amp;gt;5&amp;lt;/math&amp;gt; meters below the disk, we will use the same formula from above:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B = \frac{\mu_0 \sigma \omega}{2}\left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where in our case:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = \rho = 2 \\&lt;br /&gt;
\omega &amp;amp; = 15\pi \\&lt;br /&gt;
R &amp;amp; = 2 \\&lt;br /&gt;
z &amp;amp; = -5 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will plug these values in to get:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B &amp;amp; = \frac{\mu_0 \times 2 \times 15\pi}{2}\left[\frac{2(-5)^2 + (2)^2}{\sqrt{(-5)^2 + (2)^2}} - 2|-5| \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{50 + 4}{\sqrt{25 + 4}} - 10 \right] \\&lt;br /&gt;
&amp;amp; = \mu_0 \times 15\pi \left[\frac{54}{\sqrt{29}} - 10 \right] \\&lt;br /&gt;
B &amp;amp; = 1.63 \times 10^{-6} \text{ T}&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the rotating, charged disk is creating a magnetic field below the disk pointing &amp;quot;up&amp;quot; with a magnitude of &amp;lt;math&amp;gt;1.63 \times 10^{-6} \text{ T}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Notice the the magnitude and direction of the magnetic field at the two points are equal. This is no coincidence. In general, if &amp;lt;math&amp;gt;z_1 = a&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;z_2 = -a&amp;lt;/math&amp;gt;, then they will have the same direction and magnitude for the magnetic field (for a rotating disk).&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A uniformly charged disk of radius &amp;lt;math&amp;gt;R = 0.01 \ \text{m}&amp;lt;/math&amp;gt; is rotating at a rate of &amp;lt;math&amp;gt;\omega = 2\pi \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt; and is losing its charge to its surroundings in such a way that the charge density of the disk is increasing by &amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;, where &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt; is time in seconds.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the magnitude of the magnetic field at time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt; along the axis of rotation, if the charge density is &amp;lt;math&amp;gt;3 \ \frac{C}{A}&amp;lt;/math&amp;gt; at this time?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can immediately use our [[Magnetic Field of a Disk#Mathematical Model| Mathematical Model]] to solve for the magnetic field:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;, where&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt; is a point along the axis of rotation&lt;br /&gt;
&lt;br /&gt;
::In our case the following values have been specified for us:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma &amp;amp; = 3 \\&lt;br /&gt;
\omega &amp;amp; = 2\pi \\&lt;br /&gt;
R &amp;amp; = 0.01 \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, we can plug these values in and simplify for an answer:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
B(z) &amp;amp; = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right] \\&lt;br /&gt;
&amp;amp; =  \frac{\mu_0 \times 3 \times 2\pi}{2} \left[\frac{2z^2 + (0.01)^2}{\sqrt{z^2 + (0.01)^2}} - 2|z| \right] \\&lt;br /&gt;
B(z) &amp;amp; = \left(1.184 \times 10^{-5}\right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]\\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnetic field at time &amp;lt;math&amp;gt;t_f = 10 \ s&amp;lt;/math&amp;gt; ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::To find the magnitude of the magnetic field, we will have to find the new charge density, since we were told it is increasing with time. To do this, we will integrate &amp;lt;math&amp;gt;\frac{d \sigma}{dt}&amp;lt;/math&amp;gt; to solve for &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\frac{d \sigma}{dt} = \pi te^{-\frac{5t^2}{\pi}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;d \sigma = \left[ \pi te^{-\frac{5t^2}{\pi}} \right] dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\int d \sigma = \int \pi te^{-\frac{5t^2}{\pi}} \ dt&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&#039;&#039;&#039;or&#039;&#039;&#039;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = \int \pi te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int te^{-\frac{5t^2}{\pi}} \ dt \\&lt;br /&gt;
&amp;amp; = \pi \int -\frac{\pi}{10} e^u du \ \text{ where } u = -\frac{5t^2}{\pi} \text{ and } du = -\frac{10}{\pi} t \ dt \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \int e^u du \\&lt;br /&gt;
&amp;amp; = -\frac{\pi^2}{10} \left[e^u \right] + C\\&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + C \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find &amp;lt;math&amp;gt;C&amp;lt;/math&amp;gt; we use the original charge density of time &amp;lt;math&amp;gt;t_0 = 0&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (0) = 3&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;3 = -\frac{\pi^2}{10} \left[e^{-\frac{5(0)^2}{\pi}} \right] + C&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;C = 3 + \frac{\pi^2}{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This gives &amp;lt;math&amp;gt;\sigma (t)&amp;lt;/math&amp;gt; as:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\begin{align}&lt;br /&gt;
\sigma (t) &amp;amp; = -\frac{\pi^2}{10} \left[e^{-\frac{5t^2}{\pi}} \right] + 3 + \frac{\pi^2}{10} \\&lt;br /&gt;
\sigma (t) &amp;amp; = \frac{\pi^2}{10} \left[1 - e^{-\frac{5t^2}{\pi}} \right] + 3  \\&lt;br /&gt;
\end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find &amp;lt;math&amp;gt;\sigma (10)&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) = \frac{\pi^2}{10} \left[1 - e^{-\frac{5(10)^2}{\pi}} \right] + 3&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
::This gives:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;\sigma (10) \approx 4 \ \frac{C}{A}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Now we can find the magnetic field generated by the rotating disk:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) =  \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B(z) = \left(1.58 \times 10^{-5} \right) \left[\frac{2z^2 + \left(1 \times 10^{-4}\right)}{\sqrt{z^2 + \left(1 \times 10^{-4}\right)}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
Two concentric infinitely thin disks are touching and rotating in opposite directions in the xy-plane. Disk 1 &amp;lt;math&amp;gt;(D_1)&amp;lt;/math&amp;gt; is the inner disk, with a radius of &amp;lt;math&amp;gt;R_1 = 1 \text{ m}&amp;lt;/math&amp;gt;; it is rotating counter-clockwise at a rate of &amp;lt;math&amp;gt;\omega_1 = 10 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Disk 2 &amp;lt;math&amp;gt;(D_2)&amp;lt;/math&amp;gt; is the outer disk, with a radius of &amp;lt;math&amp;gt;R_2 = 5 \text{ m}&amp;lt;/math&amp;gt;; it is rotating clockwise at a rate of &amp;lt;math&amp;gt;\omega_2 = 3 \ \frac{\text{rads}}{s}&amp;lt;/math&amp;gt;. Their charge density is &amp;lt;math&amp;gt;\sigma = 20 \ \frac{C}{A}&amp;lt;/math&amp;gt; See [https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;ved=2ahUKEwitqJugrJ7kAhXMnOAKHfcUAXUQjRx6BAgBEAQ&amp;amp;url=https%3A%2F%2Fwww.semanticscholar.org%2Fpaper%2FA-Genetic-Algorithm-Evolved-3D-Point-Cloud-Wegrzyn-Alexandre%2F078bbb2095973f453b1dc54d1aa37f8d481e39ce%2Ffigure%2F2&amp;amp;psig=AOvVaw2PgbcgTIJOxLJT2fwkXCQc&amp;amp;ust=1566833588588092] for clarification. Ignore any frictional or electromagnetic effects the disks may have on each other.&lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;a) What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk? What is the direction of the magnetic field created by &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; at a point along the axis of rotation and above the disk?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::Using the right-hand rule, we see the two rotating disks are creating opposing magnetic fields; &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt; creates a field pointing straight up along the axis of rotation. &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; creates a magnetic field pointing straight down along the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
:&#039;&#039;&#039;b) What is the magnitude of the net magnetic field &amp;lt;math&amp;gt;0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
::We can answer this question by solving for each magnetic field separately and then summing them (taking their directions into account). The major impediment is that the general formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_z = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::will not work for the outer disk &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;. This is because in the derivation, we integrated from &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R&amp;lt;/math&amp;gt;. For &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt;, we must integrate from &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;r = R_2&amp;lt;/math&amp;gt;, since those are its radial bounds. The formula will work as usual for &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
::Going back to the third to last step of the derivation, we have:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We will correct this for &amp;lt;math&amp;gt;D_2&amp;lt;/math&amp;gt; by replacing &amp;lt;math&amp;gt;r = 0&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;r = R_1&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \int_{R_1}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Of course the integral will be the same as before, but with different bounds:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt; B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{R_1}^{R}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B =  \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - \frac{2z^2 + {R_1}^2}{\sqrt{z^2 + {R_1}^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::Therefore, the magnetic field created by &amp;lt;math&amp;gt;D_2 \ 0.13 \text{ m}&amp;lt;/math&amp;gt; along the axis of rotation and above the disks is:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} =  \frac{\mu_0 \times 20 \times 3}{2} \left[\frac{2(0.13)^2 + (5)^2}{\sqrt{(0.13)^2 + (5)^2}} - \frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_2} = 1.5 \times 10^{-4} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::To find the magnetic field due to &amp;lt;math&amp;gt;D_1&amp;lt;/math&amp;gt;, we will use the normal formula:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \sigma \omega}{2} \left[\frac{2z^2 + R^2}{\sqrt{z^2 + R^2}} - 2|z| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = \frac{\mu_0 \times 20 \times 10}{2} \left[\frac{2(0.13)^2 + (1)^2}{\sqrt{(0.13)^2 + (1)^2}} - 2|0.13| \right]&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{D_1} = 9.62 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::We can now find the net magnetic field at &amp;lt;math&amp;gt;z = 0.13 \text{ m}&amp;lt;/math&amp;gt;. Since the up direction has been the positive direction for &amp;lt;math&amp;gt;z&amp;lt;/math&amp;gt;, we will use the up direction as the positive direction for the magnetic fields:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;B_{net} = B_{D_1} - B_{D_2} = 9.62 \times 10^{-5} \text{ T} - 1.5 \times 10^{-4} = - 5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::This means there is a magnetic field with a magnitude of &amp;lt;math&amp;gt;5.38 \times 10^{-5} \text{ T}&amp;lt;/math&amp;gt; pointing in the down direction along the axis of rotation of the disks.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
::&#039;&#039;&#039;Insert Connectedness Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
::&#039;&#039;&#039;Insert History Section here&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
===Further reading===&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;br /&gt;
*[https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.47.pdf Magnetic Fields of Spinning Disk and Sphere]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
*[[Magnetic Field of a Long Straight Wire]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Magnetic Field of a Loop]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Biot-Savart Law for Currents]]&amp;lt;br&amp;gt;&lt;br /&gt;
*[[Current]]&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
*Matter &amp;amp; Interactions Vol. II&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39159</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39159"/>
		<updated>2021-04-19T02:13:03Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Edited by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
All Khan Academy content is available for free at www.khanacademy.org&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2&lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on? &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where &amp;lt;math&amp;gt;\tau&amp;lt;/math&amp;gt; is torque, I is rotational  inertia, and &amp;lt;math&amp;gt;\alpha&amp;lt;/math&amp;gt; is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
Using rotational kinematics, &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\omega = \omega_0 + \alpha \times t&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{{\omega}_{max}}{\alpha}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
Books:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Gravitation and Inertia&#039;&#039; by Ignazio Ciufolini and John Archibald Wheeler&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Focus on Inertia&#039;&#039; by Joanne Mattern &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Inertia and Gravitation&#039;&#039; by Herbert Pfister and Markus King&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass The Physics Classroom]&lt;br /&gt;
&lt;br /&gt;
[https://www.qrg.northwestern.edu/projects/vss/docs/propulsion/2-what-is-inertia.html Northwestern University]&lt;br /&gt;
&lt;br /&gt;
[https://www.thoughtco.com/inertia-2698982 ThoughtCo]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39157</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39157"/>
		<updated>2021-04-19T02:11:31Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Edited by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2&lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on? &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where &amp;lt;math&amp;gt;\tau&amp;lt;/math&amp;gt; is torque, I is rotational  inertia, and &amp;lt;math&amp;gt;\alpha&amp;lt;/math&amp;gt; is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
Using rotational kinematics, &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\omega = \omega_0 + \alpha \times t&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{{\omega}_{max}}{\alpha}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
Books:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Gravitation and Inertia&#039;&#039; by Ignazio Ciufolini and John Archibald Wheeler&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Focus on Inertia&#039;&#039; by Joanne Mattern &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Inertia and Gravitation&#039;&#039; by Herbert Pfister and Markus King&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass The Physics Classroom]&lt;br /&gt;
&lt;br /&gt;
[https://www.qrg.northwestern.edu/projects/vss/docs/propulsion/2-what-is-inertia.html Northwestern University]&lt;br /&gt;
&lt;br /&gt;
[https://www.thoughtco.com/inertia-2698982 ThoughtCo]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39156</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39156"/>
		<updated>2021-04-19T02:11:01Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2&lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on? &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where &amp;lt;math&amp;gt;\tau&amp;lt;/math&amp;gt; is torque, I is rotational  inertia, and &amp;lt;math&amp;gt;\alpha&amp;lt;/math&amp;gt; is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
Using rotational kinematics, &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\omega = \omega_0 + \alpha \times t&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{{\omega}_{max}}{\alpha}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
Books:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Gravitation and Inertia&#039;&#039; by Ignazio Ciufolini and John Archibald Wheeler&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Focus on Inertia&#039;&#039; by Joanne Mattern &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Inertia and Gravitation&#039;&#039; by Herbert Pfister and Markus King&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass The Physics Classroom]&lt;br /&gt;
&lt;br /&gt;
[https://www.qrg.northwestern.edu/projects/vss/docs/propulsion/2-what-is-inertia.html Northwestern University]&lt;br /&gt;
&lt;br /&gt;
[https://www.thoughtco.com/inertia-2698982 ThoughtCo]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39132</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39132"/>
		<updated>2021-04-18T01:31:05Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2. &lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on?&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   Where \tau is torque, I is rotational inertia, and \alpha is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
   Using rotational kinematics, &lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;\omega = \omega_0 + \alpha \times t&amp;lt;/math&amp;gt;&lt;br /&gt;
   &lt;br /&gt;
   Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational &lt;br /&gt;
   velocity.&lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{\omega_max}{\alpha}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
Books:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Gravitation and Inertia&#039;&#039; by Ignazio Ciufolini and John Archibald Wheeler&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Focus on Inertia&#039;&#039; by Joanne Mattern &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Inertia and Gravitation&#039;&#039; by Herbert Pfister and Markus King&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass The Physics Classroom]&lt;br /&gt;
&lt;br /&gt;
[https://www.qrg.northwestern.edu/projects/vss/docs/propulsion/2-what-is-inertia.html Northwestern University]&lt;br /&gt;
&lt;br /&gt;
[https://www.thoughtco.com/inertia-2698982 ThoughtCo]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39131</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39131"/>
		<updated>2021-04-18T01:25:17Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2. &lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on?&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   Where \tau is torque, I is rotational inertia, and \alpha is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
   Using rotational kinematics, &lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;\omega = \omega_0 + \alphat&amp;lt;/math&amp;gt;&lt;br /&gt;
   &lt;br /&gt;
   Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational &lt;br /&gt;
   velocity.&lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{\omega_max}{α}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
Books:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Gravitation and Inertia&#039;&#039; by Ignazio Ciufolini and John Archibald Wheeler&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Focus on Inertia&#039;&#039; Joanne Mattern &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Inertia and Gravitation&#039;&#039; by Herbert Pfister and Markus King&lt;br /&gt;
&lt;br /&gt;
===External Links===&lt;br /&gt;
&lt;br /&gt;
[https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass The Physics Classroom]&lt;br /&gt;
&lt;br /&gt;
[https://www.qrg.northwestern.edu/projects/vss/docs/propulsion/2-what-is-inertia.html Northwestern University]&lt;br /&gt;
&lt;br /&gt;
[https://www.thoughtco.com/inertia-2698982 ThoughtCo]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39130</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39130"/>
		<updated>2021-04-18T00:42:55Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: /* Difficult */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2. &lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on?&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   Where \tau is torque, I is rotational inertia, and \alpha is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
   Using rotational kinematics, &lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;\omega = \omega_0 + \alphat&amp;lt;/math&amp;gt;&lt;br /&gt;
   &lt;br /&gt;
   Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational &lt;br /&gt;
   velocity.&lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{\omega_max}{α}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult_.png]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39129</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39129"/>
		<updated>2021-04-18T00:39:50Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A motor capable of producing a constant torque of 100 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm^2. &lt;br /&gt;
&lt;br /&gt;
a) What angular acceleration will the flywheel experience as the motor is switched on?&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\tau = I\alpha&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   Where \tau is torque, I is rotational inertia, and \alpha is angular acceleration.&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{\tau}{I}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = \frac{100 Nm}{0.1 kgm^2}&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;\alpha = 1000 rad/s^2&amp;lt;\math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b) How long will the flywheel take to reach a steady speed if starting from rest?&lt;br /&gt;
&lt;br /&gt;
   Using rotational kinematics, &lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;\omega = \omega_0 + \alphat&amp;lt;/math&amp;gt;&lt;br /&gt;
   &lt;br /&gt;
   Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational &lt;br /&gt;
   velocity.&lt;br /&gt;
   &lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{\omega_max}{α}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = \frac{150 rad/s}{1000 rad/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;math&amp;gt;t = 0.15 s&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_diificult.PNG]]&lt;br /&gt;
&lt;br /&gt;
What is the rotational inertia of the object shown above?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = m_1r_1^2 + m_2r_2^2 + ... = \sum m_ir_i^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where I is rotational inertia, m is mass, and r is distance from the axis of rotation. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = (1 kg \times 1^2 m^2) + (1 kg \times 1.5^2 m^2) + (1 kg \times 0.75^2 m^2) + (2 kg \times 0.75^2 m^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = 4.9375 kgm^2&amp;lt;/math&amp;gt;&lt;br /&gt;
​	&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_diificult_.png&amp;diff=39128</id>
		<title>File:Inertia diificult .png</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_diificult_.png&amp;diff=39128"/>
		<updated>2021-04-18T00:18:30Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39127</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39127"/>
		<updated>2021-03-26T05:05:37Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39126</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39126"/>
		<updated>2021-03-26T04:58:07Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Find the mass of an object whose acceleration is 4 m/s^2 and whose translational inertia is 20 kg.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;T = ma&amp;lt;/math&amp;gt;&lt;br /&gt;
where T is translational inertia, m is mass, and a is acceleration. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{T}{a}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = \frac{20 kg}{4 m/s^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39125</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39125"/>
		<updated>2021-03-26T04:44:20Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:Actual_inertia_1.PNG]]&lt;br /&gt;
[[File:Inertia_2.PNG]]&lt;br /&gt;
[[File:Inertia_3.PNG]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_3.PNG&amp;diff=39124</id>
		<title>File:Inertia 3.PNG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_3.PNG&amp;diff=39124"/>
		<updated>2021-03-26T04:44:03Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_2.PNG&amp;diff=39123</id>
		<title>File:Inertia 2.PNG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Inertia_2.PNG&amp;diff=39123"/>
		<updated>2021-03-26T04:42:56Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Actual_inertia_1.PNG&amp;diff=39122</id>
		<title>File:Actual inertia 1.PNG</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Actual_inertia_1.PNG&amp;diff=39122"/>
		<updated>2021-03-26T04:41:06Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39121</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39121"/>
		<updated>2021-03-26T04:39:20Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are snapshots from a similar model:&lt;br /&gt;
&lt;br /&gt;
[[File:actual inertia 1.png]]&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
*https://www.physicsclassroom.com/mmedia/newtlaws/cci.cfm&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39120</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39120"/>
		<updated>2021-03-26T04:23:37Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
Here are some snapshots from this model:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39119</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39119"/>
		<updated>2021-03-26T04:19:13Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Here is a link to a computational model that can aid in better understanding inertia. [https://demonstrations.wolfram.com/Inertia/] The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
&lt;br /&gt;
[[File:&amp;lt;script crossorigin src=&amp;quot;https://www.wolframcdn.com/notebook-embedder/0.1/wolfram-notebook-embedder.js&amp;quot;&amp;gt;&amp;lt;/script&amp;gt;&lt;br /&gt;
		&amp;lt;div id=&amp;quot;3cba53e5-166a-4821-9b82-49bef55dc29b-21957-25798&amp;quot;&amp;gt;&amp;lt;/div&amp;gt;&lt;br /&gt;
		&amp;lt;script&amp;gt;&lt;br /&gt;
			WolframNotebookEmbedder.embed(&lt;br /&gt;
				&amp;quot;https://www.wolframcloud.com/obj/demonstrations/Published/Inertia&amp;quot;,&lt;br /&gt;
				document.getElementById(&amp;quot;3cba53e5-166a-4821-9b82-49bef55dc29b-21957-25798&amp;quot;), {allowInteract: true}&lt;br /&gt;
			);&lt;br /&gt;
		&amp;lt;/script&amp;gt;]]&lt;br /&gt;
&lt;br /&gt;
Here are some snapshots from this model:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39118</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39118"/>
		<updated>2021-03-26T04:13:54Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Below is a link to a computational model that can aid in better understanding inertia. The situation shown is very similar to that described in the &amp;quot;Industry Applications&amp;quot; portion of this page. &lt;br /&gt;
[https://demonstrations.wolfram.com/Inertia/]&lt;br /&gt;
&lt;br /&gt;
Here are some screenshots from this model:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39117</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39117"/>
		<updated>2021-03-26T04:01:27Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
===Industry Application===&lt;br /&gt;
&lt;br /&gt;
Using a seat belt is one of the best methods for preserving one&#039;s life in the event of a car crash. For this reason, they are a big deal in the vehicular safety industry. &lt;br /&gt;
&lt;br /&gt;
Passengers in a moving car have the same motion as the car they are in. When that car stops abruptly, as cars often do in the event of a car crash, its passengers do not stop. Instead, they continue to have the same motion that the car did before it stopped because of inertia. This motion can often propel passengers of out the car, which increases the likelihood of death. &lt;br /&gt;
&lt;br /&gt;
Seat belts are designed to fight against inertia and hold passengers in their seats, often saving their lives in the process. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
*https://rachelpetrucciano3100.weebly.com/newtons-first-law.html#:~:text=If%20you%20were%20wearing%20a,you%20from%20being%20in%20motion.&amp;amp;text=Inertia%20is%20the%20property%20of,resist%20a%20change%20in%20motion.&amp;amp;text=Because%2C%20according%20to%20Newton&#039;s%20first,unbalanced%20force%20acts%20on%20it.&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39116</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39116"/>
		<updated>2021-03-26T03:51:52Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
===Interesting Applications===&lt;br /&gt;
&lt;br /&gt;
Scenario: Tablecloth Party Trick&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
Scenario: Turning car&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Connection to Mathematics===&lt;br /&gt;
&lt;br /&gt;
Inertia is studied using Newton&#039;s Second Law, which states that &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The &amp;lt;math&amp;gt;\frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt; in this equation stands for the derivative of momentum.&lt;br /&gt;
&lt;br /&gt;
Derivatives are core concept of a branch of Mathematics called Calculus, which studies constant motion. In this way, understanding inertia gives one a deeper understanding of Calculus, which is an integral member of the Mathematics family. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39115</id>
		<title>Inertia</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Inertia&amp;diff=39115"/>
		<updated>2021-03-25T06:25:40Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;claimed by MaKenna Kelly Spring 2021&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This page defines and describes inertia.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Inertia is the tendency of matter to resist change in [[Velocity]]. It is an inherent property of matter; the inertia of an object is directly proportional to its [[Mass]] and can in fact be used to define mass. [[Newton&#039;s First Law of Motion]] states that the velocity of an object does not change unless there is an unbalanced force acting on it. This is a consequence of the object&#039;s inertia. When a net external force acts on an object, the object will accelerate, meaning its velocity will change over time. For a given force, the rate of change of velocity is inversely proportional to the mass of an object; more massive objects have more inertia and therefore experience slower changes in velocity for a given force.&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Inertia&amp;quot; is not to be confused with &amp;quot;moment of inertia&amp;quot;, a related but different topic. The moment of inertia of an object is the tendency of an object to resist change in angular velocity. It is the rotational analogue for inertia. For more information about moments of inertia, see [[The Moments of Inertia]].&lt;br /&gt;
&lt;br /&gt;
Inertia is responsible fictitious forces, which are forces that appear to act on objects in accelerating [[Frame of Reference|frames of reference]].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
According to [[Newton&#039;s Second Law: the Momentum Principle]], &amp;lt;math&amp;gt;\vec{F}_{net} = \frac{d\vec{p}}{dt}&amp;lt;/math&amp;gt;. The more massive an object is, the less its velocity needs to change in order to achieve the same change in momentum in a given time interval.&lt;br /&gt;
&lt;br /&gt;
The other form of Newton&#039;s Second Law is &amp;lt;math&amp;gt;\vec{F}_{net} = m \vec{a}&amp;lt;/math&amp;gt;. Solving for acceleration yields &amp;lt;math&amp;gt;\vec{a} = \frac{\vec{F}_{net}}{m}&amp;lt;/math&amp;gt;. This shows the inverse relationship between mass and acceleration for a given net force.&lt;br /&gt;
&lt;br /&gt;
==Inertial Reference Frames==&lt;br /&gt;
&lt;br /&gt;
In physics, properties such as objects&#039; positions, velocities, and accelerations depend on the [[Frame of Reference]] used. Different reference frames can have different origins, and can even move and accelerate with respect to one another. According to Einstein&#039;s idea of relativity, one cannot tell the position or velocity of one&#039;s reference frame by performing experiments. For example, a person standing in an elevator with no windows cannot tell if it is moving up or down by, say, dropping a ball from shoulder height and measuring how long it takes to hit the ground. The answer will be the same regardless of the velocity of the elevator, in part because when the ball is released from shoulder height, it is travelling at the same velocity as the elevator. However, one &amp;lt;b&amp;gt;can&amp;lt;/b&amp;gt; tell the &amp;lt;i&amp;gt;acceleration&amp;lt;/i&amp;gt; of one&#039;s reference frame by performing experiments. For example, if a person standing in an elevator with no windows were to drop a ball from shoulder height, if the elevator were accelerating upwards, it would take less time for the ball to hit the ground than if the elevator were moving at a constant velocity. This is because the ground accelerates towards the ball. From the point of view of the person in the elevator, the ball would appear to violate Newton&#039;s second law; it would appear to accelerate towards the ground faster than accounted for by the force of gravity. This is because Newton&#039;s second law was written for inertial reference frames. An inertial reference frame is a reference frame travelling at a constant velocity. The physics in this course is true only in inertial reference frames, and problems in this course should be solved using inertial reference frames. The surface of the earth is usually considered an inertial reference frame; its rotations about its axis and revolutions around the sun are technically forms of acceleration, but both are so small that they have negligible effects on most aspects of motion. The definition of an inertial reference frame is one in which matter exhibits inertia; in accelerating reference frames, objects may accelerate despite the absence of external forces acting on them.&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Scenario: Tablecloth Party Trick===&lt;br /&gt;
&lt;br /&gt;
A classic demonstration of inertia is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person&#039;s arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates the inertia of the dinnerware, which has an initial velocity of 0 and resists change in velocity.&lt;br /&gt;
&lt;br /&gt;
===Scenario: Turning car===&lt;br /&gt;
&lt;br /&gt;
You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of your inertia; when the car changed direction, your body&#039;s natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Before Newton&#039;s Laws of Motion came to prominence, models of motion were based on of the observation that objects on Earth always ended up in a resting state regardless of their mass and initial velocity. Today we know this belief to be the result of [[Friction]]. Since friction is present for all macroscopic motion on earth, it was difficult for academics of the time to imagine that motion could exist without it, so it was not separated from the general motion of objects. However, this posed a problem: the perpetual motion of planets and other celestial bodies. Galileo Galilei (1564-1642) was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities.&lt;br /&gt;
&lt;br /&gt;
Galileo performed an experiment with two ramps and a bronze ball.  The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height.  He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. This led to his idea of inertia.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
[[Mass]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s First Law of Motion]]&lt;br /&gt;
&lt;br /&gt;
[[Newton&#039;s Second Law: the Momentum Principle]]&lt;br /&gt;
&lt;br /&gt;
[[The Moments of Inertia]]&lt;br /&gt;
&lt;br /&gt;
[[Acceleration]]&lt;br /&gt;
&lt;br /&gt;
[[Velocity]]&lt;br /&gt;
&lt;br /&gt;
[[Galileo Galilei]]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
*http://www.scienceclarified.com/everyday/Real-Life-Chemistry-Vol-3-Physics-Vol-1/Laws-of-Motion-Real-life-applications.html&lt;br /&gt;
*http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html&lt;br /&gt;
*http://www.physicsclassroom.com/class/newtlaws/Lesson-1/Inertia-and-Mass&lt;br /&gt;
*http://etext.library.adelaide.edu.au/a/aristotle/a8ph/&lt;br /&gt;
&lt;br /&gt;
[[Category:Properties of Matter]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Fields&amp;diff=39114</id>
		<title>Magnetic Fields</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Fields&amp;diff=39114"/>
		<updated>2021-03-25T06:21:53Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
For the practice problems below, consult your professor for the solution.&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
Recall that according to Gauss&#039; law, the electric flux through any closed surface is directly proportional to the net electric charge enclosed by that surface. Given the very direct analogy which exists between an electric charge and a magnetic &#039;monopoles&#039;, we would expect to be able to formulate a second law which states that the magnetic flux through any closed surface is directly proportional to the number of magnetic monopoles enclosed by that surface. But the problem is that magnetic monopoles don&#039;t exist. It follows that the equivalent of Gauss&#039; law for magnetic fields reduces to:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\Phi_B = \oint B \cdot dA = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Realistically, the magnetic flux though any CLOSED surface is zero. The magnetic flux through an area will be its own individual value. This rule is useful when solving for a an unknown magnetic field that&#039;s coming from a side of a surface when the other fields from the other sides are known.&lt;br /&gt;
&lt;br /&gt;
==Further Description==&lt;br /&gt;
&lt;br /&gt;
Gauss&#039;s Law for magnetism tells us that magnetic monopoles do not exist. If magnetic monopoles existed, they would be sources and sinks of the magnetic field, and therefore the right-hand side could differ from zero. Gauss&#039;s Law for magnetism is one of the four Maxwell&#039;s equations, which form the foundation for the entire theory of classical electrodynamics.&lt;br /&gt;
&lt;br /&gt;
The magnitude of the magnetic flux depends on the strength of the magnetic field, the size of the surface area, and the angle between the direction in which the surface area points and the direction of the magnetic field.&lt;br /&gt;
&lt;br /&gt;
[[File:MagFlux.gif]]&lt;br /&gt;
&lt;br /&gt;
==Why It Matters==&lt;br /&gt;
Magnetic flux is interesting because it&#039;s always zero through a closed surface because magnetic fields come from dipoles not monopoles where a particle just produces a one-sided magnetic field going in one direction, unlike point charges that generate an electric field going in one direction from every side of the particle. As you read further into this wiki book, you&#039;ll find out that the most important factor concerning the production of electricity is magnetic flux. The change of magnetic flux is what creates an electric motor force (voltage) and causes current to flow. The greater the magnetic flux and the faster it changes across a wire or a coil, the more influential it will be in determining the output.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Near the end of his career, Michael Faraday proposed that electromagnetic forces can be applied to the empty space around a conductor. This idea was rejected by his fellow scientists, and Faraday did not live to see the day when is proposition was eventually accepted by the scientific community. Faraday&#039;s concept of lines of flux generated from charged bodies and magnets provided a way to display electric and magnetic fields in a new way; that conceptual model was crucial for the successful development of 19th century technology.&lt;br /&gt;
&lt;br /&gt;
Mathematician, James Maxwell had an obsession for electricity and magnetism since Faraday&#039;s lines of force was read to the Cambridge Philosophical Society in 1855. The paper presented a simplified model of Faraday&#039;s ideas and how electricity and magnetism were related. Maxwell took all of that along with what he already knew and developed a linked set of differential equations with 20 equations in 20 variables which will eventually be concatenated into the four equations known as Maxwell&#039;s Equations one of which describes magnetic flux as written in the equation displayed above.&lt;br /&gt;
&lt;br /&gt;
==Practice Problems==&lt;br /&gt;
&lt;br /&gt;
1) There is a small bar magnet with a magnetic dipole D located at the origin (0,0,0). It&#039;s aligned with the y-axis. There is a circular disk with a radius of R facing perpendicular to the yz plane and its center is 4 meters away on the +x axis from the bar magnet. What is the magnetic flux going through the disk in terms of the given variable?&lt;br /&gt;
&lt;br /&gt;
2) Referring back to the previous problem, the disk is now tilted so that the angle between the yz plane and the surface is 30 degrees. Find the new magnetic flux in terms of the given variables.&lt;br /&gt;
&lt;br /&gt;
3) A square with side length T is directly facing the xy plane 3 meters away from a current carrying 1 meter wire (from a portion of a nearby circuit powered by a battery with an emf of U. The wire is aligned with the y axis. The magnetic flux going through the square is G. Find the resistance of the wire.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Magnetic Field]]&lt;br /&gt;
*[[Ampere-Maxwell Law]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
*[http://www.nature.com/nnano/journal/v9/n5/pdf/nnano.2014.52.pdf Observation of the magnetic ﬂux and threedimensional structure of skyrmion lattices by electron holography]&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
*[http://hsfs2.ortn.edu/myschool/mperkins/09_magnetism/notes/09%20-%20magnetism%20-%20magnetic%20flux.pdf Intro to Magnetic Flux]&lt;br /&gt;
*[http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/fluxmg.html Basics of Magnetic Flux]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
* [http://www.maxwells-equations.com maxwells-equations.com] &amp;amp;mdash; An intuitive tutorial of Maxwell&#039;s equations.&lt;br /&gt;
* Mathematical aspects of Maxwell&#039;s equation are discussed on the [http://tosio.math.toronto.edu/wiki/index.php/Main_Page Dispersive PDE Wiki].&lt;br /&gt;
&lt;br /&gt;
[[Category:Maxwell&#039;s Equations]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39113</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39113"/>
		<updated>2021-03-23T09:24:25Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background:&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance &lt;br /&gt;
and application. &lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hooke’s Law: stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not bend, and fractures comparatively quickly. Ductile materials can be easily bent and are comparatively difficult to fracture.&lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39112</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39112"/>
		<updated>2021-03-23T09:06:19Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background:&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance &lt;br /&gt;
and application. &lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hooke’s Law: stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not bend, and fractures comparatively quickly. Ductile materials can be easily bent and are comparatively difficult to fracture.&lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
As a mathematics major, the jobs that will be available to me after I graduate will expand across many fields. It is important that I have a diverse bank of knowledge on different scientific topics in order to be fit for many of these jobs. Having the vocabulary to discuss properties of materials would come in handy in a multitude of careers. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39111</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39111"/>
		<updated>2021-03-23T08:54:13Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background:&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance &lt;br /&gt;
and application. &lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hooke’s Law: stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not bend, and fractures comparatively quickly. Ductile materials can be easily bent and are comparatively difficult to fracture.&lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39110</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39110"/>
		<updated>2021-03-23T08:49:52Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background:&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance &lt;br /&gt;
and application. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hooke’s Law: stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not , and fractures comparatively quickly. Ductile materials &lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
￼&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
 Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39109</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39109"/>
		<updated>2021-03-23T08:47:33Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance and application. &lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
Hooke’s Law:  stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not , and fractures comparatively quickly. Ductile materials &lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
￼&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
 Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39108</id>
		<title>Ductility</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Ductility&amp;diff=39108"/>
		<updated>2021-03-23T08:39:00Z</updated>

		<summary type="html">&lt;p&gt;Mkelly74: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Edited by Nallammai Kannan (Fall 2017)&lt;br /&gt;
&#039;&#039;&#039;Claimed by MaKenna Kelly (Spring 2021)&#039;&#039;&#039;&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
Ductility is a solids ability to deform under tensile stress. It is similar to [[malleability]], which characterizes a materials ability to deform under an applied stress. Both of these are plastic properties of materials. While they are often similar, sometimes a materials ductility is independent from its malleability[1]. Ductility is the percentage of plastic deformation right before fracture, where plastic deformation means permanent deformation or change in shape of a solid body without fracture under the action of a sustained force[10]. Materials with low ductility are defined as brittle. Materials with metallic bonds have much higher ductility due to the mobile electrons that tend to deform, rather than fracture.  Therefore, the most common ductile materials are steel, copper, gold and aluminum. Ductility is an important property in material science and metal-working industries, where solids are deformed and molded with outside forces. Ductile materials can absorb a large amount of energy before they start to show signs of deformation, whereas brittle materials tend to show deformations and cracks relatively easily.[8]&lt;br /&gt;
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]&lt;br /&gt;
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]&lt;br /&gt;
&lt;br /&gt;
Environmental factors can also affect the ductility of a material. A temperature increase causes a material to stretch, and thus increases ductility. A temperature decreases leads to brittle and fragile behavior of the material and as such decreases ductility. Generally, low temperatures adversely affect the tensile toughness of many metals. Similarly, pressure can be used to control ductile-brittle effects. Sufficiently large superimposed pressure can convert a generally brittle material into a ductile material.[11]&lt;br /&gt;
&lt;br /&gt;
Metals like aluminum, gold, silver, and copper have a face-centered cubic crystal lattice structure, and most do not experience a shift from ductile to brittle behavior. Other metals, like iron, chromium, and tungsten, have a body-centered cubic crystal structure and experience a sharp shift in ductility. [7]  &lt;br /&gt;
&lt;br /&gt;
The Ductile - Brittle Transition Temperature (DBTT) is the temperature at which the fracture energy passes below a predetermined value (typically 40 J) or the point at which the material absorbs 15 ft*lb of impact energy during fracture[7]. The Ductile - Brittle Transition Temperature is an important consideration when determining which material to select, when said material is subjected to mechanical stresses (as shown at https://www.doitpoms.ac.uk/tlplib/BD6/images/graph0.gif). A low DBTT is integral for designs which will need to function in low temperatures [7]. &lt;br /&gt;
&lt;br /&gt;
The Ductile to Brittle Transition can also occur when dislocation motion occurs. Dislocation is defined as areas where the atoms are out of position in the crystal structure[12]. The stress required to move a dislocation depends on the atomic bonding, crystal structure, and obstacles. If the stress required to move the dislocation is too high, the metal will fail instead and form cracks or other deformations instead and the failure will be brittle[6].&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Mathematically, ductility can be defined as the fracture strain, or the tensile strain along one axis that causes a fracture to occur. Fractures range from brittle fractures (Fig. 1) to fully ductile fractures (Fig. 2), resulting in very different physical appearances associated with the different types. This can be modeled on a stress/strain curve (https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Graphics/Mechanical/Brittle-Ductile.gif) showing where fracture occurs along the graph.&lt;br /&gt;
&lt;br /&gt;
Quantitatively being able to measure ductility is important with regards to comparing ductility between different materials. Ductility can be measured through two main methods: percent elongation and percent reduction of area[5]. The formulas can be found below: (http://www.engineersedge.com/material_science/ductility.htm)&lt;br /&gt;
&lt;br /&gt;
Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length &lt;br /&gt;
       = ((Lf - Lo) / Lo) * 100&lt;br /&gt;
&lt;br /&gt;
Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section&lt;br /&gt;
       = (Decrease in Area / Original Area)&lt;br /&gt;
[2]&lt;br /&gt;
==Connectedness==&lt;br /&gt;
Stress-Strain Testing Background&lt;br /&gt;
&lt;br /&gt;
When a material experiences forces such as stress and strain, understanding its performance under such conditions is critical to performance and application. &lt;br /&gt;
&lt;br /&gt;
Stress = Force / Area &lt;br /&gt;
Strain = Delta L / L &lt;br /&gt;
&lt;br /&gt;
Hooke’s Law:  stress = Modulus of Elasticity * Strain&lt;br /&gt;
&lt;br /&gt;
Modulus of elasticity is a property of the material itself, and is a material’s resistance to elastic deformation (non-permanent). From these equations, the behavior of a material undergoing deformation may be determined. &lt;br /&gt;
&lt;br /&gt;
Based on a material’s modulus of elasticity, a material can be categorized as either “brittle” or “ductile”. A brittle material does not , and fractures comparatively quickly. Ductile materials &lt;br /&gt;
&lt;br /&gt;
￼[[File:Fracture.png]]&lt;br /&gt;
[14]&lt;br /&gt;
The common points that a ductile material experiences during deformation are identified below. One important characteristic to note is that fracture strength is typically lower than the Ultimate strength of the material. A ductile material is able to withstand additional elongation (strain) even after ultimate strength has been achieved. &lt;br /&gt;
&lt;br /&gt;
[[File:Strain.jpeg ]]&lt;br /&gt;
￼&lt;br /&gt;
[15]&lt;br /&gt;
This knowledge provides a background to how stress-strain testing is conducted. &lt;br /&gt;
 Different materials are placed inside a stress-strain testing rig. A load is continuously applied to the material until material fracture. A force transducer, combined with a position sensor, allow for the determination of key characteristics. This data allows for further examination using the equations and principles explained above.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As an engineering major, determining the correct material for components can be high risk. Knowing different materials ranges of ductility, can be integral in choosing the best option. This is especially important in materials that have a high applied tensile strength.  Significant brittle fractures can cause a lot of damage. This was seen in Liberty Ships in World War 2, where ships had hull cracks and other major defects due to the cold temperatures of the water. which caused the ship&#039;s materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
Percy Williams Bridgman&#039;s findings on tensile strength and material properties led to much of what is known about ductility, including that it is highly influenced by temperature and pressure. These findings led him to win the 1946 Nobel Prize in physics.[4]&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
*[[Malleability]]&lt;br /&gt;
*[[Temperature]]&lt;br /&gt;
*[[Pressure]]&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
[1]https://en.wikipedia.org/wiki/Ductility&lt;br /&gt;
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility&lt;br /&gt;
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg&lt;br /&gt;
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman&lt;br /&gt;
[5]http://www.engineersedge.com/material_science/ductility.htm&lt;br /&gt;
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php&lt;br /&gt;
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/&lt;br /&gt;
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf&lt;br /&gt;
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html&lt;br /&gt;
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation&lt;br /&gt;
[11]http://www.failurecriteria.com/theductile-britt.html&lt;br /&gt;
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm&lt;br /&gt;
[13]https://en.wikipedia.org/wiki/Liberty_ship&lt;br /&gt;
[14]https://www.researchgate.net/figure/Engineering-Stress-Strain-curve-for-both-Brittle-and-Ductile-material-Source_fig4_326753159&lt;br /&gt;
[15]https://www.instructables.com/Steps-to-Analyzing-a-Materials-Properties-from-its/&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Properties of Matter ]]&lt;/div&gt;</summary>
		<author><name>Mkelly74</name></author>
	</entry>
</feed>