Physics Book - User contributions [en]

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T03:12:00Z

Nkannan6:

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop. Mathematically, it can be described as an object or surface generated by revolving a closed plane around an external axis that is parallel to it so it does not intersect. Toroids are commonly used as electronic compartments since they generate a magnetic field or frequency depending on the material of the ring and the type of wire and the number of loops the wire is wound around the ring.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
<br/>
[[File:ToroidalExample.JPG]]

Solution:
<br />We determine the magnitude of the magnetic B-field by using Ampere’s law
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>
We chose a circle running through the center of the hole of the toroid for Ampere’s loop. The length of this circle is
<br /> <div style="text-align: center;"><math>{l ={2πr}}</math></div>
The total current flowing through the circle bordered by the Ampere’s loop is
<br /> <div style="text-align: center;"><math>{I_{c} = NlI = {2πr}NI}</math></div>
The magnetic field vector <math>B</math> is parallel to the vector of the length element <math>dl</math> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = B{2πr}}</math></div>
The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as
<br /> <div style="text-align: center;"><math>{B2πr ={μ_{0}NI}{2πr}}</math></div>
<br /> <div style="text-align: center;"><math>{B ={μ_{0}NI}}</math></div>
The final answer is then <math>{B ={μ_{0}NI}}</math>.

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf
*http://physicstasks.eu/1784/magnetic-field-inside-a-toroid

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T03:02:56Z

Nkannan6:

Claimed by Kevin McGorrey

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
[[File:ToroidalExample.JPG]]

Solution:
<br />We determine the magnitude of the magnetic B-field by using Ampere’s law
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>
We chose a circle running through the center of the hole of the toroid for Ampere’s loop. The length of this circle is
<br /> <div style="text-align: center;"><math>{l ={2πr}}</math></div>
The total current flowing through the circle bordered by the Ampere’s loop is
<br /> <div style="text-align: center;"><math>{I_{c} = NlI = {2πr}NI}</math></div>
The magnetic field vector <math>B</math> is parallel to the vector of the length element <math>dl</math> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = B{2πr}}</math></div>
The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as
<br /> <div style="text-align: center;"><math>{B2πr ={μ_{0}NI}{2πr}}</math></div>
<br /> <div style="text-align: center;"><math>{B ={μ_{0}NI}}</math></div>

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf
*http://physicstasks.eu/1784/magnetic-field-inside-a-toroid

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T02:36:57Z

Nkannan6:

Claimed by Kevin McGorrey

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
[[File:ToroidalExample.JPG| right|frame|none|alt=Alt text| The toroidal coil]]
<br />
Solution:
We determine the magnitude of the magnetic B-field by using Ampere’s law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>
We chose a circle running through the center of the hole of the toroid (the blue line in the figure) for Ampere’s loop. The length of this circle is:
<br /> <div style="text-align: center;"><math>l = 2\pi r<math/>
The total current flowing through the circle bordered by the Ampere’s loop is:
<br /> <div style="text-align: center;"><math>I_c = NlI = N(2\pi r)I<math/>
The magnetic field vector <math>B<math/> is parallel to the vector of the length element <math>dl<math/> all the way around the integration path. Thus we can express the right-hand side of Ampere’s law as:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = B2\pi r<math/>
The left-hand and the right-hand sides of the Ampere’s law must be equal and we can express the magnitude of the <math>B</math>-field as:
<br /> <div style="text-align: center;"><math> B2\pi r = N(2\pi r)I(\mu _0)<math/>
<br /> <div style="text-align: center;"><math> B = \mu _0NI<math/>

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf
*http://physicstasks.eu/1784/magnetic-field-inside-a-toroid

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T02:19:30Z

Nkannan6: /* Another Example */

Claimed by Kevin McGorrey

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current <math>I</math> flows through it.
[[File:ToroidalExample.JPG]]
<br>
Solution:
</br>

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

File:ToroidalExample.JPG

2018-04-19T02:17:27Z

Nkannan6:

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T02:16:53Z

Nkannan6:

Claimed by Kevin McGorrey

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Derive the formula for the magnitude of the magnetic field inside a coil that has the shape of a torus whose minor radius is much smaller than the length of the central circle. The toroidal coil has <math>N</math> turns per unit length and current I flows through it.

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

File:ToroidExample2.JPG

2018-04-19T00:35:49Z

Nkannan6: Toroid Example 2

Toroid Example 2

Magnetic Field of a Toroid Using Ampere's Law

2018-04-19T00:33:09Z

Nkannan6:

Claimed by Kevin McGorrey

This page explains how to use Ampere's Law to solve for the magnetic field of a toroid.

==Magnetic Field of a Toroid using Ampere's Law==

Using Ampere's Law simplifies finding the magnetic field of a toroid since using the Biot-Savart law would be extremely difficult due to having to integrate over all the current elements in the toroid.
===Geometry of a Toroid===
[[File:Geometry Fig.JPG |frame|right|alt=Alt text| A toroid's geometry.]]
A toroid is essentially a solenoid whose ends meet. It is shaped like a doughnut (the base shape does not need to be a circle; it can be a triangle, square, quadrilateral, etc.), is symmetrical around an axis of rotation, and contains ''N'' loops around a closed, circular path with a radius of ''r'' inside of its loop.

===A Mathematical Model===
[[File:Computation Fig.JPG | left|frame|none|alt=Alt text| A toroid's view from above.]]
First we start with solving the path integral from Ampere's law:
<br /> <div style="text-align: center;"><math>{\oint\,\vec{B}•d\vec{l} = μ_{0}∑I_{inside&ensp;path}}</math></div>

<br />The magnetic field, <math>{\vec{B}}</math>, is, due to the symmetry of a toroid, constant in magnitude and always tangential to the circular path (i.e. parallel to <math>{d\vec{l}}</math>). The path of a toroid is circular, so <math>{\oint\,d\vec{l}}</math> is equal to <math>{2πr}</math>. Therefore, the path integral of the magnetic field is equal to <math>{B2πr}</math>. The amount of current piercing the soap film (i.e. <math>{∑I_{inside&ensp;path}}</math>) is <math>NI</math>, where <math>N</math> is the number of piercings (i.e. turns in the coil) and <math>I</math> is the current. Ampere's law is now this:
<br /> <div style="text-align: center;"><math>{B2πr = μ_{0}NI}</math></div>
<br /> From this, we can solve for the magnetic field for a toroid:
<br /> <div style="text-align: center;"><math>{B = \frac{μ_{0}NI}{2πr}}</math></div>

<br />
==Examples==
===Simple===
[[File:Simple Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the simple example.]]
A toroid frame is made out of plastic of small square cross section and tightly wrapped uniformly with 100 turns of wire, so that the magnetic field has essentially the same magnitude throughout the plastic (radius R of the curved part is much larger than cross section width w). With a current of 2 A and radius of 5 m, what is the magnetic field inside the plastic.
[[File:Simple Example Sol.JPG | center|frame|none|alt=Alt text| The solution for the simple example.]]

===Middling - Difficult===
[[File:Mid - Diff Example Fig.JPG | right|frame|none|alt=Alt text| Figure for the middle and difficult examples.]]
The toroid shown in the diagram has an inner radius of <math>R_{i}</math> and an outer radius of <math>R_{o}</math> and is centered at the origin in the diagram. The z-axis passes through the center of the doughnut hole. This toroid is wrapped with <math>N</math> loops of current <math>I</math> flowing up the outside surface of the toroid, radially inward, down the inner surface, and then radial outward. Assume that the magnetic field produced by this toroid has the form <math>\vec{B} = B(r,z)\hat{φ}</math> '''at every point in space''' where <math>r</math> is the perpendicular distance from the z-axis and <math>\hat{φ}</math> is a unit vector which "curls" around the z-axis, i.e., it is always tangent to any circle with rotational symmetry around the z-axis.

<br />
====Middling====
(a.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r < R_{i}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(b.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>r > R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.
[[File:Mid - Diff Example Sol ab.JPG | center|frame|none|alt=Alt text| The solution for the middle example.]]

<br />
====Difficult====
(c.) Consider a z-axis centered Amperian loop in the plane of the toroid, at <math>z = 0</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field inside the inner radius of the toroid.

<br />
(d.) Consider a z-axis centered Amperian loop far above the toroid <math>z >> R_{o}</math>, with a radius <math>R_{i} < r < R_{o}</math> and use it to find the magnitude of the magnetic field far above the toroid.
[[File:Mid - Diff Example Sol cd.JPG | center|frame|none|alt=Alt text| The solution for the difficult example.]]
<br />

====Another Example====
(a.) Consider a toroid which consists of <math>N</math> turns, as shown in the figure below. Find the magnetic field everywhere.
[[File:Example.jpg]]

== See also ==

[[Ampere's Law]]
<br />
[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]
<br />
[[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===
*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html
*http://www.phys.uri.edu/gerhard/PHY204/tsl242.pdf
*https://clas-pages.uncc.edu/phys2102/online-lectures/chapter-7-magnetism/7-3-amperes-law/example-magnetic-field-of-a-toroid/
*https://ocw.mit.edu/courses/physics/8-02t-electricity-and-magnetism-spring-2005/lecture-notes/ch9sourc_b_field.pdf

==References==
Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. Pg 888-889.
<br />
Fall 2014 Test 4 from Phys 2212.
[[Category:Maxwell's Equations]]

Ductility

2017-11-30T04:02:10Z

Nkannan6:

Edited by Nallammai Kannan (Fall 2017)
==The Main Idea==

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]
[[File:Cast iron tensile test.JPG|thumb|Fig. 1- Highly brittle fracture]]
[[File:Al tensile test.jpg|thumb| Fig. 2- Semi-ductile fracture]]

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]

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]

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].

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].

===A Mathematical Model===

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.

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)

Percent Elongation = (Final Gage Length - Initial Gage Length) / Initial Gage Length
= ((Lf - Lo) / Lo) * 100

Percent Reduction of Area = (Area of Original Cross Section - Minimum Final Area) / Area of Original Cross Section
= (Decrease in Area / Original Area)
[2]
==Connectedness==

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's materials to be brittle. Eventually, a few ships sank and were lost to these brittle fractures.[13]

==History==

Percy Williams Bridgman'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]

== See also ==

*[[Malleability]]
*[[Temperature]]
*[[Pressure]]

===Further reading===

https://en.wikipedia.org/wiki/Ductility

===External links===

[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

[1]https://en.wikipedia.org/wiki/Ductility
[2]https://en.wikibooks.org/wiki/Advanced_Structural_Analysis/Part_I_-_Theory/Materials/Properties/Ductility
[3]https://en.wikipedia.org/wiki/Ductility#/media/File:Ductility.svg
[4]https://en.wikipedia.org/wiki/Percy_Williams_Bridgman
[5]http://www.engineersedge.com/material_science/ductility.htm
[6]https://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php
[7]http://www.spartaengineering.com/effects-of-low-temperature-on-performance-of-steel-equipment/
[8]http://people.clarkson.edu/~isuni/Chap-7.pdf
[9]http://www.etomica.org/app/modules/sites/MaterialFracture/Background1.html
[10]https://www.merriam-webster.com/dictionary/plastic%20deformation
[11]http://www.failurecriteria.com/theductile-britt.html
[12]https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Structure/linear_defects.htm
[13]https://en.wikipedia.org/wiki/Liberty_ship

[[Category: Properties of Matter ]]

File:Ductility.gif

2017-11-30T03:16:22Z

Nkannan6: