Physics Book - User contributions [en]

Motional Emf using Faraday's Law

2016-11-28T04:26:39Z

Kcockerill: /* A Computational Model */

CLAIMED BY: Kasey Cockerill (Fall 2016)

Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

The first term, <math>(\frac{d}{dt}\vec{B})A</math>, represents Faraday's law and is nonzero of there is a varying magnetic field.
The second term, <math>B(\frac{d}{dt}A)</math>, represents motional emf and is nonzero if there is a change in the amount of enclosed area.

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 0.0013T</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the amount of magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \left(\vec{B} \cdot A\right)</math>
:Change in area <math>\Delta{A} = L\Delta{x}</math>
:In this case, the distance <math>x</math> is changing and resulting in a change in area, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T04:25:00Z

Kcockerill: /* A Computational Model */

CLAIMED BY: Kasey Cockerill (Fall 2016)

Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

The first term, <math>(\frac{d}{dt}\vec{B})A</math>, is nonzero of there is a varying magnetic field (Faraday's Law).
The second term, <math>B(\frac{d}{dt}A)</math>, is nonzero if there is a change in area enclosing a constant magnetic field (Motional emf).

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 0.0013T</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the amount of magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \left(\vec{B} \cdot A\right)</math>
:Change in area <math>\Delta{A} = L\Delta{x}</math>
:In this case, the distance <math>x</math> is changing and resulting in a change in area, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T04:13:33Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)

Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 0.0013T</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the amount of magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \left(\vec{B} \cdot A\right)</math>
:Change in area <math>\Delta{A} = L\Delta{x}</math>
:In this case, the distance <math>x</math> is changing and resulting in a change in area, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T04:07:01Z

Kcockerill: /* Medium */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 0.0013T</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the amount of magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \left(\vec{B} \cdot A\right)</math>
:Change in area <math>\Delta{A} = L\Delta{x}</math>
:In this case, the distance <math>x</math> is changing and resulting in a change in area, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:58:41Z

Kcockerill: /* Medium */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the amount of magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:Because v is perpendicular to B, <math>|emf| = BA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:55:41Z

Kcockerill: /* Examples */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The magnetic force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:42:49Z

Kcockerill: /* References */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

https://en.wikipedia.org/wiki/Transcranial_magnetic_stimulation#Technical_information

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:42:36Z

Kcockerill: /* Connectedness */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
: I am a biomedical engineering student, and one application of Faraday's law in the medical field is transcranial magnetic stimulation. During this procedure, magnetic coils are used to stimulate small regions of the brain through electromagnetic induction. Current is discharged from a capacitor into the coil to produce pulsed magnetic fields. This technique can be used to evaluate and diagnose various conditions affecting the connection between the brain and muscles, including strokes and motor neuron diseases. It has also been said to alleviate the symptoms of major depressive disorder.

:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:26:20Z

Kcockerill: /* References */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

https://en.wikipedia.org/wiki/Pickup_(music_technology)

http://www.physics.princeton.edu/~mcdonald/examples/guitar.pdf

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:24:51Z

Kcockerill: /* Connectedness */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux. These pickup coils often consist of magnet wrapped with a coil of copper wire, where the magnet creates a magnetic field and the vibrations of the string disturb the field, inducing a current in the coiled wire.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:21:35Z

Kcockerill: /* Connectedness */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:Believe it or not, Faraday's law can be applied to musical instruments such as the electric guitar. In many electric instruments, 'pickup coils' sense the vibration of the strings, which causes variations in magnetic flux.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:02:05Z

Kcockerill: /* History */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Based off of the results of his experiments, Faraday eventually came up with a relationship telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T03:00:42Z

Kcockerill: /* History */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered through his experiments 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. He observed that when he held a bar magnet was held stationary with respect to the loop, the galvanometer did not read a current. However, when he moved the bar magnet towards or away from the loop, the galvanometer read a non-zero current. If a current is flowing, that means there must be some emf. Eventually Faraday was able to form a relationship that related his experimental results, telling us that the emf generated in a loop of wire in some magnetic field is proportional to the rate of change of the magnetic flux through the loop. This is what we know today as Faraday's law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:48:57Z

Kcockerill: /* References */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:48:42Z

Kcockerill: /* History */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>. However, Michael Faraday discovered 2 ways in which current could be induced in a closed loop of wire in the absence of a battery: by changing the magnetic field around the loop, or by moving the loop through a constant magnetic field. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:31:59Z

Kcockerill: /* History */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

We know from the loop rule that around a closed loop, <math>V = emf = \oint \vec{E} \cdot d\vec{l} = 0</math>

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:29:47Z

Kcockerill: /* History */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

We know from the loop rule that around a closed loop, <math>V = emf = \oint E</math>

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. We know from the loop rule that around a closed loop, <math>V = emf = Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:16:38Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encloses a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the enclosed magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:12:43Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the Non-Coulomb electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:11:33Z

Kcockerill: /* The Main Idea */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Motional emf results when the area enclosing a constant magnetic field changes. Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:03:34Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

Let's observe a specific scenario in which a bar of length L slides along two frictionless bars. We can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T02:00:36Z

Kcockerill: /* A Mathematical Model */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

As a bar of length L slides along two frictionless bars, we can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>. We already know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>. In the case that v is perpendicular to B, we combine these to get: <math>\frac{\Delta{\Phi_m}}{\Delta{t}} = BLv </math>.

Emf is said to be the work done per unit charge: <math>emf = \frac{FL}{q} = \frac{qvBL}{q} = vBL</math> (again, we are assuming v is perpendicular to B).

Comparing the above two formulas, we can clearly see that <math>|{emf}| = |\frac{d\Phi_m}{dt}|</math>. This is exactly what Faraday's Law tells us!

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:35:38Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===
.
As a bar of length L slides along two frictionless bars, we can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>.

We know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

Faraday's law also tells us that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:33:19Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===
.
As a bar of length L slides along two frictionless bars, we can observe the change in area over a short time as <math>\Delta{A} = L\Delta{x} = Lv\Delta{t}</math>.

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:24:33Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===
.
As a bar of length L slides along two frictionless bars, we can observe the changes that occurs over a short time <math>\Delta{t}<\math>.

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:15:16Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===
.
As a bar of length L slides along two frictionless bars, we can observe the changes that occurs over a short time <math>&delta</math>

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:14:16Z

Kcockerill: /* A Mathematical Model */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

As a bar of length L slides along two frictionless bars, we can observe the changes that occurs over a short time <math>&delta

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:05:17Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math> where v is the velocity of the bar and L is the bar length. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T01:00:49Z

Kcockerill: /* The Main Idea */

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math>. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T00:48:27Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

Motional emf can be calculated in terms of magnetic flux, where motional emf is quantitatively equal to the rate of change of the magnetic flux. If an enclosed magnetic field remains constant but the loop changes shape or orientation, the resulting change in area leads to a change in magnetic flux.

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math>. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Motional Emf using Faraday's Law

2016-11-28T00:38:56Z

Kcockerill:

CLAIMED BY: Kasey Cockerill (Fall 2016)
Chelsea Calhoun

==The Main Idea==

When a wire moves through an area of magnetic field, a current begins to flow along the wire as a result of magnetic forces. Originally, we learned to calculate the motional emf in a moving bar by using the equation <math>{\frac{q(\vec{v} \times \vec{B})L}{q}}</math>. However, there's an easier way to do this: by writing an equation for emf in terms of magnetic flux.

===A Mathematical Model===

We know that the magnitude of motional emf is equal to the rate of change of magnetic flux. <math>|emf| = \left|\frac{d\Phi_m}{dt}\right|</math>

We also know that magnetic flux is defined by the formula: <math>\Phi_m = \int\! \vec{B} \cdot\vec{n}dA</math>

When a metal bar or wire is moving across an area, creating a closed circuit, the magnetic field contained inside the area of the circuit is no longer constant. When there is a non-constant magnetic field, we must use Faraday's law (a combination of the two equations above) to solve for motional emf.

<div align="center">'''Faraday's Law is defined as: <math>emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>'''

where <math>\vec{E}</math> is the electric field along the path, <math>l</math> is the length of the path you're integrating on, <math>\vec{B}</math> is the magnetic field inside the area enclosed, and <math>\vec{n}</math> is the unit vector perpendicular to area A.

===A Computational Model===

<div align="center">[[File:ExamplePic1.jpg]]</div>

In the image shown above, a bar of length <math>L</math> is moving along two other bars from right to left. The blue circles containing "x"s represent a magnetic field directed into the page. As the bar moves to the right, the system encircles a greater amount of magnetic field. To explain this concept more clearly, take a look at the figures below. This image shows a bar moving in a magnetic field at two different times. In the first picture, at time <math>t_1</math>, the system encircles half of two individual magnetic field circles. However, in the second picture taken at time <math>t_2</math>, the system now encircles 6 full magnetic field circles. Of course, this explanation isn't using technical terms, but the point still stands: the magnetic field is increasing as time increases.

<div align="center">[[File:ExamplePic2.jpg]]</div>

Returning to the scenario in the first image, because the magnetic field is not constant, we can use Faraday's Law to solve for the motional emf.

As stated above, the formula is as follows: <div align="center"><math>emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math></div>

First, integrate the integral with respect to the area of the rectangle enclosed.

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A)</math></div>

We have the dimensions of the bar in variables: length <math>L</math> and width <math>x</math>.
Substitute these values for the area, <math>A</math>

<div align="center"><math>emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x))</math></div>

Now that we have this formula, we have to figure out how to take its derivative with respect to <math>t</math>. Which of the magnitudes of these values is changing?
:::The magnitude of the magnetic field is constant. (More "circles" are added as time increases, but the magnitude of each "circle" does not change.
:::The magnitude of the normal vector is constant.
:::The length, <math>L</math>, of the bar is constant.
:::The width of the surface enclosed, '''<math>x</math>''', '''changes'''.

As a result, the formula now becomes:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right)</math></div>

In this case, <math>\frac{dx}{dt} = \vec{v}</math> because <math>x</math> is a function of time, where <math>\vec{v}</math> is the velocity of the moving bar. Substituting that in, we get:

<div align="center"><math>emf = (\vec{B} \cdot \vec{n}(L))\vec{v}</math></div>

Plugging in these values, we can solve for the motional emf of the bar.

Because the magnetic field is changing with time, however, there is also an induced current flowing through the circuit. We can find the direction of the current using the right hand rule. To do this, we can use 2 different methods:
: '''1.''' We can use the equation <math>\vec{F} = q\vec{v} \times \vec{B}</math>, where <math>\vec{F}</math> is the force on the bar, and <math>\vec{v}</math> is the velocity of the bar. Using the right hand rule, we can point our fingers in the direction of the velocity of the bar and curl them in the direction of the magnetic field. The direction that our thumb points is the direction of the force on a positive charge. In this case, <math>\vec{F}</math> points upward, so the positive charges in the bar will move to the top, causing it to polarize with positive charges at the top and negative charges at the bottom. We can now visualize the bar as a battery that causes a current <math>I</math> to run out of the positive end. In this case, since the bar is polarized with the positive charges at the top, the current will flow out of the top of the bar and continue around the circuit.

:'''2.''' We can use the negative direction of the change in magnetic field, <math>-\frac{dB}{dt}</math> to find the direction of the current. To do this, make a diagram comparing the magnitude of the magnetic field enclosed at time <math>t_1</math> and at time <math>t_2</math>. Then, draw an arrow representing the direction of change of the magnetic field. Now, flip the arrow to take the negative of that vector's direction. Using the right hand rule, point your thumb in the direction of <math>-\frac{dB}{dt}</math>, and the curl of your fingers will give you the direction of the induced current, <math>I</math>.

If the magnetic field is NOT constant, meaning it changes with time, the derivative <math>\frac{d}{dt}</math> will be distributed to both <math>\vec{B}</math> and <math>x</math> in the formula. In this case, we must use the product rule to be able to set up the equation and continue solving for <math>emf</math>.

<div align="center"><math>emf = -(\frac{d}{dt} \vec{B})A \cdot B(\frac{d}{dt}A)</math></div>

==Examples==

===Simple===
Using the figure below, identify the following.

:a) Direction of magnetic field
:b) Direction of change in magnetic field, <math>\frac{d\vec{B}}{dt}</math>
:c) Direction of negative change in magnetic field, <math>-\frac{d\vec{B}}{dt}</math>
:d) Direction of current, <math>I</math>
:e) Polarization of moving bar
:f) Direction of electric field inside bar due to polarization
:g) Direction of force on bar

[[File:Example1.png]]

SOLUTION:
:a) Into the page
:: A circle with an 'x' inside of it represents a vector into the page. A circle with a dot inside represents a vector out of the page.
:b) Into the page
:: Initially, at the time of the image, there are 4 circles representing magnetic field enclosed by the bars. However, as the bar moves, at some time t, the number of circles enclosed by the bar will increase; therefore, there is more magnetic field inside the loop. This means that the change in magnetic field is in the direction of the magnetic field.
:c) Out of the page
:: The negative change in magnetic field is in the opposite direction as change in magnetic field.
:d) Counterclockwise
:: Point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>. Your fingers will curl in the direction of current.
:e) Positive charges at the top, negative charges at the bottom
::The electric force on a particle is <math>\vec{F} = q\vec{v} \times \vec{B} </math>, so point your fingers in the direction of the velocity of the bar and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on a positive particle.
:f) Down
::Positive charges have an electric field that points away from them while negative particles have an electric field that point towards them. If the top of the bar is positively charged, the field will point downward toward the negative particles.
:g) Left
::When a current is involved, <math>\vec{F} = I\vec{l} \times \vec{B}</math>, so point your fingers in the direction of the length of the bar (in the direction of current) and curl them in the direction of magnetic field. The direction of your thumb is the direction of force on the bar.

===Medium===
A bar of length <math>L = 2</math> is moving across two other bars in a region of magnetic field, <math>B = 1.3E-3</math> directed into the page. The bar is moving with a velocity of 10 m/s, and <math>x</math> is the width of the area enclosed. What is the magnitude of the <math>emf</math> produced?

[[File:Example1.png]]

SOLUTION:
:Because the magnetic field enclosed by the system is changing with time, we must use Faraday's Law: <math>|emf| = \frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA</math>
:First, integrate through the formula: <math>|emf| = \frac{d}{dt} \vec{B} \cdot A</math>
:Area <math>A = Lx</math>
:Next, distribute <math>\frac{d}{dt}</math>. In this case, distance <math>x</math> is what is changing, so the formula becomes: <math>|emf| = \vec{B} \cdot L\frac{d}{dt}x</math>
:The derivative of distance is velocity. <math>\frac{dx}{dt} = v</math>
:Therefore, |emf| in this problem is equal to <math>BLv = .026 V </math>

===Difficult===
A long straight wire carrying current I = .3 A is moving with speed v = 5 m/s toward a small circular coil of radius R = .005 and 10 turns. The long wire is in the plane of the coil. The coil is very small, so that, at any fixed moment in time, you can neglect the spatial variation of the wire's magnetic field over the area of the coil.
[[File:Example2.png]]

:a) Is the induced current in the coil flowing clockwise or counterclockwise?
:b) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

Now consider the case where the wire is stationary and the coil is moving down parallel to the wire with a constant speed, <math>v = 2 m/s</math>.
:c) At the instant when the long wire is a distance x = 4 m from the center of the coil, determine the magnitude of the induced emf in the coil.

[[File:Exemploo3.png]]

SOLUTION:
:a) Counterclockwise
:: Using the right hand rule, if you point your thumb in the direction of current (+y), your fingers will curl in the direction of magnetic field. In this case, magnetic field is pointing into the page at the coil. At the location of the coil, the magnitude of the magnetic field due to the wire is increasing as the wire moves closer; therefore, <math>\frac{d\vec{B}}{dt}</math> is pointing into the page, and <math>-\frac{d\vec{B}}{dt}</math> is pointing out of the page. If you point your thumb in the direction of <math>-\frac{d\vec{B}}{dt}</math>, your fingers curl in the direction of the induced current.
:b) <math> |emf| = 1.47E-11 V</math>
::After integrating Faraday's Law, we get <math>|emf| = \frac{d}{dt} (\vec{B} \cdot A)</math>
::Notice that distance <math>x</math> is changing with time.
::After doing this derivative, we get <math>|emf| = \frac{\mu_0IR^2v}{2x^2}</math>
::This is the magnitude of emf for '''one''' loop in the coil, so we have to multiply it by the number of loops, <math>N</math>.
::<math>|emf| = \frac{N\mu_0IR^2v}{2x^2}</math>
:c) |emf| = 0
::Remember that the emf relies on a changing magnetic field, which was dependent on a changing <math>x</math> in the previous example. Now, however, the coil is moving parallel to the wire, meaning there is no change in <math>x</math>, and no change in magnetic field.

==Connectedness==
1. How is this topic connected to something that you are interested in?
:In physics 1, we focused on large scale, physical pieces of matter, kinetics and rotation, which I personally find fairly interesting. Physics 2, however, focuses on new, small-scale concepts that can often be more challenging to envision since topics like electric and magnetic fields are not something we can see with the naked eye. With motional emf, though, problems now become more visual, and many times can be conceptually solved using the right hand rule. It's interesting to learn about how something as simple as moving a bar through space filled with a magnetic field can cause a current to flow. A system like this could even potentially be used in the place of a battery.
2. How is it connected to your major?
:I am currently majoring in mechanical engineering, and in this field, we are required to work with both mechanics and circuit-like scenarios. Personally, I am interested in going into the car manufacturing industry, where motional emf plays a very important role. When you move an object through a magnetic field, it resists movement and generates electricity in the loop. If this is done with enough force, it could be used to stop a small car or roller-coaster.

==History==

Prior to 1831, the only known way to make an electric current flow through a conducting wire was to connect the ends of the wire to the positive and negative terminals of a battery. Michael Faraday, however, in his experiments, managed to find another way of generating an electromotive force around a loop of wire: to keep the magnetic field constant around area of the loop, and move the loop. In his first experiment, Faraday wrapped two wires around opposite sides of an iron ring and plugged one wire into a galvanometer and the other into a battery. After this, he noticed a current, which he called a "wave of electricity". Within two months, Faraday had discovered several other manifestations of electromagnetic induction, formulating Faraday's Law.

== See also ==

You may want to explore the process of calculating motional emf before the use of Faraday's Law. Maxwell's equations and circuits with resistance are also relevant and may be worth looking into.

Motional emf problems can be pretty tricky depending on what the question is asking you to do. It's always a good idea to know how each formula came about, and how it can change bases on different scenarios. This includes the formula for resistance in a circuit, <math>V = IR</math>. A problem could go as far as to give you a resistance for a circuit, ask you to solve for the potential difference, <math>V</math>, or <math>emf</math>, and then ask you to solve for the current as well.

Lastly, I advise you to become familiar with Lenz's law because it gives the direction of the induced emf and current resulting from electromagnetic induction.

===Further reading===

:SparkNotes: SAT Physics
:Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4nd Edition by R. Chabay & B. Sherwood (John Wiley & Sons 2015)

===External links===

:'''Video Explanation:''' https://www.youtube.com/watch?v=Wgtw5lPKFXI
:'''Text Explanation:''' https://www.boundless.com/physics/textbooks/boundless-physics-textbook/induction-ac-circuits-and-electrical-technologies-22/magnetic-flux-induction-and-faraday-s-law-161/motional-emf-570-6257/

==References==

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

http://farside.ph.utexas.edu/teaching/em/lectures/node43.html

[[Category: Faraday's Law]]

Current in an LC Circuit

2016-11-28T00:38:16Z

Kcockerill:

CLAIMED BY: Harrison Cowart
4/11/2016

An LC circuit contains an inductor and a capacitor, and because of the inductor's sluggishness and resistance to change the current, the charge in this circuit can oscillate back and forth forever.

==The Main Idea==

A circuit containing an inductor and a capacitor is called an LC circuit. "L" and "C" are conventional identifiers used in circuit analysis, used for inductors and capacitors respectively. "L" is the symbol used almost exclusively in circuitry as the variable for inductance (measured in henries), or the property of a coil of wire to resist change in current. "C" is the symbol used to denote capacitance (measured in farads), the property of a capacitor, two metal plates separated by a small distance, to store electrical charge. Here is the basic idea behind the process of an LC circuit:

A capacitor is charged initially, and then a switch is closed. This connects the capacitor to the inductor. At first, because of the nature of an inductor, it is difficult for charge on the capacitor to flow through the inductor. This is because the inductor opposes all attempts to change the current. However, since an inductor cannot completely prevent a current from changing, there is more current little by little. This drains the capacitor of its charge.

At the moment the capacitor runs out of charge, there is a current in the inductor, and because inductors are sluggish by nature, the current can't immediately change to zero. Therefore, the system does not come to equilibrium and instead increases the charge in the capacitor. When the capacitor is fully charged, it starts to discharge back through the inductor, and the process repeats. Since the circuit will never reach equilibrium and current never reaches zero, this oscillating process repeats forever. Oscillations could reach a stopping point if there is some resistance, but it may still go through multiple cycles before equilibrium is reached.

A series LC circuit is classified by having an initially charged capacitor and an inductor along the same length of wire. This category of circuitry has special properties that vary from non-inducting circuits. R and RC circuits charge and discharge in an exponential fashion, which prompts shorter time periods between initial current flow and steady-state equilibrium. LC circuits “discharge” in a pattern that resembles the graph of a trigonometric function. This circuit can oscillate continuously if the resistance is small. Accordingly, the connecting wires in an LC circuit are low-resistance thick copper wires.

===A Mathematical Model===

An LC circuit has an energy conservation rule associated with it. The energy conservation loop rule for an LC circuit is:
<math> \Delta V_{capacitor} + \Delta V_{inductor} = \frac{Q}{C}-L\frac{dI}{dt} = 0 </math> where <math>Q</math> is the charge on the upper plate of the capacitor and <math>I</math> is the conventional current leaving the upper plate and going through the inductor.

<math>dQ/dt</math> is the amount of charge flowing off the capacitor every second. This is the same as the current. Because the charge is leaving the capacitor, <math> I = -\frac{dQ}{dt}</math>.

Energy conservation can be re-written as <math>\frac{1}{C}Q+L\frac{d^{2}Q}{dt^{2}} = 0 </math>

In addition, by substituting Q and its second derivative into the equation, a possible solution of the rewritten energy conservation equation is <math> Q = Q_{i}cos(\frac{1}{\sqrt{LC}}t) </math>

Therefore, the current is given by: <math> I = -\frac{dQ}{dt} = \frac{Q_{i}}{\sqrt{LC}}sin(\frac{1}{\sqrt{LC}}t) </math>

===A Computational Model===

[[File:LCcircuit.jpg]]

[[File:Tuned_circuit_animation_3_300ms.gif]]

==Examples==

Use the given circuit and information to solve for the following quantities:

L = 4.0 mH

C = 5.0 μF

[[File:WikiQ.jpg|400x400px]]

===Simple===

Find the resonant frequency in radians per second and Hertz and the period in seconds:

[[File:WikiA1.jpg|400x400px]]

===Middling===

Find the initial charge on the fully charged capacitor and the maximum current between oscillations:

[[File:WikiA2.jpg|400x400px]]

==Connectedness==

This topic is connected to RC circuits, which approach equilibrium slowly, as well as RL circuits, which approach steady state current slowly. However, an LC circuit is an example of a circuit that oscillates, where charge moves back and forth forever, and never settles to an equilibrium or steady state.

As an engineer, these circuits can be useful in many areas, particularly to electrical and mechanical engineers. A common applications is tuning radio transmitters and receivers.When we tune a radio to a particular station, the LC circuits are set at a resonance for that particular carrier frequency.

LC circuits can also be applied to voltage magnification and current magnification. LC circuit can additionally be used in induction heating.

==History==

In 1826, French scientist Felix Savary was the first to discover evidence that a capacitor and an inductor could produce electrical oscillations. In his experiment, he discharged a Leyden jar through a wire wound around an iron needle. He found that sometimes the needle was left magnetized in one direction and other times it was in the opposite direction. He concluded that this was due to a damped oscillating discharge current in the wire, which reversed the magnetization of the needle back and forth until it was too small to have an effect, leaving the needle magnetized in a random direction.

== See also ==

===External links===
Helpful websites for furthering your understanding of LC circuits, including some mathematical applications:
[http://farside.ph.utexas.edu/teaching/315/Waves/node5.html]
[http://physics.info/circuits-rlc/]

A helpful video for further information:
[https://www.youtube.com/watch?v=v3-HwZMThzQ]

==References==

Matter & Interactions 4th Edition: Electric and Magnetic Interactions

http://www.chegg.com/homework-help/questions-and-answers/x-i5rong-please-help-part-g-learning-goal-understand-processes-series-circuit-containing-i-q74127

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

http://farside.ph.utexas.edu/teaching/315/Waves/node5.html

http://physics.info/circuits-rlc/

https://www.youtube.com/watch?v=v3-HwZMThzQ

https://www.youtube.com/watch?v=_8d4LQMFc1o

[[Category:Which Category did you place this in?]]

Current in an LC Circuit

2016-11-28T00:31:17Z

Kcockerill:

Claimed by: Kasey Cockerill (Fall 2016)
CLAIMED BY: Harrison Cowart
4/11/2016

An LC circuit contains an inductor and a capacitor, and because of the inductor's sluggishness and resistance to change the current, the charge in this circuit can oscillate back and forth forever.

==The Main Idea==

A circuit containing an inductor and a capacitor is called an LC circuit. "L" and "C" are conventional identifiers used in circuit analysis, used for inductors and capacitors respectively. "L" is the symbol used almost exclusively in circuitry as the variable for inductance (measured in henries), or the property of a coil of wire to resist change in current. "C" is the symbol used to denote capacitance (measured in farads), the property of a capacitor, two metal plates separated by a small distance, to store electrical charge. Here is the basic idea behind the process of an LC circuit:

A capacitor is charged initially, and then a switch is closed. This connects the capacitor to the inductor. At first, because of the nature of an inductor, it is difficult for charge on the capacitor to flow through the inductor. This is because the inductor opposes all attempts to change the current. However, since an inductor cannot completely prevent a current from changing, there is more current little by little. This drains the capacitor of its charge.

At the moment the capacitor runs out of charge, there is a current in the inductor, and because inductors are sluggish by nature, the current can't immediately change to zero. Therefore, the system does not come to equilibrium and instead increases the charge in the capacitor. When the capacitor is fully charged, it starts to discharge back through the inductor, and the process repeats. Since the circuit will never reach equilibrium and current never reaches zero, this oscillating process repeats forever. Oscillations could reach a stopping point if there is some resistance, but it may still go through multiple cycles before equilibrium is reached.

A series LC circuit is classified by having an initially charged capacitor and an inductor along the same length of wire. This category of circuitry has special properties that vary from non-inducting circuits. R and RC circuits charge and discharge in an exponential fashion, which prompts shorter time periods between initial current flow and steady-state equilibrium. LC circuits “discharge” in a pattern that resembles the graph of a trigonometric function. This circuit can oscillate continuously if the resistance is small. Accordingly, the connecting wires in an LC circuit are low-resistance thick copper wires.

===A Mathematical Model===

An LC circuit has an energy conservation rule associated with it. The energy conservation loop rule for an LC circuit is:
<math> \Delta V_{capacitor} + \Delta V_{inductor} = \frac{Q}{C}-L\frac{dI}{dt} = 0 </math> where <math>Q</math> is the charge on the upper plate of the capacitor and <math>I</math> is the conventional current leaving the upper plate and going through the inductor.

<math>dQ/dt</math> is the amount of charge flowing off the capacitor every second. This is the same as the current. Because the charge is leaving the capacitor, <math> I = -\frac{dQ}{dt}</math>.

Energy conservation can be re-written as <math>\frac{1}{C}Q+L\frac{d^{2}Q}{dt^{2}} = 0 </math>

In addition, by substituting Q and its second derivative into the equation, a possible solution of the rewritten energy conservation equation is <math> Q = Q_{i}cos(\frac{1}{\sqrt{LC}}t) </math>

Therefore, the current is given by: <math> I = -\frac{dQ}{dt} = \frac{Q_{i}}{\sqrt{LC}}sin(\frac{1}{\sqrt{LC}}t) </math>

===A Computational Model===

[[File:LCcircuit.jpg]]

[[File:Tuned_circuit_animation_3_300ms.gif]]

==Examples==

Use the given circuit and information to solve for the following quantities:

L = 4.0 mH

C = 5.0 μF

[[File:WikiQ.jpg|400x400px]]

===Simple===

Find the resonant frequency in radians per second and Hertz and the period in seconds:

[[File:WikiA1.jpg|400x400px]]

===Middling===

Find the initial charge on the fully charged capacitor and the maximum current between oscillations:

[[File:WikiA2.jpg|400x400px]]

==Connectedness==

This topic is connected to RC circuits, which approach equilibrium slowly, as well as RL circuits, which approach steady state current slowly. However, an LC circuit is an example of a circuit that oscillates, where charge moves back and forth forever, and never settles to an equilibrium or steady state.

As an engineer, these circuits can be useful in many areas, particularly to electrical and mechanical engineers. A common applications is tuning radio transmitters and receivers.When we tune a radio to a particular station, the LC circuits are set at a resonance for that particular carrier frequency.

LC circuits can also be applied to voltage magnification and current magnification. LC circuit can additionally be used in induction heating.

==History==

In 1826, French scientist Felix Savary was the first to discover evidence that a capacitor and an inductor could produce electrical oscillations. In his experiment, he discharged a Leyden jar through a wire wound around an iron needle. He found that sometimes the needle was left magnetized in one direction and other times it was in the opposite direction. He concluded that this was due to a damped oscillating discharge current in the wire, which reversed the magnetization of the needle back and forth until it was too small to have an effect, leaving the needle magnetized in a random direction.

== See also ==

===External links===
Helpful websites for furthering your understanding of LC circuits, including some mathematical applications:
[http://farside.ph.utexas.edu/teaching/315/Waves/node5.html]
[http://physics.info/circuits-rlc/]

A helpful video for further information:
[https://www.youtube.com/watch?v=v3-HwZMThzQ]

==References==

Matter & Interactions 4th Edition: Electric and Magnetic Interactions

http://www.chegg.com/homework-help/questions-and-answers/x-i5rong-please-help-part-g-learning-goal-understand-processes-series-circuit-containing-i-q74127

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

http://farside.ph.utexas.edu/teaching/315/Waves/node5.html

http://physics.info/circuits-rlc/

https://www.youtube.com/watch?v=v3-HwZMThzQ

https://www.youtube.com/watch?v=_8d4LQMFc1o

[[Category:Which Category did you place this in?]]