Motional Emf using Faraday's Law

Claimed by 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]\displaystyle{ {\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]\displaystyle{ |emf| = \left|\frac{d\Phi_m}{dt}\right| }[/math]

We also know that magnetic flux is defined by the formula: [math]\displaystyle{ \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.

Faraday's Law is defined as: [math]\displaystyle{ emf = \int\! \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA }[/math]

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

A Computational Model

In the image shown above, a bar of length [math]\displaystyle{ 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]\displaystyle{ t_1 }[/math], the system encircles half of two individual magnetic field circles. However, in the second picture taken at time [math]\displaystyle{ 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.

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:

[math]\displaystyle{ emf = -\frac{d}{dt} \int\! \vec{B} \cdot \vec{n}dA }[/math]

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

[math]\displaystyle{ emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}A) }[/math]

We have the dimensions of the bar in variables: length [math]\displaystyle{ L }[/math] and width [math]\displaystyle{ x }[/math]. Substitute these values for the area, [math]\displaystyle{ A }[/math]

[math]\displaystyle{ emf = -\frac{d}{dt} (\vec{B} \cdot \vec{n}(L)(x)) }[/math]

Now that we have this formula, we have to figure out how to take its derivative with respect to [math]\displaystyle{ 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]\displaystyle{ L }[/math], of the bar is constant.

The width of the surface enclosed, [math]\displaystyle{ x }[/math], changes.

As a result, the formula now becomes:

[math]\displaystyle{ emf = (\vec{B} \cdot \vec{n}(L))\left(-\frac{d}{dt}(x)\right) }[/math]

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

[math]\displaystyle{ emf = (\vec{B} \cdot \vec{n}(L))\vec{v} }[/math]

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]\displaystyle{ \vec{F} = q\vec{v} \times \vec{B} }[/math], where [math]\displaystyle{ \vec{F} }[/math] is the force on the bar, and [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ -\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]\displaystyle{ t_1 }[/math] and at time [math]\displaystyle{ 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]\displaystyle{ -\frac{dB}{dt} }[/math], and the curl of your fingers will give you the direction of the induced current, [math]\displaystyle{ I }[/math].

If the magnetic field is NOT constant, meaning it changes with time, the derivative [math]\displaystyle{ \frac{d}{dt} }[/math] will be distributed to both [math]\displaystyle{ \vec{B} }[/math] and [math]\displaystyle{ 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]\displaystyle{ emf }[/math].

[math]\displaystyle{ emf = -(\frac{d}{dt} \vec{B})A \cdot (B\frac{d}{dt}A) }[/math]

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Using the figure below, identify the following.

a) Direction of magnetic field

b) Direction of change in magnetic field, [math]\displaystyle{ \frac{d\vec{B}}{dt} }[/math]

c) Direction of negative change in magnetic field, [math]\displaystyle{ -\frac{d\vec{B}}{dt} }[/math]

d) Direction of current, [math]\displaystyle{ I }[/math]

e) Polarization of moving bar

f) Direction of electric field inside bar due to polarization

g) Direction of force on bar

SOLUTION:

a) Into the page

b) Into the page

c) Out of the page

d) Counterclockwise

e) Positive charges at the top, negative charges at the bottom

f) Down

g) Left

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.

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