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.
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, 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. For example, the following figure shows a bar moving an 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.
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]
Now, we can use the right hand rule to figure out which way the normal vector is pointing based on the current. First, though, we need to figure out which way the current is flowing. To do this, we can use the equation [math]\displaystyle{ \vec{F} = I\vec{L} \times \vec{B} }[/math], where [math]\displaystyle{ \vec{F} }[/math] is the force on the bar, and [math]\displaystyle{ \vec{I} }[/math] is the induced current. Using the right hand rule, we can point our thumb in the direction of the force and
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