A square loop of wire of side L with total resistance R moves at constant speed v into a region of uniform magnetic field B pointing perpendicular to the plane of the loop. What is the average current that is induced in the square loop between the time it starts entering the region of magnetic field and the time it has completely entered the region? Do not worry about the sign of the current.
Prob. 9
http://www-physics.ucsd.edu/students/courses/winter2010/physics2b/hw/hw8.pdf
To find the average current induced in the square loop, we can use Faraday's law of electromagnetic induction. According to this law, the induced EMF (electromotive force) in a closed loop is equal to the rate of change of magnetic flux through the loop.
The magnetic flux (Φ) is given by the product of the magnetic field (B) and the area (A) enclosed by the loop. In this case, since the loop is a square:
A = L^2
Therefore, the magnetic flux (Φ) is:
Φ = B * A = B * L^2
Now, as the loop enters the region of the uniform magnetic field, the magnetic flux changes. The rate of change of magnetic flux (dΦ/dt) can be calculated by considering the change in the area of the loop with respect to time. Since the loop moves at a constant speed (v), the change in the area is given by:
dA/dt = v * L
Using this, we can express the rate of change of magnetic flux as:
dΦ/dt = B * dA/dt = B * v * L
According to Faraday's law, the induced EMF (ε) is equal to the negative of the rate of change of magnetic flux. Therefore:
ε = -dΦ/dt = -B * v * L
The induced current (I) can be calculated by dividing the induced EMF by the resistance (R) of the loop:
I = ε / R = (-B * v * L) / R
Thus, the average current induced in the square loop between the time it starts entering the region of the magnetic field and the time it has completely entered is (-B * v * L) / R. Note that we do not need to worry about the sign of the current for this question.