Part of a single rectangular loop of wire with dimensions .35 m by .75 m (.35 is the height and .75 is the length) is situated inside a region of uniform magnetic field of 0.395 T. The total resistance of the loop is 0.690 Ω.

Question

Calculate the force required to pull the loop from the field (to the right) at a constant velocity of 6.36 m/s. Neglect gravity.

Answer 1

Calculate the rate of magnetic flux decrease when withdrawing the loop from the B-field at that velocity. The field will be out of the loop when it has moved 0.75 m, which will take t = 0.118 s if the loop is moved along the length dimension. Calculate the energy disipated in the resistor during that time. It is
(V^2/R)*t
The voltage is
V = A*B/t, where A is the loop area.

The energy disipated in the resistor while moving the loop equals the force times the distance it moves. Solve for the force.

Answer 2

The induced emf in the loop is
ε =dΦ/dt =d(BwL)/dt=Bwv,
The current is
I = ε/R = Bwv/R
The force is
F =(Bw)^2v/R
(w =0.35 m, B = 0.395 T, v =6.36 m/s,
R= 0.690 Ω)