Question

Part of a single rectangular loop of wire with dimensions shown in the (Figure 1) is situated inside a region of uniform magnetic field of 0.470T . The total resistance of the loop is 0.610? .

Part A

Calculate the force required to pull the loop from the field (to the right) at a constant velocity of 6.20m/s

. Neglect gravity.

Answer 1

Te motional emf induced in the loop is via the velocity "v" of the back leg "length L" as it is pulled thru the "B" field, and is;

E = LvB

this creates a current "i" in the loop of resistance "R" ,whic from Ohm's law is;

i = E/R = LvB/R

The induced current now interacts with the B-field to produce the force;

P = iLB = vL^2B^2/R

From Lenz' law ,this force will be opposite the pulling force. So you have the general force eq, from Newton's 2nd law;

F - P = ma

For constant velocity the acceleration is zero and you finally have;

F = P = vL^2B^2/R

F = 6.2*0.35^2*0.47^2/0.61 = 0.275 N