Question

A long, straight wire carries a current of I1 = 13.2 A. Next to the wire is a square loop with sides L = 1.20 m in length. d = 1.47 m, as shown in the figure below.


The loop carries a current of I2 = 2.48 A in the direction indicated. Calculate the magnitude of the net force acting on the loop.

Answer 1

the wire will induct field b which will imply force on the parts of the loop which are in parallel to the wire.
F=i2LB

so we need to calculate two forces (one on each side)

F1=i2*L*B=i2*L*uo*i1/(2*pi*(d-L))=2.9e-5 N

F2=i2*L*B=i2*L*uo*i1/(2*pi*d)=5.34e-6 N

Two force will acto on opposite directions (because the current flows in different directions)

So the net force will be the difference (not the sum)

F total = F1-F2=2.36e-5 N