Q. A rigid square loop of side 'a' and carrying current I$_2$ is lying on a horizontal surface near a long current I$_1$ carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be :
Solution:
F$_3$ & F$_4$ cancel each other
Force on PQ will be F$_1 = I_2 B_1$ a
$ \, \, \, \, \, \, \, \, \, \, \, =I_2 \frac{\mu_0 I_1}{2 \pi a}a$
$ \, \, \, \, \, \, \, \, \, \, \, = \frac{\mu_0 I_1 I_2}{2\pi}$
Force on RS will be F$_2$ = I$_2 B_2$ a
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = I_2 \frac{\mu_0 I_1}{2\pi 2a}a$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{\mu_0 I_1 I_2}{4 \pi}$
Net force = $F_1 - F_2 \, = \, \frac{\mu_0 I_1 I_2}{4 \pi}$ repulsion
Option (2)
