A square loop A B C D , carrying a current I_2 , is placed near and...

Force on arm $A B$ due to current in conductor $X Y$ is $F_1=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(L / 2)}=\frac{\mu_0 I_1 I_2}{\pi}$ acting towards the wire in the plane of loop. Force on arm $C D$ due to current in conductor $X Y$ is $F_2=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2 L}{(3 L / 2)}=\frac{\mu_0 I_1 I_2}{3 \pi}$ away from the wire the plane of loop. $\therefore \quad$ Net force on the loop $=F_1-F_2$ $=\frac{\mu_0 I_1 I_2}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_0 I_1 I_2}{\pi}$