Physics - Magnetism Question with Solution | TestHub

Force on arm $AB$ due to current in conductor $XY$ is $F_1=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_{2}L}{(L/2)}=\frac{{\mu }_0{I}_1{I}_2}{\pi }$ acting towards the wire in the plane of loop. Force on arm $CD$ due to current in conductor $XY$ is $F_2=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_{2}L}{(3L/2)}=\frac{{\mu }_0{I}_1{I}_2}{3\pi }$ away from the wire the plane of loop. $\therefore$ Net force on the loop $=F_1-F_2$ $=\frac{{\mu }_0{I}_1{I}_2}{\pi }\left[1-\frac{1}{3}\right]=\frac{2}{3}\frac{{\mu }_0{I}_1{I}_2}{\pi }$