Fig. $\left(a\right)$ shows a square loop, $20 \mathrm{cm}$ on each side in the $x-y$ plane with its Centre at the origin. The loop carries a current of $78 \mathrm{A}$. above it at $y=0, z=12 \mathrm{cm}$ is an infinitely long wire parallel to the $\mathrm{x}$$-$axis carrying a current of 10 A. Find the force on each side of loop.

The figure shows side view, looking toward positive $y-axis$

Force on wire $AB, CD :$ We take a differential element $dx$ on the wire, see $Fig. \left(b\right).$

$x=d\tan \theta$

$dx=d{\sec }^2\theta d\theta$

Magnetic force on it,

$\left|d\overrightarrow{F}\right|=I_{2}dxB\sin \theta$

         $=I_2\left(d{\sec }^2\theta d\theta \right)\times \frac{{\mu }_0{I}_1}{2\pi (d/\cos \theta )}\sin \theta$

or $\left|\overrightarrow{F}\right|={\int }_{-\theta }^{+\theta }\left(\frac{{\mu }_0{I}_1{I}_2}{2\pi }\right)\frac{\sin \theta }{\cos \theta }d\theta$

     $\left|\overrightarrow{\mathbf{F}}\right|=\frac{{\mu }_0{\text{I}}_1{\text{I}}_2}{2\pi }{\int }_{-\theta }^{+\theta }\tan \theta d\theta =0$

Thus resultant force on wire AB is zero, similarly force on CD is also zero. we can check this result by symmetry. We tae two similar differential elements at angles $\theta$ and $-\theta$ and note the direction of magnetic force ; they cancel out

Force on wires $BC, DA$

Force on wire $BC=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}\times a$, along $OB$

Force on wire $DA=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}\times a$, along $DO$