A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\vec{F}$, the net force on the remaining three arms of the loop is

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by $\tau=\mathrm{NBiA} \sin \theta$. Torque is maximum when $\theta=90^{\circ}$, i. e., the plane of the coil is parallel to the field $\tau_{\max }=\mathrm{NBiA}$
Forces $\overrightarrow{\mathrm{F}}_1$ and $\overrightarrow{\mathrm{F}}_2$ acting on the coil are equal in magnitude and opposite in direction. As the forces $\vec{F}_1$ and $\vec{F}_2$ have the same line of action, their resultant effect on the coil is zero. The two forces $\overrightarrow{\mathrm{F}}_3$ and $\overrightarrow{\mathrm{F}}_4$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is $\overrightarrow{\mathrm{F}}$, the net force on the remaining three arms of the loop is $-F$.