Question

A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the y-axis. Take the vertically upward direction as the z-axis. A uniform magnetic field $\overrightarrow{B}=(3\hat{i}+4\hat{k})B_0$ exists in the region. The loop is held in the x-y plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium.
If the torque acting on this loop is $\overrightarrow{{\tau }_B}=X\mu B_0\hat{j}$. Find X?

Answer 1

Torque due to weight of coil, $\overrightarrow{\tau }=\left(\frac{a}{q}\hat{i}\right)\times (-mg\hat{k})=mg\frac{a}{2}\left(\hat{j}\right)$
For the equilibrium of loop, torque on it must be along negative y-axis. Let the magnetic moment of loop be $\mu \hat{k}$.

As the loop lies in x-y plane, its magnetic moment vector (from right hand thumb rule) either points up or down.
Torque due to magnetic force, $\overrightarrow{{\tau }_B}=\overrightarrow{\mu }\times \overrightarrow{B}=\mu \hat{k}\times (3\hat{i}+4\hat{k})B_0=3\mu B_0\hat{j}$
If it is in negative direction, $\overrightarrow{\mu }$ must point downward. So, the current in the coil must be from P to Q.