Question 4

The loop hangs from a horizontal axis PQ. When it rotates by an angle $\theta$, the center of mass (at the center of the loop) is displaced horizontally.

The gravitational torque about PQ is $\tau_g = (mg) r \sin \theta$.

The magnetic moment of the loop is $M = I A = I (\pi r^2)$. Its direction is perpendicular to the plane of the loop. Initially horizontal, after rotation by $\theta$ it makes an angle $(90^\circ - \theta)$ with the vertical magnetic field $B_0$.

The magnetic torque magnitude is $\tau_B = M B_0 \sin(90^\circ - \theta) = (I \pi r^2) B_0 \cos \theta$. At equilibrium:

$\tau_g = \tau_B$

$mgr \sin \theta = I \pi r^2 B_0 \cos \theta$

$\frac{\sin \theta}{\cos \theta} = \frac{I \pi r^2 B_0}{mgr}$

$\tan \theta = \frac{\pi r I B_0}{mg}$