$\text{A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from}$

$\text{the same point of the loop. }\text{The wire loop has mass}\mathrm{ m}\text{ and radius}\mathrm{ r}\text{ and it is in a uniform vertical magnetic}$

$\text{field }{\overrightarrow{B}}_0,\text{as shown in the figure.}\text{ Initially, it hangs vertically downwards, because of acceleration due to}$

$\text{gravity}g,\text{on two conducting supports at}\mathrm{ P }\text{and }Q.\text{When a current}\mathrm{ I}\text{ is passed through the loop, the loop}$

$\text{turns about the line }PQ\text{ by an angle}\mathrm{ \theta }\text{ given by}$

Correct Answer:
$\tan \theta =\frac{\pi rI{B}_0}{mg}$

Step-by-Step Explanation:

1. Magnetic Moment of the Circular Loop:
A circular loop of radius r carrying a current I forms a magnetic dipole with area:

$A=\pi r^2$

The magnitude of the magnetic dipole moment $\overrightarrow{M}$ is given by:

$M=IA=I\left(\pi r^2\right)$

The magnetic moment vector $\overrightarrow{M}$ points perpendicular to the plane of the circular loop. Initially, since the loop hangs vertically downwards, $\overrightarrow{M}$ points horizontally.

2. Magnetic Torque:
When the loop rotates by an angle θ about the horizontal axis line PQ, the magnetic dipole moment vector $\overrightarrow{M}$ also tilts by an angle θ from the horizontal plane.
As a result, the angle between $\overrightarrow{M}$ and the vertically upward magnetic field ${\overrightarrow{B}}_0$ becomes (90° - θ).
The magnitude of the deflection magnetic torque ${\tau }_M$ about the axis of rotation PQ is:

${\tau }_M=M{B}_0\sin (90°-\theta )=M{B}_0\cos \theta$

Substituting $M=I\pi r^2$:

${\tau }_M=I\pi r^2{B}_0\cos \theta$

3. Gravitational Torque:
The center of mass of the circular loop lies at its geometric center, which is at a distance r from the top tangent axis PQ.
When the loop turns by an angle θ, the horizontal distance of the center of mass from the axis line PQ becomes $r\sin \theta$.
The downward gravitational force mg acting at the center of mass produces a restoring torque ${\tau }_g$ given by:

${\tau }_g=mg\left(r\sin \theta \right)$

4. Rotational Equilibrium Condition:
In the equilibrium rotated position, the magnetic torque balancing the restoring gravitational torque:

${\tau }_g={\tau }_M$

$mgr\sin \theta =I\pi r^2{B}_0\cos \theta$

Dividing both sides by $mgr\cos \theta$:

$\frac{\sin \theta }{\cos \theta }=\frac{I\pi r^2{B}_0}{mgr}$

Simplifying the right hand side:

$\tan \theta =\frac{\pi rI{B}_0}{mg}$