A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass 𝑚 and radius 𝑟 and it is in a uniform vertical magnetic field 𝐵0, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity 𝑔, on two conducting supports at P and Q. When a current 𝐼 is passed through the loop, the loop turns about the line PQ by an angle 𝜃 given by (A) tan 𝜃 = 𝜋𝑟𝐼𝐵0/(𝑚𝑔) (B) tan 𝜃 = 2𝜋𝑟𝐼𝐵0/(𝑚𝑔) (C) tan 𝜃 = 𝜋𝑟𝐼𝐵0/(2𝑚𝑔) (D) tan 𝜃 = 𝑚𝑔/(𝜋𝑟𝐼𝐵0)

To solve this problem, we need to analyze the forces and torques acting on the loop when a current $I$ is passed through it in the presence of a uniform vertical magnetic field $B_0$.

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Step 1: Magnetic Torque on the Loop

The loop is a circular wire of radius $r$, and when a current $I$ flows through it, it experiences a magnetic torque due to the magnetic field $B_0$. The magnetic torque ${\tau }_{\text{mag}}$ on a current-carrying loop is given by:

$${\tau }_{\text{mag}}=\vec{\mu }\times \overrightarrow{B_0}$$

where $\vec{\mu }$ is the magnetic moment of the loop. For a circular loop of radius $r$ carrying current $I$, the magnetic moment is:

$$\mu =I\cdot A=I\cdot \pi r^2$$

The magnetic torque is then:

$${\tau }_{\text{mag}}=\mu B_0\sin \theta =I\pi r^2{B}_0\sin \theta$$

Here, $\theta$ is the angle between the magnetic moment $\vec{\mu }$ and the magnetic field $\overrightarrow{B_0}$.

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Step 2: Gravitational Torque on the Loop

The loop has mass $m$, and it hangs vertically downward due to gravity. When the loop rotates by an angle $\theta$, the gravitational torque ${\tau }_{\text{grav}}$ about the line PQ is:

$${\tau }_{\text{grav}}=mg\cdot r\sin \theta$$

Here, $r$ is the radius of the loop, and $\sin \theta$ accounts for the component of the gravitational force perpendicular to the line PQ.

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Step 3: Equilibrium Condition

At equilibrium, the magnetic torque balances the gravitational torque:

$${\tau }_{\text{mag}}={\tau }_{\text{grav}}$$

Substituting the expressions for ${\tau }_{\text{mag}}$ and ${\tau }_{\text{grav}}$:

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

Canceling $\sin \theta$ (since $\theta \mathrm{\ne }0$):

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

Solving for $\theta$:

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

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Final Answer:

The angle $\theta$ is given by:

$$\boxed{{\displaystyle \text{(A) }\tan \theta =\frac{\pi rI{B}_0}{mg}}}$$