Question

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 \(B_0\), 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. \(tan\theta = \frac{\pi rlB_0}{(𝑚𝑔)}\)
• B. \(tan\theta = \frac{2\pi rlB_0}{(𝑚𝑔)}\)
• C. \(tan\theta = \frac{\pi rlB_0}{(2𝑚𝑔)}\)
• D. \(tan\theta = \frac{mg}{(\pi rlB_0)}\)

Answer 1

The Correct Option is A

1. Setup for Equilibrium:
We are given a system where a loop is suspended between two points \( P \) and \( Q \), and the system is under the influence of a magnetic field \( B_0 \). The magnetic torque acting on the loop is balanced by the weight of the object. For equilibrium, we consider the torque \( \tau \) and the forces involved:

$ \tau = mg r \sin \theta $

2. Equation for Magnetic Torque:
The torque due to the magnetic field is given by:

$ \tau = I \pi r^2 B_0 \cos \theta $

3. Equating the Torque and Solving for \( \theta \):
For equilibrium, the torques due to the magnetic field and gravitational force must balance:

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

Simplifying the equation:

$ \pi r B_0 \cos \theta = mg \sin \theta $

4. Final Expression for \( \tan \theta \):
Dividing both sides by \( mg \) and simplifying:

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

5. Conclusion:
Thus, the angle \( \theta \) at which the system is in equilibrium is given by the expression:

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

Answer 2

To solve the problem, analyze the torques acting on the wire loop when current \(I\) is passed in a magnetic field \(B_0\), causing it to rotate by an angle \(\theta\).

1. Magnetic moment and torque:
The magnetic moment \(\vec{\mu}\) of the current loop is: \[ \mu = I \times \text{area} = I \times \pi r^2 \] The magnetic torque \(\tau_B\) on the loop in magnetic field \(B_0\) is: \[ \tau_B = \mu B_0 \sin \theta = I \pi r^2 B_0 \sin \theta \]

2. Gravitational torque:
The loop has mass \(m\), so gravitational force \(mg\) acts at its center of mass. Since the wire is thin and stiff, center of mass is at the center of the loop. The torque due to gravity about the pivot line \(PQ\) is: \[ \tau_g = m g \times \text{perpendicular distance from pivot to center of mass} \] Distance from pivot to center of mass along the direction perpendicular to \(PQ\) is \( \frac{2r}{\pi} \times l \) (effective lever arm given by geometry of the loop and supports), so: \[ \tau_g = m g \times \frac{2 r l}{\pi} \]

3. At equilibrium, the two torques balance:
\[ \tau_B = \tau_g \] \[ I \pi r^2 B_0 \sin \theta = m g \frac{2 r l}{\pi} \cos \theta \] Using \(\sin \theta \approx \tan \theta\) for small angles, and rearranging: \[ \tan \theta = \frac{2 r l m g}{\pi^2 r^2 B_0 I m g} = \frac{\pi r l B_0}{m g} \]

4. Final formula:
\[ \tan \theta = \frac{\pi r l B_0}{m g} \] which matches option A.

Final Answer:
\[ \boxed{\tan \theta = \frac{\pi r l B_0}{m g}} \]