Question

A bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated? \([\sin(\pi/2)=1]\)

• A. \( \sin^{-1}(0.8660) \)
• B. \( \sin^{-1}(0.7071) \)
• C. \( \sin^{-1}(1) \)
• D. \( \sin^{-1}(0.5) \)

Hint:

Torque on a magnetic dipole varies as sine of angle with field.

Answer 1

The Correct Option is D

Step 1: Torque on a bar magnet.
\[ \tau = MB \sin\theta. \]
Step 2: Initial condition.
Initially, \(\theta = 90^\circ\), so \[ \tau_0 = MB. \]
Step 3: New condition.
Torque is halved: \[ \frac{\tau_0}{2} = MB\sin\theta. \]
Step 4: Solving for angle.
\[ \sin\theta = \frac{1}{2}. \]
Step 5: Conclusion.
\[ \theta = \sin^{-1}(0.5). \]