Question

A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved, by rotating it, the angle by which it is to be rotated is

• A. 90°
• B. 30°
• C. 60°
• D. 45°

Hint:

When a magnet is placed perpendicular to the magnetic field, the torque is maximized. To reduce the torque by half, the angle should be rotated to 30° from the original position.

Answer 1

The Correct Option is B

The torque (\(\tau\)) acting on a bar magnet placed in a uniform magnetic field is given by the equation: \[ \tau = M B \sin \theta \] Where: - \(M\) is the magnetic moment of the bar magnet, - \(B\) is the magnetic field strength, - \(\theta\) is the angle between the magnetic moment and the magnetic field. When the bar magnet is perpendicular to the magnetic field, \(\theta = 90^\circ\), and the torque is maximum, given by: \[ \tau_{\text{max}} = M B \sin 90^\circ = M B \] If the couple acting on the magnet is to be halved, the new torque must be half of the maximum torque. Therefore, we require: \[ \frac{1}{2} M B = M B \sin \theta \] Solving for \(\theta\), we get: \[ \sin \theta = \frac{1}{2} \]
Thus, \[ \theta = 30^\circ \] Therefore, the angle by which the bar magnet must be rotated to halve the torque is \( 30^\circ \), and the correct answer is B.