Question

A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is:

• A. \( 30^\circ \)
• B. \( 45^\circ \)
• C. \( 60^\circ \)
• D. \( 75^\circ \)

Hint:

Torque on a magnetic dipole in a magnetic field is maximum when it’s perpendicular to the field. If torque becomes half, solve \( \sin \theta = \frac{1}{2} \).

Answer 1

The Correct Option is C

The torque \( \tau \) on a magnetic dipole in a magnetic field is given by: \[ \tau = MB \sin\theta \] where: - \( M \) is magnetic moment - \( B \) is magnetic field - \( \theta \) is the angle between \( \vec{M} \) and \( \vec{B} \) Initially, \( \theta = 90^\circ \Rightarrow \tau_0 = MB \sin 90^\circ = MB \)
We are told: \[ \tau = \frac{1}{2} \tau_0 = \frac{1}{2} MB \] Then: \[ MB \sin \theta = \frac{1}{2} MB \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = 30^\circ \] So, the magnet is rotated from \( 90^\circ \) to \( 60^\circ \), hence angle rotated is: \[ 90^\circ - 30^\circ = 60^\circ \] Final answer: \( 60^\circ \)