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 \( \tau \) experienced by a magnetic dipole within a magnetic field \( B \) is defined by the equation: \( \tau = MB \sin\theta \). Here, \( M \) represents the magnetic moment, \( B \) denotes the magnetic field strength, and \( \theta \) is the angle between the magnetic moment vector \( \vec{M} \) and the magnetic field vector \( \vec{B} \). Initially, the angle is \( \theta = 90^\circ \), resulting in an initial torque of \( \tau_0 = MB \sin 90^\circ = MB \). We are given that the new torque is half of the initial torque: \( \tau = \frac{1}{2} \tau_0 = \frac{1}{2} MB \). Substituting this into the torque equation yields: \( MB \sin \theta = \frac{1}{2} MB \). This simplifies to \( \sin \theta = \frac{1}{2} \), which means \( \theta = 30^\circ \). Therefore, the magnet has been rotated from an initial angle of \( 90^\circ \) to a final angle of \( 30^\circ \). The angle of rotation is calculated as \( 90^\circ - 30^\circ = 60^\circ \). Final answer: \( 60^\circ \)