Question

A magnetic dipole aligned parallel to a uniform magnetic field requires a work of W units to rotate it through 60°. The torque exerted by the field on the dipole in this new position is:

• A. 2W
• B. W
• C. \( \frac{\sqrt{3}}{2} W \)
• D. 3​W

Answer 1

The Correct Option is C

The work (W) required to rotate a magnetic dipole from an initial angle \( \theta_1 \) to a final angle \( \theta_2 \) within a magnetic field B is expressed as:

\( W = mB (\cos \theta_1 - \cos \theta_2) \)

Given that the dipole starts parallel to the field (\( \theta_1 = 0^\circ \)) and is rotated by 60° (\( \theta_2 = 60^\circ \)), the work done is:

\( W = mB (\cos 0^\circ - \cos 60^\circ) = mB (1 - \frac{1}{2}) = \frac{mB}{2} \)

This implies that \( mB = 2W \).

The torque (\( \tau \)) acting on a magnetic dipole in a magnetic field is calculated using:

\( \tau = mB \sin \theta \)

At the new position, where \( \theta = 60^\circ \), the torque is:

\( \tau = mB \sin 60^\circ = mB \cdot \frac{\sqrt{3}}{2} \)

Substituting \( mB = 2W \) into the torque equation yields:

\( \tau = 2W \cdot \frac{\sqrt{3}}{2} = \sqrt{3} W \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} W \)

Consequently, the torque exerted by the field on the dipole in this final orientation is \( \frac{\sqrt{3}}{2} W \).