A magnetic dipole aligned parallel to a uniform magnetic field requires a work of W to rotate it through 60°. The torque exerted by the field on the dipole in this new position is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → √3 W
The work done in turning a magnetic dipole slowly equals the increase in its potential energy, \( U = -mB\cos\theta \). Starting from the aligned position (\( \theta = 0 \)):
\( \displaystyle W = mB(\cos 0^\circ - \cos 60^\circ) = mB\left(1 - \frac{1}{2}\right) = \frac{mB}{2} \)
\( \displaystyle mB = 2W \)
The torque in the new position is
\( \displaystyle \tau = mB\sin 60^\circ = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}\,W \)
Note: A similar question was also asked in CUET (UG) 2024 on 19-07-2024 (Slot 1), and in CUET (UG) 2023 on 30-05-2023 (Slot 2).