A magnetic dipole aligned parallel to a uniform magnetic field requires a work of W units to rotate it through \( 60^\circ \). 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) → \( \sqrt{3}\,W \)
The potential energy of a magnetic dipole in a field is \( U = -mB\cos\theta \). The work needed to rotate it equals the increase in this energy, and the torque at angle \( \theta \) is \( \tau = mB\sin\theta \).
Rotating from \( \theta = 0^\circ \) to \( \theta = 60^\circ \):
\( \displaystyle W = U_{60} - U_{0} = -mB\cos60^\circ + mB\cos0^\circ = mB\left(1 - \frac{1}{2}\right) = \frac{mB}{2} \)
So \( mB = 2W \). In the new position,
\( \displaystyle \tau = mB\sin60^\circ = 2W\times\frac{\sqrt{3}}{2} = \sqrt{3}\,W \)
Note: A similar question was also asked in CUET (UG) 2026 on 20-05-2026 (Slot 1), and in CUET (UG) 2023 on 30-05-2023 (Slot 2).