A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in its position, will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{3}\,W \)
A magnetic needle is a magnetic dipole of moment \( m \). In a field \( B \) its potential energy is \( U = -mB\cos\theta \), and the field exerts a torque \( \tau = mB\sin\theta \) on it. The work done in turning it from \( \theta = 0 \) to \( \theta = 60^\circ \) is the increase in potential energy.
\( \displaystyle W = mB(\cos 0^\circ - \cos 60^\circ) = mB\left(1 - \frac{1}{2}\right) = \frac{mB}{2} \)
\( mB = 2W \)
To hold the needle at \( 60^\circ \), the applied torque must balance the torque of the field.
\( \displaystyle \tau = mB\sin 60^\circ = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}\,W \)
Note: A similar question was also asked in CUET (UG) 2026 on 20-05-2026 (Slot 1), and in CUET (UG) 2024 on 19-07-2024 (Slot 1).