1. Potential energy of a magnetic dipole in a magnetic field

A magnetic needle behaves like a magnetic dipole of magnetic moment $m$ in a uniform magnetic field $B$.

Its potential energy at angle $\theta$ with the field is

$U = -mB\cos\theta$

2. Work done to rotate from $0^\circ$ to $60^\circ$

Initially, the needle is parallel to the field, so

$\theta_1 = 0^\circ$

Finally,

$\theta_2 = 60^\circ$

The work required equals the increase in potential energy:

$W = U(60^\circ) - U(0^\circ)$

So,

$W = (-mB\cos 60^\circ) - (-mB\cos 0^\circ)$

$W = -mB\left(\frac12\right) + mB$

$W = \frac{mB}{2}$

Hence,

$mB = 2W$

3. Torque required to maintain the needle at $60^\circ$

The magnetic torque on a dipole at angle $\theta$ is

$\tau = mB\sin\theta$

At $\theta = 60^\circ$,

$\tau = mB\sin 60^\circ$

$\tau = (2W)\left(\frac{\sqrt{3}}{2}\right)$

$\tau = \sqrt{3}\,W$

This is the external torque needed to maintain the needle in that position.

4. Option check

• A: $\sqrt{3}W$ ✅
• B: $W$ ❌
• C: $\dfrac{\sqrt{3}}{2}W$ ❌
• D: $2W$ ❌

Therefore, the correct answer is A.