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:

$\sqrt{3}W$

The correct answer is Option (3) → $\sqrt{3}W$

The work done W to rotate a magnetic dipole is,

$W=\int\limits_{θ_1}^{θ_2}τ\,dθ$

and,

Torque, $τ=mB\sin θ$

$∴W=\int\limits_{0}^{60}mB\sin θ\,dθ$

$=mB[-\cos θ]_{0}^{60}$

$=mB\left[-\frac{1}{2}+1\right]=\frac{mB}{2}$

Now, to find the new torque at $θ=0$

$τ=mB\sin 60°$

$=mB\frac{\sqrt{3}}{2}=2W×\frac{\sqrt{3}}{2}=\sqrt{3}W$