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:
Answer & explanation
Correct answer: option 3
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 $θ=0$
$τ=mB\sin 60°$
$=mB\frac{\sqrt{3}}{2}=2W×\frac{\sqrt{3}}{2}=\sqrt{3}W$