A magnetic dipole experiences a torque of $80 \sqrt{3} \mathrm{~N} \mathrm{~m}$ when placed in uniform…

A magnetic dipole experiences a torque of $80 \sqrt{3} \mathrm{~N} \mathrm{~m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of $60^{\circ}$ with magnetic field. The potential energy of the dipole is :
  1. $80 \mathrm{~J}$
  2. $-40 \sqrt{3} \mathrm{~J}$
  3. $-60 \mathrm{~J}$
  4. $-80 \mathrm{~J}$

Solution

$\tau=\mathrm{M} \times \mathrm{B}=\mathrm{MB} \sin 60=\frac{\sqrt{3}}{2} \mathrm{MB}=80 \sqrt{3}$
$\begin{aligned} & \mathrm{MB}=160 \\ & \mathrm{U}=-\mathrm{M} \cdot \mathrm{B}=-\mathrm{MB} \cos 60 \\ & \mathrm{U}=-160 \times 1 / 2=-80 \mathrm{~J}\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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