JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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 :
- A \(80 \mathrm{~J}\)
- B \(-40 \sqrt{3} \mathrm{~J}\)
- C \(-60 \mathrm{~J}\)
- D \(-80 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(-80 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(\tau=\mathrm{M} \times \mathrm{B}=\mathrm{MB} \sin 60=\frac{\sqrt{3}}{2} \mathrm{MB}=80 \sqrt{3}\) \(\begin{aligned} & \mathrm{MB}=160 \\ & \mathrm{U}=-\mathrm{M} \cdot \mathrm{B}=-\mathrm{MB} \cos 60 \\ & \mathrm{U}=-160 \times 1 / 2=-80 \mathrm{~J}\end{aligned}\)
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