Magnetic Effects & Magnetism

JEE Main level, single correct, medium

263 of 276

Question 263 of 276

A magnetic dipole experiences a torque of 803 N m80 \sqrt{3} \mathrm{~N} \mathrm{~m} when placed in uniform magnetic field in such a way that dipole moment makes angle of 60∘60^{\circ} with magnetic field. The potential energy of the dipole is :

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Answer: (D) −80 J-80 \mathrm{~J}

τ=M×B=MBsin⁡60=32MB=803\tau=\mathrm{M} \times \mathrm{B}=\mathrm{MB} \sin 60=\frac{\sqrt{3}}{2} \mathrm{MB}=80 \sqrt{3}
MB=160U=−M⋅B=−MBcos⁡60U=−160×1/2=−80 J\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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