JEE Main 2024 — Magnetic Properties of Matter Question with Solution
JEE Main 2024 (04 Apr Shift 2)
Question
The magnetic moment of a bar magnet is . It is suspended in a uniform magnetic field of . The work done in rotating it from its most stable to most unstable position is:
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Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
At stable equilibrium
At unstable equilibrium $\begin{aligned} & \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\ & \mathrm{W}=\Delta \mathrm{U} \\ & \text { W.D. }=2 \mathrm{mB} \\ & =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J} \end{aligned}$
At unstable equilibrium $\begin{aligned} & \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\ & \mathrm{W}=\Delta \mathrm{U} \\ & \text { W.D. }=2 \mathrm{mB} \\ & =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J} \end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Magnetic Properties of Matter chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.