JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2024
- Hard
- MCQ
The magnetic moment of a bar magnet is \(0.5 \mathrm{Am}^2\). It is suspended in a uniform magnetic field of \(8 \times 10^{-2} \mathrm{~T}\). The work done in rotating it from its most stable to most unstable position is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(8 \times 10^{-2} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
At stable equilibrium
\(\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB}\)
At unstable equilibrium
\(\mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB}\)
\(\mathrm{W}=\Delta \mathrm{U}\)
\(\text { W.D. }=2 \mathrm{mB}\)
\(=2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}\)
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