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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 _______.

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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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