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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

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 _______.

  1. A \(16 \times 10^{-2} \mathrm{~J}\)
  2. B  \(8 \times 10^{-2} \mathrm{~J}\)
  3. C \(4 \times 10^{-2} \mathrm{~J}\)
  4. D Zero
Verified Solution

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 \(\mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB}\) \(\mathrm{W}=\Delta \mathrm{U}\) \(\text { W.D. }=2 \mathrm{mB}\)…
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