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