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:
JEE MAIN 2024, Diffcult
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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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