Question 10 of 57
The magnetic moment of a bar magnet is . It is suspended in a uniform magnetic field of . The work done in rotating it from its most stable to most unstable position is:
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Answer: (A)
At stable equilibrium
At unstable equilibrium
$\begin{aligned}
& \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\
& \mathrm{W}=\Delta \mathrm{U} \\
& \text { W.D. }=2 \mathrm{mB} \\
& =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}
\end{aligned}$
At unstable equilibrium
$\begin{aligned}
& \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\
& \mathrm{W}=\Delta \mathrm{U} \\
& \text { W.D. }=2 \mathrm{mB} \\
& =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}
\end{aligned}$
