The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of $8 imes 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:

Question

The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of $8 	imes 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:

Aicharya · Accepted Answer

At stable equilibrium $\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB}$
At unstable equilibrium $\begin{aligned} & \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\ & \mathrm{W}=\Delta \mathrm{U} \\ & \text { W.D. }=2 \mathrm{mB} \\ & =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J} \end{aligned}$

The magnetic moment of a bar magnet is $0.5 \mathrm{Am}^2$. It is suspended in a uniform magnetic field of…

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