The magnetic moment of a bar magnet is $0.5 \mathrm{~A} \mathrm{~m}^{2}$. It is suspended in a unifo

Question

The magnetic moment of a bar magnet is $0.5 \mathrm{~A} \mathrm{~m}^{2}$. 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\

Professor PC Thomas Classes · Answer

(b) : Here: $M=0.5 \mathrm{~A} \mathrm{~m}^{2}, B=8 	imes 10^{-2} \mathrm{~T}$ For stable equilibrium, $	heta=0^{\circ}$\ For unstable, equilibrium, $	heta=180^{\circ}$\ $W=\Delta U=U_{f}-U_{i}$\ $=-M B \cos 180^{\circ}-\left(-M B \cos 0^{\circ}ight)$\ $W=2 M B=2 	imes 0.5 	imes 8 	imes 10^{-2}=8 	imes 10^{-2} \mathrm{~J}$\