A short bar magnet placed with its axis at $30^{\circ}$ with an external field of 800 Gauss, experie

Question

A short bar magnet placed with its axis at $30^{\circ}$ with an external field of 800 Gauss, experiences a torque of $0.016 \mathrm{~N} . \mathrm{m}$. The work done in moving it from most stable to most unstable position is $\alpha 	imes 10^{-3} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

The torque experienced by a magnetic dipole in an external magnetic field is $	au = MB \sin	heta$. Given $	au = 0.016 	ext{ N}\cdot	ext{m}$ and $	heta = 30^\circ$, we have $0.016 = MB \sin(30^\circ) = 0.5 MB \implies MB = 0.032 	ext{ J}$. The work done in rotating the magnet from its most stable position ($	heta_1 = 0^\circ$) to its most unstable position ($	heta_2 = 180^\circ$) is given by $W = -MB(\cos	heta_2 - \cos	heta_1) = -MB(\cos 180^\circ - \cos 0^\circ) = -MB(-1 - 1) = 2MB$. Therefore, $W = 2 	imes 0.032 = 0.064 	ext{ J} = 64 	imes 10^{-3} 	ext{ J}$. Thus, $\alpha = 64$.