A short bar magnet placed with its axis inclined at `30^@` to the external magnetic field of 800 G acting horizontally experiences torque of 0.016 Nm. Calculate :
(i) the magnetic moment of the magnet.
(ii) the work done by an external force in moving it from most stable to most unstable position.
(iii) what is the work done by the force due to the external magnetic field in the process mentioned in (ii)?

Explanation

It given that `B = 800 G = 8 xx 10^(-2) T , tau =0.016 N m and theta =30^@`
(i) Magnetic moment of the magnet m`=(tau)/(B sin theta) = (0.016)/(8 xx 10^(-2) xx sin 30^@) = 0.40 A m^2`
(ii) In most stable position `theta_1 = 0^@` and in most unstable position `theta_2 = 180^@ `. hence , work done by an external force in moving the magnet from stable to unstable position
`W = mB[sin theta_1 - sin theta_2] = 0.40 xx 8 xx 10^(-2) [ sin 0^@ - sin 180^@] = +0.064 J`
(iii) work done by the force due to magnetic field B in moving the magnet from stable to unstable position.
`W. =-W =-0.064J`
[Since now force and displacement are in mutually opposite directions]