A short bar magnet with its axis at `30^@` with a uniform magnetic field of 160 gauss experiences a torque of 0.032 Nm.
(a) Estimate the magnetic moment of the magnet.
(b) If we rotate the magnet from its most stable to its most unstable position, what will be the work done by an external agent on the magnet?
(c) With reference to part (b), calculate the work done by the force due to the magnetic field in the process.

Explanation

(a) Torque experienced by a bar magnet in an external magnetic field is
` tau = MB sin theta `
`rArr M= (tau)/(B sin theta)`
here ,` tau = 0.032 Nm , B = 160 xx 10^(-4) T = 0.016 T, theta = 30^@`
`rArr M = (0.032)/(0.016 xx sin 30^@) = (0.032 xx 2)/(0.016)`
` = 4Am^2`
(b) Angle between `vecM` and `vecB` is `0^@` in stable equilibrium and is `180^@` in an unstable equilibrium. Required work done by external agent will be:
` W = - MB (cos theta_2 - cos theta_1)`
here , ` theta_2 = 180^@ " and " theta_1 = 0^@`
` rArr W = -MB (cos 180^@ - cos 0^@)`
` = -MB (-1-1)`
` = 2MB = 2 xx 4 xx 0.016 = 0.128 J`
(c) Work done by force due to magnetic field = -W =-0.128 J