A bar magnet of magnetic moment 6 `J//T` is aligned at `60^(@)` with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Explanation

Formula of work done is given by .
`W = MIB (cos theta f - cos thetai)`
M = magnetic moment
`betaf` = magnetic field
`thetaf` = Final angle-substanded by bar magnet.
`thetai` = Initial angle substanded by bar magnet.
(a) (i) Normal to the magnetic field `theta f = 90^(@)` `thetai = 60^(@)`
`M = 6J//T and B = 0.44T`
So, `W =- 6xx 0.44 [ cos 90^(@) - cos 60^(@)]`
`=- 2.46[ 0-1//2] = 1.32J`
(ii) Opposite to the magnetic filed.
`thetaf = 180^(@)" "thetai = 60^(@)`
`W =- 6 xx 0.44 [ cos 180^(@) - cos 60^(@)] = - 3.64 xx (-1-1//2) = 3.96J`
(b) `T = MB sin theta`
`=- 6 xx 0.44 sin (180^(@)) =0`