A bar magnet of magnetic moment `1.5JT^-1` lies aligned with the direction of a uniform magnetic field of `0.22T`.
(a) What is the amount of work done to turn the magnet so as to align its mangetic moment
(i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Explanation

Given, magnetic moment of the bar magnet
`M = 1.5 JT^(-1)`
External magnetic field, B = 0.22 T
Work done in rotating a bar magnet from angle `theta_1` with the external magnetic field to angle `theta_2` with the magnetic field is:

`W = -MB (cos theta_2 - cos theta_1)`
(a) (i) In this case `theta_1 = 0`
` theta_2 = pi/2 ` radian
` therefore W = -MB (cos pi/2 - cos0^@) = MB`
` = - 1.5 xx 0.22 = 0.33 J`
(ii) `thete_1 = 0 , theta_2 = pi ` radian
`W = -MB (-1-1) = 2MB`
` = 2 xx 0.33 = 0.66 J`
(b) (i) Torque acting on the bar magnet will be:
` tau = MB sin pi/2 = MB xx `
` = 1.5 xx 0.20 = 0.33 Nm`
It will tend to align with the magnetic moment vector along the direction of the magnetic field.
(ii) ` tau = MB sin pi = MB xx 0 = 0`