A bar magnet of magnetic moment `1.5 JT^(-1)` lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of worl required by an external torque to turn the magnet so as to align its magnetic moment : (i) normal to the field direction, (ii) opposite to the field direction ?

Explanation

Here `m = 1.5 J T^(-1), B = 0.22 T, W = ?`
(a) Here `theta_(1) = 0^(@)` (along the field)
`theta_(2) = 90^(@)` (`_|_` to the field)
As `W = -mB (cos theta_(2) - cos theta_(1))`
`W =-1.5 xx 0.22 (cos 90^(@) - cos 0^(@)) = -0.33 (0-1) = 0.33f`
(ii) Here `theta_(1) = 0^(@), theta_(2) = 180^(@)`
`W = -1.5 xx 022 (cos 180^(@) - cos 0^(@))`
`= -0.33 (-1-1) = 0.66 J`.