A bar magnet having a magnetic moment of `1.0 xx 10^4 J T^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B = 4 xx 10^(-5) T` exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction `60^@` from the field.

Explanation

The work done by the external agent = change in potential energy.
`= (-MB costheta_2) - (-MB costheta_1)`
`= -MB (cos 60^2 - cos 0^@) = 1/2 MB`
`= 1/2 xx (1.0 xx 10^4 JT^(-1) (4 xx 10^(-5) T) = 0.2 J`.