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 cos theta_(2))-(-MB cos theta_(1))`
`=-MB (cos 60^(@)-cos^(@))`
`=1/2MB`
`=1/2xx(1.0 xx 10^(4) J//T)(4xx10^(-5)T)=0.2 J`