A magnet having magnetic moment of `2.0 xx 10^(4) (J)/(T)` is free to rotate in a horizontal plane where magnetic field is `6 xx 10^(-5)` T . Find the work done to rotate the magnet slowly from a direction parallel to field to a direction `60^(@)` from the field .

Explanation

`W = MB [ cos theta_(1) - cos theta_(2)]`
`= 2.0 xx 10^(4) xx 6 xx 10^(-5) [ cos 0^(@) - cos 60^(@)]`
`= 2 xx 10^(4) xx 6 xx 10^(-5) [ 1 - (1)/(2)]`
`= (12)/(2) xx 10^(-1)`
`= 0.6 J`