A bar magnet having a magnetic moment of `2xx10^(4) JT^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B=6xx10^(-4)T` exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction `60^(@)` from the field is

Options

A. 12 J
B. 6J
C. 2J
D. `0.6J`

Answer

b

Explanation

(b) Work done `=MB(cos theta_(1)-cos theta_(2))=MB (cos 0^(@)- cos 60^(@))`
`=MB(1-(1)/(2))=(2xx10^(4)xx6xx10^(-4))/(2)=6J`