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.

Options

A. `0.2 J`
B. `0.1 J`
C. `0.4 J`
D. `0.3 J`

Answer

A

Explanation

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