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. `0.2J`
B. `2.0J`
C. `4.18J`
D. `2xx10^(2)J`

Answer

A

Explanation

Magnetic moment of bar `M=10^(4)J//T`
`B=mu_(0)/(4pi).(2sqrt(2)M)/d^(3)`
Hence, work done `W=vec(M).vec(B)`
`=10^(4)xx4xx10^(-5)xxcos 60^(@) =0.2J`