A bar magnet placed in a uniform magnetic field of strength `0*3T` with its axis at `30^@` to the field experiences a torque of `0*06N-m`. What is the magnetic moment of the bar magnet?

Answer

`0*4Am^2`

Explanation

Here, `B=0*3T`, `theta=30^@`, `tau=0*06N-m`, `M=?`
As `tau=Mbsintheta`, `M=(tau)/(Bsintheta)=(0*06)/(0*3sin30^@)`