A short bar magnet placed with its axis at `30^@` with a uniform magnetic field of 0.16 T experiences a torque of magnitude 0.032 J. The magnetic dipole moment of the bar magnet is

Options

A. `0.23 J T^(-1)`
B. `0.40 J T^(-1)`
C. `0.80 J T^(-1)`
D. 0

Answer

B

Explanation

`vec m = (tau)/(B sin theta) = (0.032)/(0.16xxsin 30^@) = 0.40 J T^(-1)`