Question

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?

Answer 1

Given: The angle between magnetic field and magnetic moment is 30°, magnitude of the external magnetic field B=0.25 T , and the torque experienced by the magnet is 4.5× 10 −2  J.

The torque is given as,

T=MBsinθ

Where, M is the magnitude of the magnetic moment.

By substituting the given values in the above equation, we get.

4.5× 10 −2 =M( 0.25sin30° ) M=0.36  JT −1

Thus, magnitude of the magnetic moment is 0.36  JT −1 .