A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is:

1. MB

2. \(\frac{\sqrt{3}~MB}{2}\)

3. MB/2

4. zero

(d) Hint: The change in potential energy gives the work done.
The rotation of the loop by 30° about an axis perpendicular to its plane makes no change in the angle made by the axis of the loop with the direction of the magnetic field, therefore, the work done to rotate the loop is zero.