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

Options

A. MB
B. `sqrt(3)(MB)/(2)`
C. `(MB)/(2)`
D. zero

Explanation

To find the work done to rotate a circular current loop of magnetic moment \( M \) in an external magnetic field \( \vec{B} \) by \( 30^\circ \) about an axis perpendicular to its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Moment and External Field**:
- The magnetic moment \( \vec{M} \) of the loop is directed along the axis perpendicular to the plane of the loop. In this case, let's assume it is along the \( z \)-axis.
- The external magnetic field \( \vec{B} \) is assumed to be along the \( y \)-axis.

2. **Initial and Final Orientation**:
- Initially, the angle \( \theta \) between \( \vec{M} \) and \( \vec{B} \) is \( 0^\circ \) (since both are aligned).
- After rotating the loop by \( 30^\circ \) about an axis perpendicular to its plane (which we can assume is the \( y \)-axis), the magnetic moment remains in the same plane, and the angle between \( \vec{M} \) and \( \vec{B} \) remains \( 0^\circ \).

3. **Potential Energy Calculation**:
- The potential energy \( U \) of a magnetic moment in a magnetic field is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta \]
- Initially, when \( \theta = 0^\circ \):
\[ U_{\text{initial}} = -MB \cos(0^\circ) = -MB \]
- After the rotation, since the angle \( \theta \) remains \( 0^\circ \):
\[ U_{\text{final}} = -MB \cos(0^\circ) = -MB \]

4. **Change in Potential Energy**:
- The work done \( W \) is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} = (-MB) - (-MB) = 0 \]

5. **Conclusion**:
- Since the change in potential energy is zero, the work done to rotate the loop by \( 30^\circ \) about an axis perpendicular to its plane is:
\[ W = 0 \]

### Final Answer:
The work done to rotate the loop by \( 30^\circ \) is \( 0 \).