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 \( M \) by \( 30^\circ \) about an axis perpendicular to its plane in an external magnetic field \( \vec{B} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Moment and Magnetic Field Orientation**:
- The magnetic moment \( \vec{M} \) of the loop is oriented at an angle \( \alpha \) with respect to the magnetic field \( \vec{B} \).

2. **Identify the Rotation Axis**:
- The loop is rotated about an axis that is perpendicular to its plane. This means that the angle between the magnetic moment and the magnetic field does not change as the loop is rotated.

3. **Determine the Initial and Final Angles**:
- Before rotation, the angle between the magnetic moment and the magnetic field is \( \theta_1 = \alpha \).
- After rotating the loop by \( 30^\circ \), the angle remains \( \theta_2 = \alpha \) because the rotation does not affect the orientation of \( \vec{M} \) with respect to \( \vec{B} \).

4. **Calculate the Work Done**:
- The work done \( W \) in rotating the loop in a magnetic field is given by the formula:
\[ W = -M B (\cos \theta_2 - \cos \theta_1) \]
- Since \( \theta_1 = \theta_2 = \alpha \), we have:
\[ W = -M B (\cos \alpha - \cos \alpha) = -M B (0) = 0 \]

5. **Conclusion**:
- The work done to rotate the loop by \( 30^\circ \) about an axis perpendicular to its plane is \( 0 \).

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

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