A ciruclar current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field `vecB`. The work done to rotate the loop by `30^@` about an axis perpendicular to its plane is

Options

A. `MB`
B. `sqrt3(MB)/(2)`
C. `(MB)/(2)`
D. zero

Explanation

To solve the problem of finding 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 Configuration**:
- We have a circular current loop with a magnetic moment \( M \).
- The loop is placed in an external magnetic field \( \vec{B} \).
- The loop is rotated by \( 30^\circ \) about an axis that is perpendicular to the plane of the loop.

2. **Identify the Initial and Final Angles**:
- Let the initial angle between the magnetic moment \( M \) and the magnetic field \( \vec{B} \) be \( \theta_1 \).
- After rotating the loop by \( 30^\circ \), the final angle will be \( \theta_2 = \theta_1 + 30^\circ \).

3. **Formula for Work Done**:
- The work done \( W \) in rotating a magnetic moment in a magnetic field is given by the formula:
\[ W = -M B (\cos \theta_2 - \cos \theta_1) \]
- Here, \( M \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta_1 \) and \( \theta_2 \) are the initial and final angles respectively.

4. **Evaluate the Change in Cosine**:
- Since the rotation is about an axis perpendicular to the plane of the loop, the direction of the magnetic moment relative to the magnetic field does not change significantly in this case.
- Thus, if \( \theta_1 \) and \( \theta_2 \) are such that the angle between the magnetic moment and the magnetic field remains effectively the same, we can conclude that:
\[ \cos \theta_2 = \cos \theta_1 \]

5. **Calculate Work Done**:
- Substituting into the work done formula:
\[ W = -M B (\cos \theta_1 - \cos \theta_1) = -M B (0) = 0 \]
- Therefore, the work done to rotate the loop by \( 30^\circ \) is \( 0 \).

### Final Answer:
The work done to rotate the loop by \( 30^\circ \) about an axis perpendicular to its plane is \( 0 \).