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

Options

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

Answer

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 \( B \), we can follow these steps:

### Step 1: Understand the Magnetic Moment and External Magnetic Field
The magnetic moment \( M \) of the loop is defined as:
\[ M = I \cdot A \]
where \( I \) is the current flowing through the loop and \( A \) is the area of the loop.

### Step 2: Determine the Initial and Final Orientation
When the loop is rotated by \( 30^\circ \) about an axis perpendicular to its plane, the angle \( \theta \) between the magnetic moment vector \( M \) and the magnetic field vector \( B \) does not change. This is because the rotation does not affect the alignment of \( M \) with respect to \( B \).

### Step 3: Calculate the Work Done
The work done \( W \) in rotating the magnetic moment in a magnetic field is given by:
\[ W = -M \cdot B \cdot \cos(\theta) \]
However, since the angle \( \theta \) remains \( 90^\circ \) (as the rotation does not change the angle between \( M \) and \( B \)), we have:
\[ W = -M \cdot B \cdot \cos(90^\circ) \]
Since \( \cos(90^\circ) = 0 \), we find:
\[ W = -M \cdot B \cdot 0 = 0 \]

### Step 4: Conclusion
Thus, 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 is \( 0 \).

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