As loop of magnetic moment `M` is placed in the orientation of unstable equilbirum position in a uniform magnetic field `B`. The external work done in rotating it through an angle `theta` is

Options

A. `-MB(1-costheta)`
B. `-Mbcostheta`
C. `Mbcostheta`
D. `MB(1-costheta)`

Explanation

To find the external work done in rotating a loop of magnetic moment \( M \) through an angle \( \theta \) in a uniform magnetic field \( B \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Work Done**:
The work done \( W \) in rotating the magnetic moment in a magnetic field can be expressed as the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]

2. **Potential Energy in a Magnetic Field**:
The potential energy \( U \) of a magnetic moment \( M \) in a magnetic field \( B \) is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field.

3. **Identifying Initial and Final Angles**:
Since the loop is initially in an unstable equilibrium position, the initial angle \( \theta_{\text{initial}} \) is \( 180^\circ \) (or \( \pi \) radians). When the loop is rotated through an angle \( \theta \), the final angle \( \theta_{\text{final}} \) becomes:
\[ \theta_{\text{final}} = 180^\circ - \theta \]

4. **Calculating Initial and Final Potential Energies**:
- The initial potential energy \( U_{\text{initial}} \) when \( \theta = 180^\circ \):
\[ U_{\text{initial}} = -MB \cos(180^\circ) = -MB(-1) = MB \]
- The final potential energy \( U_{\text{final}} \) when \( \theta = 180^\circ - \theta \):
\[ U_{\text{final}} = -MB \cos(180^\circ - \theta) = -MB(-\cos \theta) = MB \cos \theta \]

5. **Finding the Work Done**:
Now, substituting the potential energies into the work done equation:
\[ W = U_{\text{final}} - U_{\text{initial}} = MB \cos \theta - MB \]
\[ W = MB (\cos \theta - 1) \]

6. **Rearranging the Expression**:
To express the work done in a more standard form, we can factor out \( -MB \):
\[ W = -MB (1 - \cos \theta) \]

### Final Answer:
The external work done in rotating the loop through an angle \( \theta \) is:
\[ W = -MB (1 - \cos \theta) \]