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. `-MB(costheta)`
C. `MBcos theta`
D. `MB(1-costheta)`

Answer

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. **Understand the Initial and Final Positions**:
- The loop is initially in an unstable equilibrium position, which means it is aligned with the magnetic field. Thus, the initial angle \( \theta_i \) is \( 180^\circ \) (opposite to the field direction).
- After rotation through an angle \( \theta \), the final angle \( \theta_f \) becomes \( 180^\circ - \theta \).

2. **Calculate the Potential Energy**:
- The potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos(\theta) \]
- For the initial position (unstable equilibrium):
\[ U_i = -MB \cos(180^\circ) = -MB(-1) = MB \]
- For the final position after rotation:
\[ U_f = -MB \cos(180^\circ - \theta) = -MB(-\cos(\theta)) = MB \cos(\theta) \]

3. **Calculate the Work Done**:
- The work done \( W \) in rotating the loop is the change in potential energy:
\[ W = U_f - U_i \]
- Substituting the values of \( U_f \) and \( U_i \):
\[ W = MB \cos(\theta) - MB \]
- Factor out \( MB \):
\[ W = MB (\cos(\theta) - 1) \]
- Since \( \cos(180^\circ) = -1 \), we can rewrite this as:
\[ W = -MB (1 - \cos(\theta)) \]

4. **Final Expression**:
- Therefore, the external work done in rotating the loop through an angle \( \theta \) is:
\[ W = -MB (1 - \cos(\theta)) \]