Work done in rotating a bar magnet from 0 to angle `theta` is:

Options

A. `MH(1-cos theta)`
B. `M/H(1- cos theta)`
C. `M/H (cos theta -1)`
D. `MH(cos theta -1)`

Explanation

To find the work done in rotating a bar magnet from an angle of 0 degrees to an angle of θ, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The work done (W) in rotating a bar magnet is given by the formula:
\[ W = M B (\cos \theta_i - \cos \theta_f) \]
where:
- \( M \) is the magnetic moment of the bar magnet,
- \( B \) is the magnetic field strength,
- \( \theta_i \) is the initial angle,
- \( \theta_f \) is the final angle.

2. **Identify Initial and Final Angles**: In this case, we are rotating the magnet from:
- Initial angle \( \theta_i = 0^\circ \)
- Final angle \( \theta_f = \theta \)

3. **Substitute the Angles into the Formula**: Plugging the angles into the work done formula gives:
\[ W = M B (\cos 0^\circ - \cos \theta) \]

4. **Calculate Cosine Values**: We know that:
- \( \cos 0^\circ = 1 \)
Therefore, the equation simplifies to:
\[ W = M B (1 - \cos \theta) \]

5. **Final Expression for Work Done**: Thus, the work done in rotating the bar magnet from 0 degrees to θ degrees is:
\[ W = M B (1 - \cos \theta) \]

### Final Answer:
The work done in rotating a bar magnet from 0 to angle θ is:
\[ W = M B (1 - \cos \theta) \]