A magnet of magnetic moment `M` is situated with its axis along the direction of a magnetic field of strength `B`. The work done in rotating it by an angle of `180^(@)` will be

Options

A. ` - MB`
B. `+ MB`
C. zero
D. ` + 2 MB`

Explanation

To solve the problem of calculating the work done in rotating a magnet of magnetic moment \( M \) by an angle of \( 180^\circ \) in a magnetic field of strength \( B \), we can follow these steps:

### Step 1: Understand the Initial and Final Positions
Initially, the magnetic moment \( M \) is aligned with the magnetic field \( B \). When the magnet is rotated by \( 180^\circ \), the magnetic moment \( M \) will be opposite to the direction of the magnetic field.

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

For the initial position, where the angle \( \theta = 0^\circ \):
\[ U_{\text{initial}} = -MB \cos(0^\circ) = -MB \cdot 1 = -MB \]

### Step 3: Calculate the Final Potential Energy
For the final position, where the angle \( \theta = 180^\circ \):
\[ U_{\text{final}} = -MB \cos(180^\circ) = -MB \cdot (-1) = MB \]

### Step 4: Calculate the Work Done
The work done \( W \) in rotating the magnet is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} = MB - (-MB) = MB + MB = 2MB \]

### Conclusion
Thus, the work done in rotating the magnet by \( 180^\circ \) is:
\[ \boxed{2MB} \]

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