The work done in rotating a bar magnet f magnetic moment M from its unstable equilibrium positon to its stable equilibrium position in a uniform magnetic field B is

Options

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

Answer

To find the work done in rotating a bar magnet from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Positions**:
- The unstable equilibrium position occurs when the magnetic moment \( \vec{M} \) of the bar magnet is anti-parallel to the magnetic field \( \vec{B} \). This means \( \theta = 180^\circ \).
- The stable equilibrium position occurs when the magnetic moment \( \vec{M} \) is parallel to the magnetic field \( \vec{B} \). This means \( \theta = 0^\circ \).

2. **Magnetic Energy Formula**:
- The magnetic potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta \]

3. **Calculate Initial Magnetic Energy \( U_i \)**:
- For the unstable equilibrium position (\( \theta = 180^\circ \)):
\[ U_i = -MB \cos(180^\circ) = -MB \cdot (-1) = MB \]

4. **Calculate Final Magnetic Energy \( U_f \)**:
- For the stable equilibrium position (\( \theta = 0^\circ \)):
\[ U_f = -MB \cos(0^\circ) = -MB \cdot 1 = -MB \]

5. **Calculate Work Done**:
- The work done \( W \) in rotating the magnet from the unstable to the stable position is equal to the change in magnetic energy:
\[ W = U_f - U_i \]
- Substituting the values:
\[ W = (-MB) - (MB) = -MB - MB = -2MB \]

6. **Final Result**:
- The work done in rotating the bar magnet from its unstable equilibrium position to its stable equilibrium position is:
\[ W = -2MB \]