A bar magnet of magnetic moment `2Am^(2)` is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east-west position. Then the KE of the magnet as it takes N-S position is
`(B_(H)=25muT)`

Options

A. `25muJ`
B. `50muJ`
C. `100muJ`
D. `12.5muJ`

Explanation

To solve the problem, we need to calculate the kinetic energy of a bar magnet as it rotates from the east-west position to the north-south position in the presence of a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Initial and Final Positions**:
- The bar magnet is initially in the east-west position, which means it is perpendicular to the magnetic field.
- When it rotates to the north-south position, it aligns with the magnetic field.

2. **Magnetic Moment and Magnetic Field**:
- Given:
- Magnetic moment \( M = 2 \, \text{Am}^2 \)
- Magnetic field \( B_H = 25 \, \mu T = 25 \times 10^{-6} \, T \)

3. **Calculating Initial Potential Energy**:
- The potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -M \cdot B \cdot \cos(\theta) \]
- Initially, when the magnet is in the east-west position, \( \theta = 90^\circ \):
\[ U_{\text{initial}} = -M \cdot B_H \cdot \cos(90^\circ) = 0 \]

4. **Calculating Final Potential Energy**:
- When the magnet is in the north-south position, \( \theta = 0^\circ \):
\[ U_{\text{final}} = -M \cdot B_H \cdot \cos(0^\circ) = -M \cdot B_H \]
- Substituting the values:
\[ U_{\text{final}} = -2 \, \text{Am}^2 \cdot 25 \times 10^{-6} \, T = -50 \times 10^{-6} \, \text{J} = -50 \, \mu J \]

5. **Calculating Change in Potential Energy**:
- The change in potential energy \( \Delta U \) as the magnet moves from the initial to the final position is:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = -50 \, \mu J - 0 = -50 \, \mu J \]

6. **Relating Change in Potential Energy to Kinetic Energy**:
- The change in potential energy is converted into kinetic energy \( KE \):
\[ KE = -\Delta U = 50 \, \mu J \]

### Final Answer:
The kinetic energy of the magnet as it takes the north-south position is \( 50 \, \mu J \).