A bar magnet has a magnetic moment of 2.5 J/T and is placed in a magnetic field of 0.2 T. find the work done in turning the magnet from parallel to anti-parallel position relative to field direction.

Answer

To find the work done in turning a bar magnet from a parallel position to an anti-parallel position relative to a magnetic field, we can use the formula for work done (W) in a magnetic field, which is given by:

\[ W = -M \cdot B \cdot (\cos \theta_f - \cos \theta_i) \]

Where:
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field,
- \( \theta_i \) is the initial angle (0 degrees for parallel),
- \( \theta_f \) is the final angle (180 degrees for anti-parallel).

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic moment, \( M = 2.5 \, \text{J/T} \)
- Magnetic field, \( B = 0.2 \, \text{T} \)
- Initial angle, \( \theta_i = 0^\circ \)
- Final angle, \( \theta_f = 180^\circ \)

2. **Calculate the cosine values:**
- \( \cos(0^\circ) = 1 \)
- \( \cos(180^\circ) = -1 \)

3. **Substitute the values into the work done formula:**
\[ W = -M \cdot B \cdot (\cos(180^\circ) - \cos(0^\circ)) \]
\[ W = -2.5 \cdot 0.2 \cdot (-1 - 1) \]

4. **Simplify the expression:**
\[ W = -2.5 \cdot 0.2 \cdot (-2) \]
\[ W = -2.5 \cdot 0.2 \cdot -2 = 2.5 \cdot 0.2 \cdot 2 \]

5. **Calculate the final result:**
\[ W = 2.5 \cdot 0.2 \cdot 2 = 1.0 \, \text{J} \]

### Final Answer:
The work done in turning the magnet from parallel to anti-parallel position is **1.0 Joule**.