A bar magnet having a magnetic moment of `2xx10^(4) JT^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B=6xx10^(-4)T` exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction `60^(@)` from the field is

Answer

To solve the problem, we need to calculate the work done in rotating a bar magnet from a position parallel to a magnetic field to a position at an angle of 60 degrees from the field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic moment, \( M = 2 \times 10^4 \, \text{JT}^{-1} \)
- Magnetic field, \( B = 6 \times 10^{-4} \, \text{T} \)
- Initial angle, \( \theta_1 = 0^\circ \) (parallel to the field)
- Final angle, \( \theta_2 = 60^\circ \)

2. **Use the Formula for Work Done:**
The work done \( W \) in rotating a magnetic moment in a magnetic field is given by:
\[ W = -M B (\cos \theta_2 - \cos \theta_1) \]

3. **Calculate \( \cos \theta_1 \) and \( \cos \theta_2 \):**
- \( \cos \theta_1 = \cos 0^\circ = 1 \)
- \( \cos \theta_2 = \cos 60^\circ = \frac{1}{2} \)

4. **Substitute the Values into the Formula:**
\[ W = -M B \left( \frac{1}{2} - 1 \right) \]
\[ W = -M B \left( -\frac{1}{2} \right) = \frac{1}{2} M B \]

5. **Substitute the Values of \( M \) and \( B \):**
\[ W = \frac{1}{2} \times (2 \times 10^4) \times (6 \times 10^{-4}) \]

6. **Calculate the Work Done:**
\[ W = \frac{1}{2} \times 2 \times 6 \times 10^4 \times 10^{-4} \]
\[ W = \frac{1}{2} \times 12 \, \text{J} = 6 \, \text{J} \]

### Final Answer:
The work done in taking the magnet from a direction parallel to the field to a direction \( 60^\circ \) from the field is **6 Joules**.