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

Options

A. `0.6J`
B. `12J`
C. `6J`
D. `2J`

Explanation

To solve the problem, we need to calculate the work done in rotating the bar magnet from a direction parallel to the magnetic field to a direction that is 60 degrees from the magnetic 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 Work Done Formula:**
The work done \( W \) in rotating the magnet in a magnetic field is given by the formula:
\[ W = M \cdot B \cdot (\cos \theta_1 - \cos \theta_2) \]

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 Values into the Formula:**
Now we can substitute the values into the work done formula:
\[ W = (2 \times 10^4) \cdot (6 \times 10^{-4}) \cdot \left(1 - \frac{1}{2}\right) \]

5. **Simplify the Expression:**
- Calculate \( 1 - \frac{1}{2} = \frac{1}{2} \)
- Substitute this back into the equation:
\[ W = (2 \times 10^4) \cdot (6 \times 10^{-4}) \cdot \frac{1}{2} \]

6. **Calculate the Work Done:**
- First, calculate \( (2 \times 10^4) \cdot (6 \times 10^{-4}) \):
\[ = 12 \, \text{J} \]
- Now multiply by \( \frac{1}{2} \):
\[ W = 12 \cdot \frac{1}{2} = 6 \, \text{J} \]

### 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 \( 6 \, \text{J} \).

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