A magnet having magnetic moment of `2.0 xx 10^(4) (J)/(T)` is free to rotate in a horizontal plane where magnetic field is `6 xx 10^(-5)` T . Find the work done to rotate the magnet slowly from a direction parallel to field to a direction `60^(@)` from the field .

Explanation

To find the work done to rotate the magnet from a direction parallel to the magnetic field to a direction 60 degrees from the field, we can use the formula for work done in the context of magnetic moments:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Magnetic moment, \( M = 2.0 \times 10^4 \, \text{J/T} \)
- Magnetic field, \( B = 6 \times 10^{-5} \, \text{T} \)
- Initial angle, \( \theta_1 = 0^\circ \) (parallel to the field)
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