Magnetic moment \( M = 10^4 \, \text{J T}^{-1} \).
Magnetic field strength \( B = 4 \times 10^{-5} \, \text{T} \).
The potential energy of a bar magnet in a magnetic field is \( U = -MB\cos\theta \), where \( \theta \) is the angle between the magnetic moment \( \vec{M} \) and the magnetic field \( \vec{B} \).
The magnet is rotated from a direction parallel to the field to a direction \( 60^\circ \) to the field.
Initial angle \( \theta_1 = 0^\circ \) (parallel to the field).
Final angle \( \theta_2 = 60^\circ \).
Initial potential energy \( U_1 = -MB\cos(0^\circ) = -MB(1) = -MB \).
Final potential energy \( U_2 = -MB\cos(60^\circ) = -MB\left(\frac{1}{2}\right) = -\frac{1}{2}MB \).
The work done in rotating the magnet slowly is equal to the change in its potential energy:
\( W = U_2 - U_1 = \left(-\frac{1}{2}MB\right) - (-MB) = -\frac{1}{2}MB + MB = \frac{1}{2}MB \).
Substitute the values of M and B:
\[ W = \frac{1}{2} (10^4 \, \text{J T}^{-1}) (4 \times 10^{-5} \, \text{T}) \]
\[ W = \frac{1}{2} \times 10^4 \times 4 \times 10^{-5} \, \text{J} \]
\[ W = \frac{1}{2} \times 4 \times 10^{4-5} \, \text{J} = 2 \times 10^{-1} \, \text{J} \]
\[ W = 0.
2 \, \text{J} \]
This matches option (1).