A bar magnet having a magnetic moment of $$2.0 \times 10^5$$ J T$$^{-1}$$, is placed along the direction of uniform magnetic field of magnitude $$B = 14 \times 10^{-5}$$ T. The work done in rotating the magnet slowly through $$60°$$ from the direction of field is

We need to find the work done in rotating a bar magnet from 0° to 60° in a uniform magnetic field.

Magnetic moment: $$M = 2.0 \times 10^5$$ J T$$^{-1}$$

Magnetic field: $$B = 14 \times 10^{-5}$$ T

Initial angle: $$\theta_1 = 0°$$ (along the field)

Final angle: $$\theta_2 = 60°$$

The work done in rotating a magnetic dipole in a uniform magnetic field is:

$$W = -MB(\cos\theta_2 - \cos\theta_1)$$

$$W = MB(\cos\theta_1 - \cos\theta_2)$$

$$W = 2.0 \times 10^5 \times 14 \times 10^{-5} \times (\cos 0° - \cos 60°)$$

$$W = 28 \times \left(1 - \frac{1}{2}\right)$$

$$W = 28 \times \frac{1}{2} = 14 \text{ J}$$

Hence, the correct answer is Option A.