1. Given data

• Magnetic moment of bar magnet: $m = 2.0 \times 10^5\ \text{J T}^{-1}$
• Magnetic field: $B = 14 \times 10^{-5}\ \text{T}$
• Initial position: along the field, so $\theta_1 = 0^\circ$
• Final position: rotated through $60^\circ$, so $\theta_2 = 60^\circ$

2. Potential energy of a magnetic dipole in a uniform magnetic field

The potential energy is $U = -mB\cos\theta$

For a slow rotation, the work done by the external agent is equal to the increase in potential energy: $W = \Delta U = U_2 - U_1$

3. Compute initial and final energies

Initial energy: $U_1 = -mB\cos 0^\circ = -mB$

Final energy: $U_2 = -mB\cos 60^\circ = -mB\left(\frac{1}{2}\right)$

So, $\Delta U = -\frac{mB}{2} - (-mB) = \frac{mB}{2}$

Thus, $W = \frac{mB}{2}$

4. Substitute values

First calculate $mB$: $mB = (2.0 \times 10^5)(14 \times 10^{-5}) = 2.0 \times 14 = 28$

Therefore, $W = \frac{28}{2} = 14\ \text{J}$

5. Option check

• A: $14\ \text{J}$ ✅
• B: $8.4\ \text{J}$ ❌
• C: $4\ \text{J}$ ❌
• D: $1.4\ \text{J}$ ❌

Therefore, the correct option is A.