There exists a horizontal magnetic field in space of value $B = 9 × 10^{-4}T$. A magnetic bar of magnetic moment $9 × 10^3 J/T$ is placed at an angle 60° to the field and is free to rotate. Finally, it comes to a position parallel to the field. Work done by magnetic field on the magnetic bar will be

$-4.05 J$

The correct answer is Option (3) → $-4.05 J$

Work done by magnetic field in rotating a magnetic dipole from angle $\theta_1$ to $\theta_2$:

$W = M B (\cos\theta_1 - \cos\theta_2)$

Given: $B = 9 \times 10^{-4} \ \text{T}$, $M = 9 \times 10^3 \ \text{J/T}$, $\theta_1 = 60^\circ$, $\theta_2 = 0^\circ$

$W = (9 \times 10^3)(9 \times 10^{-4}) (\cos 60^\circ - \cos 0^\circ)$

$W = 8.1 \times (0.5 - 1) = 8.1 \times (-0.5) = -4.05 \ \text{J}$