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A bar magnet having a magnetic moment of $2\times10^4$ J $T^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6\times10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^\circ$ from the field is
A
2 J
B
0.6 J
C
12 J
D
6 J
Detailed Solution
Work done in rotating a magnetic dipole from angle $\theta_1$ to $\theta_2$ with the field: $W = MB(\cos\theta_1 - \cos\theta_2)$
Here $\theta_1 = 0^\circ$ (parallel to the field) and $\theta_2 = 60^\circ$.
$W = MB(\cos0^\circ - \cos60^\circ) = MB\left(1 - \frac{1}{2}\right) = \frac{MB}{2}$
$W = \frac{2\times10^4\times6\times10^{-4}}{2} = \frac{12}{2}$
W = 6 J
Here $\theta_1 = 0^\circ$ (parallel to the field) and $\theta_2 = 60^\circ$.
$W = MB(\cos0^\circ - \cos60^\circ) = MB\left(1 - \frac{1}{2}\right) = \frac{MB}{2}$
$W = \frac{2\times10^4\times6\times10^{-4}}{2} = \frac{12}{2}$
W = 6 J