A bar magnet having a magnetic moment of $2 \times 10^{4} {JT}^{- 1}$ is free to rotate in a horizontal plane.
A horizontal magnetic field $B = 6 \times 10^{- 4} T$ exists in the space.
The work done in taking the magnet slowly from a direction parallel to the field to a direction $60 °$ from the field is

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$12 J$

$6 J$

$2 J$

$0.6 J$

answer is B.

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Detailed Solution

Work done

$\begin{array}{l} = MB (\cos θ_{1} - \cos θ_{2}) \\ = MB (\cos 0^{\circ} - \cos 60^{\circ}) \\ = MB (1 - \frac{1}{2}) = \frac{2 \times 10^{4} \times 6 \times 10^{- 4}}{2} = 6 J \end{array}$

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