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

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4 J

8.4 J

14 J

1.4 J

answer is A.

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Detailed Solution

Work done = MB $(\cos θ_{1} - \cos θ_{2})$
$\begin{array}{l} θ_{1} = 0^{\circ}, θ_{2} = 60^{\circ} \\ = 2 \times 10^{5} \times 14 \times 10^{- 5} (1 - 1 / 2) \\ = 14 J \end{array}$

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