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A magnet is parallel to a uniform magnetic field. If it is rotated by $60^{0}$ the work done is 0.8 J. How much work is done in moving it $30^{0}$ further?

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$0.8 \times 10^{7} erg$

0.4 J

8 J

0.8erg

answer is A.

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Detailed Solution

$W = MB (c os θ_{1} - c os θ_{2})$

First it is rotated from $0^{0} to 60^{0}$

$\Rightarrow 0.8 = M B (\cos 0^{0} - \cos 60^{0}) = \frac{M B}{2}$

Then it is rotated from $60^{0} to 90^{0}$

$W' = M B (\cos 60^{0} - \cos 90^{0}) = \frac{M B}{2} = 0.8 J$

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