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A bar magnet with magnetic moment 2.5 X 10³JT^-1 is rotating in horizontal plane in the space containing magnetic induction B = 4 x 10^-5 T. The work done in rotating the magnet slowly from a direction parallel to the field to a direction 45° from the field, is (in joule) [UP CPMT 2015]

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0.02

0.0

0.03

0.2

answer is C.

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Detailed Solution

The work done (in rotating) by external agent

= change in potential energy

$= - M B \cos θ_{2} - (- M B \cos θ_{1})$

$= - M B (\cos 45 ° - \cos 0 °)$

$= (1 - \frac{1}{\sqrt{2}}) M B = \frac{0.41}{\sqrt{2}} \times \frac{1}{10} \approx 0.03 J$

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