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A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of 0.1 weber/ $m^{2}$ . The amount of work done in rotating it through 180° from its equilibrium position will be

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0.2 J

0.4 J

0.8 J

0.1 J

answer is A.

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Detailed Solution

$Magnetic moment due to a circular coil of n turns is M = niA$

$and work done to rotate by 180^{0} will be MB (1 - Cos 180^{0}) = 2 MB$

$W = 2 {niπr}^{2} B$

$= 2 x 50 x 2 xπx 16 x 10^{- 4} x 0.1 = 0.1 J$

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