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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/m2. The amount of work done in rotating it through 180° from its equilibrium position will be

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a

0.1 J

b

0.2 J

c

0.4 J

d

0.8 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 1800 will be MB1-Cos1800= 2MB W= 2 niπr2B   =2x50 x 2xπx16x10-4x0.1= 0.1 J
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