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A rectangular coil of 100 turns has length 5cm and width 4cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. If the torque acting on the coil is 0.2 N – m, The magnitude of the magnetic field B is

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0.4T

0.5 T

40 T

50T

answer is D.

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Detailed Solution

$B = \frac{T}{NIA} = \frac{0.2}{100 \times 2 \times 20 \times 10^{- 4}} = 0.5 T$

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