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A circular coil of 500 turns of wire has an enclosed area of $0.1 m^{2}$ per turn. It is kept perpendicular to a magnetic field of induction 0.2T and rotated by $180^{0}$ About a diameter perpendicular to the field in 0.1 sec. How much charge will pass when the coil is connected to a galvanometer with a Combined resistance of 50 ohms.

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0.2C

0.4C

answer is B.

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Detailed Solution

$N = 500, A = 0.1 m^{2} p e r t u r n, B = 0.2 T, t = 0.1 s$

It is rotated by $18 0^{0}$ about a diameter

$R = 50 Ω$

$q = \frac{2 B A N}{R} = \frac{2 \times 0.2 \times 0.1 \times 500}{50}$

=0.4C

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