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A closed coil consists of 500 turns has area $4 c m ²$ and a resistance of $50 Ω$ . The coil is kept with its plane perpendicular to a uniform magnetic field of $0.2 W b / m ²$ . Calculate the amount of charge flowing through the coll if it is rotated through $180 °$ .

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$2.6 \times 10^{- 3} C$

$4.6 \times 10^{- 3} C$

$5.6 \times 10^{- 3} C$

$1.6 \times 10^{- 3} C$

answer is D.

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Detailed Solution

$∆ ϕ = 2 (N B S) ∆ q = \frac{∆ ϕ}{R} = \frac{2 N B S}{R} = \frac{2 \times 500 \times 0.2 \times 4 \times 10^{- 4}}{50} = 1.6 \times 10^{- 3} C$

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