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A hundred turns of insulated copper wire are wrapped around an iron cylinder of cross sectional area $1 \times 10^{- 3} m^{2}$ and are connected to a resistor. The total resistance in the circuit is 10 ohms. If the longitudinal magnetic induction in the iron changes from $1 Wb m^{- 2}$ in one direction to $1 Wb m^{- 2}$ in the opposite direction, how much charge flows through the circuit?

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$2 \times 10^{- 2} C$

$2 \times 10^{- 3} C$

$2 \times 10^{- 4} C$

$2 \times 10^{- 5} C$

answer is A.

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Detailed Solution

$dQ = \frac{dϕ}{R} = \frac{nAdB}{R} = \frac{100 \times 1 \times 10^{- 3} \times 2}{10} = 2 \times 10^{- 2} C$

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