Q. A hundred turns of insulated copper wire are wrapped around an iron cylinder of 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 $1\ weber\ m^{- 2}$, in one direction to $1\ weber\ m^{- 2}$ in the opposite direction, how much charge flows through the circuit
A) $2 \times 10^{- 2}C$
B) $2 \times 10^{- 3}C$
C) $2 \times 10^{- 4}C$
D) $2 \times 10^{- 5}C$
Explanation
(a) $dQ = \frac{d\phi}{R} = \frac{nAdB}{R} = \frac{100 \times 1 \times 10^{- 3} \times 2}{10} = 2 \times 10^{- 2}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.