A long solenoid of diameter $0.1 \text{ m}$ has $2 \times 10^4$ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius $0.01 \text{ m}$ is placed with its axis coinciding with the solenoid's axis. The current in the solenoid reduces at a constant rate to $0 \text{ A}$ from $4 \text{ A}$ in $0.05 \text{ s}$. If the resistance of the coil is $10\pi^2 \Omega$, the total charge flowing through the coil during this time is:

1. Magnetic Field of Solenoid: The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length .
2. Magnetic Flux: The magnetic flux linked with the small coil placed at the center is $\Phi = N_{coil} B A_{coil} = N_{coil} (\mu_0 n I) (\pi r^2)$, where $r$ is the radius of the coil .
3. Induced Charge: The induced current is $I_{ind} = \frac{1}{R} \frac{d\Phi}{dt}$. The total charge $Q$ flowing through the circuit is obtained by integrating the current over time: $Q = \int I_{ind} dt = \int \frac{1}{R} \frac{d\Phi}{dt} dt = \frac{\Delta \Phi}{R}$. Note that the time interval does not influence the total charge, only the total change in flux.
4. Calculation: $n = 2 \times 10^4 \text{ m}^{-1}$ $N_{coil} = 100$ $r = 0.01 \text{ m} \Rightarrow r^2 = 10^{-4} \text{ m}^2$ $\Delta I = 4 - 0 = 4 \text{ A}$ $R_{coil} = 10\pi^2 \Omega$ $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$

The change in flux $\Delta \Phi = N_{coil} \cdot (\pi r^2) \cdot \mu_0 n \Delta I$ $\Delta \Phi = 100 \times (\pi \times 10^{-4}) \times (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4$ $\Delta \Phi = (100 \times 4 \times 2 \times 4) \times \pi^2 \times (10^{-4} \times 10^{-7} \times 10^4)$ $\Delta \Phi = 3200 \pi^2 \times 10^{-7} = 32 \pi^2 \times 10^{-5} \text{ Wb}$ (or $3.2\pi^2 \times 10^{-4} \text{ Wb}$)

Charge $Q = \frac{\Delta \Phi}{R_{coil}} = \frac{32 \pi^2 \times 10^{-5}}{10\pi^2} = 3.2 \times 10^{-5} \text{ C}$ $Q = 32 \times 10^{-6} \text{ C} = 32 \mu \text{C}$.