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A long solenoid of diameter 0.1 m has $2 \times 10^{4}$ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is $10\pi^2\ \Omega$, the total charge flowing through the coil during this time is
Explanation
Induced charge = $N\Delta\phi/R$, independent of time taken.
Detailed Solution
$\varepsilon = -N\frac{d\phi}{dt}$
$\left|\frac{\varepsilon}{R}\right| = \frac{N}{R}\frac{d\phi}{dt} \Rightarrow dq = \frac{N}{R}d\phi$
$\Delta Q = \frac{N(\Delta\phi)}{R} = \frac{N(BA)}{R} = \frac{N\mu_0ni\pi r^2}{R}$
Putting values: $\Delta Q = \frac{4\pi\times10^{-7}\times100\times2\times10^{4}\times4\times\pi\times(0.01)^2}{10\pi^2}$
$\Delta Q = 32\ \mu C$