43. A long solenoid of diameter $0.1 m$ has $2 \times 10^4$ turns per metre. At the centre of 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: (2017-Delhi)
Solution:
Total charge flowing is $q = \frac{\Delta \Phi}{R} = \frac{N_{coil} (\Delta B) A_{coil}}{R}$.
Change in magnetic field is $\Delta B = \mu_0 n \Delta I = (4\pi \times 10^{-7}) \times (2 \times 10^4) \times 4 = 32\pi \times 10^{-3} T$.
$q = \frac{100 \times (32\pi \times 10^{-3}) \times [\pi (0.01)^2]}{10 \pi^2} = \frac{100 \times 32\pi^2 \times 10^{-7}}{10 \pi^2} = 32 \times 10^{-6} C = 32 \mu C$.
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