A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the… | NEET 2017

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Magnetic Flux & Faraday's Law

A long solenoid of diameter 0.1 m has 2 $\times$ 104 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

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Answer: (b)

Solenoid field $B = \mu_0 nI = 4\pi \times 10^{-7} \times 2 \times 10^4 \times 4 = 32\pi \times 10^{-3}$ T.
Flux through the coil: $\phi = NB\pi r^2 = 100 \times 32\pi \times 10^{-3} \times \pi \times 10^{-4} = 32\pi^2 \times 10^{-5}$ Wb; it falls to $0$ .
$q = \frac{\Delta\phi}{R} = \frac{32\pi^2 \times 10^{-5}}{10\pi^2} = 3.2 \times 10^{-5}$ C
So the answer is (b) 32 $\mu$ C.

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