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 $10 π^{2} Ω$ . the total charge flowing through the coil during this time is :-

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$16 μ C$

$32 μ C$

$16 π μ C$

$32 π μ C$

answer is B.

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Detailed Solution

Given, no. of turns N=100 radius, $, r = 0.01 m$

resistance, $R = 10 π^{2} Ω, n = 2 \times 10^{4}$

As we know,

$\begin{array}{l} ε = - N \frac{d ϕ}{dt} \\ \frac{ε}{R} = - \frac{N}{d} \frac{d ϕ}{dt} \\ ΔI = - \frac{N}{R} \frac{d ϕ}{dt} \\ \frac{Δq}{Δt} = - \frac{N}{R} \frac{Δ ϕ}{Δt} \\ Δq = - [\frac{N}{R} (\frac{Δ ϕ}{Δt})] Δt \end{array}$

$'_'$ ve sign shows that induced emf opposes the change of flux.

$\begin{array}{l} Δq = [μ_{0} nN π r^{2} (\frac{Δi}{Δt})] \frac{1}{R} Δt = \frac{μ_{0} n N π T^{2} Δi}{R} \\ Δq = \frac{4 π \times 10^{- 7} \times 100 \times 4 \times π \times (0.01)^{2} \times 2 \times 10^{4}}{10 π^{2}} \\ Δq = 32 μ C \end{array}$