Electromagnetic Induction - Result Question 2
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2. 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 :- [2017]
======= ####2. 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 :- [2017]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/electromagnetic-induction/electromagnetic-induction—result-question-2.md (a) $16 \mu C$
(b) $32 \mu C$
(c) $16 \pi \mu C$
(d) $32 \pi \mu C$
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Answer:
Correct Answer: 2. (b)
Solution:
- (b) Given, no. of turns $N=100$
radius, $r=0.01 m$
resistance, $R=10 \pi^{2} \Omega, n=2 \times 10^{4}$
As we know,
$\varepsilon=-N \frac{d \phi}{dt} \Rightarrow \frac{\varepsilon}{R}=-\frac{N}{R} \frac{d \phi}{dt}$
$\Delta I=-\frac{N}{R} \frac{d \phi}{dt} \Rightarrow \frac{\Delta q}{\Delta t}=-\frac{N}{R} \frac{\Delta \phi}{\Delta t}$
$\Delta q=-[\frac{N}{R}(\frac{\Delta \phi}{\Delta t})] \Delta t$
‘-’ ve sign shows that induced emf opposes the change of flux.
$ \begin{aligned} \Delta q & =[\mu_0 n N \pi r^{2}(\frac{\Delta i}{\Delta t})] \frac{1}{R} \Delta t=\frac{\mu_0 n N \pi r^{2} \Delta i}{R} \\ \Delta q & =\frac{4 \pi \times 10^{-7} \times 100 \times 4 \times \pi \times(0.01)^{2} \times 2 \times 10^{4}}{10 \pi^{2}} \\ \Delta q & =32 \mu C \end{aligned} $
