SATHEE: Electromagnetic Induction

Electromagnetic Induction - Result Question 2

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 $10 π^{2} Ω$ . the total charge flowing through the coil during this time is :- [2017]

(a) $16 μ C$

(b) $32 μ C$

(d) $32 π μ C$

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Answer:

Correct Answer: 2. (b)

Solution:

(b) Given, no. of turns $N = 100$

radius, $r = 0.01 m$

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

As we know,

$ε = - N \frac{d ϕ}{d t} \Rightarrow \frac{ε}{R} = - \frac{N}{R} \frac{d ϕ}{d t}$

$Δ I = - \frac{N}{R} \frac{d ϕ}{d t} \Rightarrow \frac{Δ q}{Δ t} = - \frac{N}{R} \frac{Δ ϕ}{Δ t}$

$Δ q = - [\frac{N}{R} (\frac{Δ ϕ}{Δ t})] Δ t$

‘-’ ve sign shows that induced emf opposes the change of flux.

$\begin{aligned} Δ q & = [μ_{0} n N π r^{2} (\frac{Δ i}{Δ t})] \frac{1}{R} Δ t = \frac{μ_{0} n N π r^{2} Δ i}{R} \\ Δ q & = \frac{4 π \times 10^{- 7} \times 100 \times 4 \times π \times (0.01)^{2} \times 2 \times 10^{4}}{10 π^{2}} \\ Δ q & = 32 μ C \end{aligned}$

Electromagnetic Induction - Result Question 2

Answer:

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Electromagnetic Induction - Result Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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