Electromagnetic Induction
- An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current if any, induced in the coil?
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Current will be induced, when e– comes closer the induced current will be anticlockwise when e– comes farther induced current will be clockwise
Correct Option: B
Current will be induced, when e– comes closer the induced current will be anticlockwise when e– comes farther induced current will be clockwise
- A long solenoid of diameter 0.1 m has 2 × 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 0A from 4 A in 0.05 s. If the resistance of the coil is 10π2Ω. the total charge flowing through the coil during this time is :-
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Given, no. of turns N = 100 radius,
r = 0.01 m
resistance, R = 10π2Ω, n = 2 × 104
As we know,ε = -N dφ dt ε = - N dφ R R dt ΔI = - N dφ R dt Δq = - N Δφ Δt R Δt Δ q = - 
N 
Δφ 
Δt R Δt
'–' ve sign shows that induced emf opposes the change of flux.Δ q = - μ0nNπr2 Δi 1 Δt Δt R = μ0nNπr2Δi R Δq = 4π × 10-7 × 100 × 4 × π (0.01)2 × 2 × 104 10π2
Δq = 32μCCorrect Option: B
Given, no. of turns N = 100 radius,
r = 0.01 m
resistance, R = 10π2Ω, n = 2 × 104
As we know,ε = -N dφ dt ε = - N dφ R R dt ΔI = - N dφ R dt Δq = - N Δφ Δt R Δt Δ q = - N Δφ Δt R Δt
'–' ve sign shows that induced emf opposes the change of flux.Δ q = - μ0nNπr2 Δi 1 Δt Δt R = μ0nNπr2Δi R Δq = 4π × 10-7 × 100 × 4 × π (0.01)2 × 2 × 104 10π2
Δq = 32μC
- The current in self inductance L = 40 mH is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in the induct or during the process is
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e = L di dt
Given that L = 40 × 10–3 H,
di = 11 A – 1 A = 10 A
and dt = 4 × 10–3 s∴ e = 40 × 10–3 × 10 = 100 V 4 × 10–3 Correct Option: A
e = L di dt
Given that L = 40 × 10–3 H,
di = 11 A – 1 A = 10 A
and dt = 4 × 10–3 s∴ e = 40 × 10–3 × 10 = 100 V 4 × 10–3
- A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced in the frame will be proportional to
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Emf induced in side 1 of frame e1 = B1Vℓ
B1 = μ0l 2π(x - a/2)
Emf induced in side 2 of frame e2 = B2VℓB2 = μ0l 2π(x + a/2) 
Emf induced in square frame
e = B1Vℓ – B2Vℓ= μ0l ℓv - μ0l ℓv 2π(x - a/2) 2π(x + a/2) or, e ∝ 1 (2x - a)(2x + a) Correct Option: C
Emf induced in side 1 of frame e1 = B1Vℓ
B1 = μ0l 2π(x - a/2)
Emf induced in side 2 of frame e2 = B2VℓB2 = μ0l 2π(x + a/2)
Emf induced in square frame
e = B1Vℓ – B2Vℓ= μ0l ℓv - μ0l ℓv 2π(x - a/2) 2π(x + a/2) or, e ∝ 1 (2x - a)(2x + a)
- A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 ×10–3 Wb. The self- inductance of the solenoid is
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Total number of turns in the solenoid, N = 500
Current, I = 2A.
Magnetic flux linked with each turn
= 4 × 10–3 Wb
As, φ = LI or N φ = LI⇒ L = Nφ 1 = 500 × 4 × 10-3 henry = 1 H. 2 Correct Option: C
Total number of turns in the solenoid, N = 500
Current, I = 2A.
Magnetic flux linked with each turn
= 4 × 10–3 Wb
As, φ = LI or N φ = LI⇒ L = Nφ 1 = 500 × 4 × 10-3 henry = 1 H. 2
