Electromagnetic Induction
- 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
- 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 thin semicircular conducting ring (PQR) of radius ‘r’ is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is :
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Rate of decreasing of area of semicircular ring
= dA = (2r)V dt
From Faraday’s law of electromagnetic inductione = - dθ = - B dA = -B(2rV) dt dt 
As induced current in ring produces magnetic field in upward direction hence R is at higher potential.Correct Option: D
Rate of decreasing of area of semicircular ring
= dA = (2r)V dt
From Faraday’s law of electromagnetic inductione = - dθ = - B dA = -B(2rV) dt dt
As induced current in ring produces magnetic field in upward direction hence R is at higher potential.
- A coil of resistance 400Ω is placed in a magnetic field. If the magnetic flux φ (wb) linked with the coil varies with time t (sec) as φ = 50t2 + 4. The current in the coil at t = 2 sec is :
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According, to Faraday’s law of induction
Induced e.m.f.ε = - dφ = -(100t) dt
Induced current i at t = 2 sec.= 
ε = + 100 × 2 = + 0.5 A R 400 Correct Option: A
According, to Faraday’s law of induction
Induced e.m.f.ε = - dφ = -(100t) dt
Induced current i at t = 2 sec.= ε = + 100 × 2 = + 0.5 A R 400
- In a coil of resistance 10 Ω, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in Weber is :
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The charge through the coil = area of current-time(i – t) graph
q = 1 × 0.1 × 4 = 0.2 C 2 q = Δφ R
∵ Change in flux (Δφ) = q × Rq = 0.2 = Δφ 10
Δφ = 2 WeberCorrect Option: B
The charge through the coil = area of current-time(i – t) graph
q = 1 × 0.1 × 4 = 0.2 C 2 q = Δφ R
∵ Change in flux (Δφ) = q × Rq = 0.2 = Δφ 10
Δφ = 2 Weber
