Solved Examples on Electromagnetic Induction and Alternating Current

Question 1:-

A copper disc 20 cm in diameter rotates with an angular velocity of 60 rev s^-1 about its axis. The disc is placed in a magnetic field of induction 0.2 T acting parallel to the axis of rotation of the disc. Calculate the magnitude of the e.m.f. induced between the axis of rotation and the rim of the disc.

Solution:-

As the disc rotates, any of its radii cuts the lines of force of magnetic field.

Area swept by radius vector during one revolution

= πr²

= π(10 cm)²

= 100π cm²

= π×10^-2 m²

Area swept in one second,

A = (area swept in one revolution) × (Number of revolutions per second)

   = π×10^-2×60

A = 0.6 πm²

Rate of change of magnetic flux

= dϕ_B/dt

= BA

= 0.2×0.6π

= 0.12π Wb

According to Faraday’s law, magnitude of induced e.m.f is,

E = dϕ_B/dt

Therefore, magnitude of e.m.f. is

E = 0.12 πV

Or, E = 0.377 V

Thus from the above observation we conclude that, the magnitude of the e.m.f. induced between the axis of rotation and the rim of the disc would be 0.377 V.

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Question 2:-

A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity ω about Y-axis. The loop lies in a uniform magnetic field B in the direction of X-axis. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current Induced in the loop. What is the magnitude of torque required on the loop to keep it moving with constant ω?

Solution:-

As ϕ is maximum at t = 0,

ϕ(t) = BA cos ωt

Magnitude of induced emf = N|dϕ/dt|

= BAωN |sinωt|

Magnitude of induced current = [BAωN/R] |sin ωt|

So, peak value of emf = BAωN

peak value of induced current = BAωN/R

To obtain the magnitude of torque required on the loop to keep it moving with constant ω, we have to equate power input is equal to heat dissipation per second.

So, power input = heat dissipation per second

Or, ω = I²R

Or,  = [(BAωN)²/R] |sin²ωt|

From the above observation we conclude that,  the magnitude of torque required on the loop to keep it moving with constant ω would be [(BAωN)²/R] |sin²ωt|.

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Question 3:-

An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1W and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit.

Solution:-

In case of an ac, the voltage leads the current in phase by angle,

ϕ = tan^-1 (X_L/R)

Here, X_L = ωL

= (2πfL)  

= (2π) (50) (0.01)

=  πΩ

and

R = 1Ω

So, ϕ = tan^-1(π)

≈ 72.3°

Further, i_rms = V_rms/|Z|

= V_rms/√R²+X_L²

Substituting the values we have,

 i_rms = 200/√(1)²+(π)²

= 60.67 amp

From the above observation we conclude that, the virtual current in the circuit would be 60.67 amp.

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Question 4:-

A long solenoid of length 1 m, cross sectional area 10 cm², having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s?

Solution:-

Let N₁ = number of turns in solenoid

N2 = number of turns in coil

A₁ nad A₂ be their respective areas of cross-section. 

(A₁ = A₂ in this problem)

Flux ϕ₂ through coil crated by current i₁ in solenoid is ϕ₂ = N₂(B₁A₂)

ϕ₂ = N₂ (µ₀i₁N₁/l)A₂

Or, ϕ₂ = (µ₀N₁N₂A₂/l) i₂

Comparing with  ϕ₂ = Mi₁, we get,

Mutual inductance, M = µ₀N₁N₂A₂/l

= [4π×10^-7×1000×20×10×10^-4]/1

= 2.51×10^-5 H

So magnitude of induced emf = E₂ = M|di₁/dt|