Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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 = πr^{2} = π(10 cm)^{2} = 100π cm^{2} = π×10^{-2} m^{2} Area swept in one second, A = (area swept in one revolution) × (Number of revolutions per second) = π×10^{-2}×60 A = 0.6 πm^{2} 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. _____________________________________________________________________________________ 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 = I^{2}R Or, = [(BAωN)^{2}/R] |sin^{2}ωt| 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] |sin^{2}ωt|. _________________________________________________________________________________________ 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}/√R^{2}+X_{L}^{2} 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. _______________________________________________________________________________________________ Question 4:- A long solenoid of length 1 m, cross sectional area 10 cm^{2}, 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 N_{1} = number of turns in solenoid N2 = number of turns in coil A_{1} nad A_{2} be their respective areas of cross-section. (A_{1} = A_{2} in this problem) Flux ϕ_{2} through coil crated by current i_{1} in solenoid is ϕ_{2} = N_{2}(B_{1}A_{2}) ϕ_{2} = N_{2} (µ_{0}i_{1}N_{1}/l)A_{2} Or, ϕ_{2} = (µ_{0}N_{1}N_{2}A_{2}/l) i_{2} Comparing with ϕ_{2} = Mi_{1}, we get, Mutual inductance, M = µ_{0}N_{1}N_{2}A_{2}/l = [4π×10^{-7}×1000×20×10×10^{-4}]/1 = 2.51×10^{-5} H So magnitude of induced emf = E_{2} = M|di_{1}/dt|
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.
As the disc rotates, any of its radii cuts the lines of force of magnetic field.
Area swept by radius vector during one revolution
= πr^{2}
= π(10 cm)^{2}
= 100π cm^{2}
= π×10^{-2} m^{2}
Area swept in one second,
A = (area swept in one revolution) × (Number of revolutions per second)
= π×10^{-2}×60
A = 0.6 πm^{2}
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.
_____________________________________________________________________________________
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?
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 = I^{2}R
Or, = [(BAωN)^{2}/R] |sin^{2}ωt|
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] |sin^{2}ωt|.
_________________________________________________________________________________________
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.
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}/√R^{2}+X_{L}^{2}
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.
_______________________________________________________________________________________________
A long solenoid of length 1 m, cross sectional area 10 cm^{2}, 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?
Let N_{1} = number of turns in solenoid
N2 = number of turns in coil
A_{1} nad A_{2} be their respective areas of cross-section.
(A_{1} = A_{2} in this problem)
Flux ϕ_{2} through coil crated by current i_{1} in solenoid is ϕ_{2} = N_{2}(B_{1}A_{2})
ϕ_{2} = N_{2} (µ_{0}i_{1}N_{1}/l)A_{2}
Or, ϕ_{2} = (µ_{0}N_{1}N_{2}A_{2}/l) i_{2}
Comparing with ϕ_{2} = Mi_{1}, we get,
Mutual inductance, M = µ_{0}N_{1}N_{2}A_{2}/l
= [4π×10^{-7}×1000×20×10×10^{-4}]/1
= 2.51×10^{-5} H
So magnitude of induced emf = E_{2} = M|di_{1}/dt|
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.