Science > Physics > Electromagnetic Induction > Numerical Problems on Electromagnetic Induction
In this article, we shall learn to solve numerical problems to find e.m.f. induced, change in flux, and change in magnetic induction.
Example 01:
A circular coil having 1000 insulated turns each of area 1.274 m2 is kept in a uniform magnetic field of induction 5 x 10-4 T. If the plane of the coil makes an angle of 60o with the direction of the magnetic field, find magnetic flux flowing through the coil.
Given: Number of turns of coil = n = 1000, Area of coil = A = 1.274 m2, Magnetic induction = B = 5 x 10-4 T,Angle between plane of coil and magnetic field = θ = 60o.
To Find: Magnetic Flux = Φ =?
Solution:
Magnetic flux Φ = nAB cos θ
∴ Φ = 1000 x 1.274 x 5 x 10-4 x cos 60o
∴ Φ = 0.55 Wb
Ans: Magnetic flux flowing through the coil = 0.55 Wb
Example – 02:
The magnetic flux associated with a coil changes from zero to 6 x 10-2 Wb in 0.6 s. Find the average e.m.f. induced in the coil
Given: Initial flux Φ1 = 0, Final flux Φ2 = 6 x 10-2 Wb, Time in which change took place = dt = 0.6 s.
To Find: e.m.f. induced = e =?
Solution:
Change in flux dΦ = Φ2 – Φ1 = 6 x 10-2 – 0
∴ dΦ = 6 x 10-2 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 6 x 10-2 /0.6
∴ e = 0.1 V
Ans: e.m.f. induced = 0.1 V
Example – 03:
A coil of effective area 4 m2 is placed at right angles to a magnetic field of induction 0.05 Wb m-2. If the field is decreased by 20% of its original value in 10 seconds, find the e.m.f. induced in the coil
Given: Effective area of coil = nA = 4m2, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic induction = B1 = 0.05 Wb m-2, Decrease in magnetic induction = 20%, Time in which change took place = dt = 10 s.
To Find: e.m.f. induced = e =?
Solution:
Final magnetic induction = B2 = B1 – 20% of B1
∴ B2 = 80% B1 = 0.80 x 0.05 = 0.04 Wb m-2
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 4 x 0.05 x cos 90o = 0.2 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2 = 4 x 0.04 x cos 90o = 0.16 Wb
Change in flux dΦ= Φ1 – Φ2 = 0.2 – 0.16
∴ dΦ = 0.04 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 0.04 /10 = 0.004 V
∴ e = 4 x 10-3 V = 4 mV
Ans: e.m.f. induced = 0.4 mV
Example – 04:
A coil of effective area 4 m2 is placed at right angles to a magnetic field of induction 0.05 Wb m-2. If the field is decreased to 20% of its original value in 10 seconds, find the e.m.f. induced in the coil
Given: Effective area of coil = nA = 4m2, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic field = B1 = 0.05 Wb m-2, New magnetic induction (B2) = 20% of original induction (B1), Time in which change took place = dt = 10 s.
To Find: e.m.f. induced = e =?
Solution:
Final magnetic induction = B2 = 20% of B1 = 0.20 x 0.05 = 0.01 Wb m-2
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 4 x 0.05 x cos 90o = 0.2 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2 = 4 x 0.01 x cos 90o = 0.04 Wb
Change in flux dΦ = Φ1 – Φ2 = 0.2 – 0.04
∴ dΦ = 0.16 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 0.16 /10 = 0.016 V
∴ e = 16 x 10-3 V = 16 mV
Ans: e.m.f. induced = 16 mV
Example – 05:
A coil of effective area 2 m2 is placed at right angles to a magnetic field of induction 0.08 Wb m-2. If the field is decreased to 10% of its original value in 0.6 seconds, find the e.m.f. induced in the coil
Given: Effective area of coil = nA = 2m2, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic field = B1 = 0.08 Wb m-2, New magnetic induction (B2) = 10% of original induction (B1), Time in which change took place = dt = 0.6 s.
To Find: e.m.f. induced = e =?
Solution:
Final magnetic induction = B2 = 10% of B1
= 0.10 x 0.08 = 0.008 Wb m-2
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 2 x 0.08 x cos 90o = 0.16 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2 = 2 x 0.008 x cos 90o = 0.016 Wb
Change in flux dΦ = Φ1 – Φ2 = 0.16 – 0.016
∴ dΦ = 0.144 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 0.144 /0.6
∴ e = 0.24 V
Ans: e.m.f. induced = 0.24 V
Example – 06:
A coil of 50 turns each of area 800 cm2 is held with its plane perpendicular to a uniform magnetic field of induction 5 x 10-5 T. If it is pulled out of the field in 2 s, find the e.m.f. induced in the coil.
Given: Number of turns of coil = n = 50, area of each turn of coil = A = 800m2= 800 x 10-4 m2, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic field = B1 = 5 x 10-5 T, New magnetic induction = B2 = 0, Time in which change took place = dt = 2 s.
To Find: e.m.f. induced = e =?
Solution:
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 50 x 800 x 10-4 x 5 x 10-5 x cos 90o
∴ Φ1 = 2 x 10-4 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2= 50 x 800 x 10-4 x 0 x cos 90o
∴ Φ2= 0 Wb
Change in flux dΦ = Φ1 – Φ2 = 2 x 10-4 – 0
∴ dΦ = 2 x 10-4 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 2 x 10-4 /2
∴ e = 1 x 10-4 V
Ans: e.m.f. induced = 10-4 V
Example – 07:
A rectangular coil of length 0.5 m and breadth 0.4 m has a resistance of 5 ohm is placed with its plane perpendicular to a uniform magnetic field of induction 0.05 T. The magnetic induction is uniformly reduced to zero in 5 x 10-3 s. What is the e.m.f. induced and the current induced in the coil?
Given: Length of coil = l = 0.5 m, Breadth of coil = b = 0.4 m, Resistance of coil = R = 5 ohm, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic induction = B1 = 0.05 T, Final magnetic induction = B2 = 0, Time in which change took place = dt = 5 x 10-3s.
To Find: e.m.f. induced = e =?, Current induced = I =?
Solution:
Area of the coil = A = l x b = 0.5 x 0.4 = 0.2 m2
Assuming the coil has single turn i.e. n = 1
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 1 x 0.2 x 0.05 x cos 90o = 0. 0.01 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2 = 1 x 0.2 x 0 x cos 90o = 0 Wb
Change in flux dΦ = Φ1 – Φ2 = 0.01 – 0 6 = 0.01 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 0.01 /5 x 10-3
∴ e = 2 V
By Ohm’s law e = IR
I = e/R = 2/5 = 0.4 A
Ans: e.m.f. induced in coil = 2 V and current induced in coil = 0.4 A
Example – 08:
A coil of 50 turns each of length 0.5 m, and breadth 0.1 m is placed with its plane perpendicular to a uniform magnetic field of induction 10-3 T. The magnetic induction is uniformly reduced to zero in 10-2 s. What is the current induced in the coil if its resistance is 5 ohm.
Given: Number of turns of coil = n = 50, Length of coil = l = 0.5 m, Breadth of coil = b = 0.1 m, Resistance of coil = R = 5 ohm, Angle between plane of coil and the magnetic field = θ = 90o, Initial magnetic induction = B1 = 10-3 T, Final magnetic induction = B2 = 0, Time in which change took place = dt = 10-2s.
To Find: Current induced = I =?
Solution:
Area of the coil = A = l x b = 0.5 x 0.1 = 0.05 m2
Assuming the coil has single turn i.e. n = 50
Initial flux Φ1 = nAB1 cos θ
∴ Φ1 = 50 x 0.05 x 10-3 x cos 90o = 2.5 x 10-3 Wb
Final Flux Φ2 = nAB1 cos θ
∴ Φ2 = 50 x 0.05 x 0 x cos 90o = 0 Wb
Change in flux dΦ = Φ1 – Φ2 = 2.5 x 10-3 – 0 6 = 2.5 x 10-3 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 2.5 x 10-3 / 10-2
∴ e = 0.25 V
By Ohm’s law e = IR
I = e/R = 0.25/5 = 0.05 A
Ans: The current induced in coil = 0.05 A
Example – 09:
A rectangular loop having length 0.1 m and breadth 0.2 m and resistance 8 Ω is kept perpendicular to a magnetic field, what change in the magnetic induction should be brought about in 0.5 s so as to induce a current of 4 mA in it.
Given: For a loop = n = 1, Length of coil = l = 0.1 m, Breadth of coil = b = 0.2 m, Resistance of coil = R = 8 Ω Angle between plane of coil and the magnetic field = θ = 90o, Time in which change took place = dt = 0.5s, I = 4 mA = 4 x 10-3 A.
To Find: Change in magnetic induction = dB =?
Solution:
Area of the coil = A = l x b = 0.1 x 0.2 = 0.02 m2
By Ohm’s law e = IR = 4 x 10-3 x 8 = 0.032 V
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt
∴ dΦ = e. dt = 0.032 x 0.5 = 0.016 Wb
∴ dΦ = f1 – f2 = nAB1 cos θ – nAB2 cos θ
∴ dΦ = nA(B1 –B2) cos θ
∴ dΦ = nA dB cos θ
∴ dB = dΦ/ nA cos θ = 0.016 / (1 x 0.02 x cos 90o)
∴ dB= 0.8 T
Ans: Change in magnetic induction = 0.8 T
Example – 10:
A coil having an area of 0.5 m2 and 1000 turns is kept perpendicular to a magnetic field of 0.05 T. Calculate the change in magnetic flux through the coil when it is rotated through an angle of 60o about its diameter in 5 s. Find the e.m.f. induced in the coil.
Given: Number of turns = n = 1000, Area of coil = A = 0.5 m2, Magnetic Induction = B = 0.05 T, Initial angle between plane of coil and the magnetic field = θ1 = 90o, New angle between plane of coil and the magnetic field = θ2 = 60o, Time in which change took place = dt = 5s.
To Find: e.m.f. induced = e =?
Solution:
Initial flux Φ1 = nAB cos θ1
∴ Φ1 = 1000 x 0.5 x 0.05 x cos 90o = 25 Wb
Final Flux Φ2 = nAB cos θ2
∴ Φ2 = 1000 x 0.5 x 0.05 x cos 60o = 12.5 Wb
Change in flux dΦ = Φ1 – Φ2 = 25 – 12.5 = 12.5 Wb
By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by
e = dΦ/dt = 12.5 /5
∴ e = 2.5 V
Ans: Induced e.m.f. = 2.5 V