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# Problems on Electromagnetic Induction

Electromagnetism is a combination of two different phenomena i.e. Electricity and magnetism. Electricity and magnetism are interrelated, when electric charges move through a conductor they produce magnetic fields. The converse of this is also possible, In 1830, Michael Faraday in England and Joseph Henry in the USA conducted an experiment and demonstrated effectively that electric currents were induced in closed coils when placed in a changing magnetic field.

The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction.

When an electric current is produced in a closed conducting loop when the flux of the magnetic field through the surface enclosed by this loop changes. This phenomenon is called electromagnetic induction, and the current produced an induced current. Electromagnetic induction is also called magnetic induction, as the principle is the same whether the process is carried out through electromagnetic or static magnets.

Michael Faraday, in the year 1830 discovered electromagnetic induction and demonstrated it with a copper coil around a toroidal piece of iron, a galvanometer (a gauge-based device used to show currently), and a magnet. When the magnet was moved towards the coil, an EMF is created, and moved the gauge on the galvanometer was. If the north end of the magnet is drawn closer, the current flows one way and if the south end of the magnet is drawn closer, then the current flow in the opposite direction. This discovery of electromagnetic induction was a fundamental principle in understanding and harnessing electricity.

The principle of electromagnetic induction is required in electronic components such as inductors and transformers. Electromagnetic induction is the basis of all types of electric generators and motors.

## Solved Problems on Electromagnetic Induction

Problem 1: A short loop with an area of 4.0 cm2 is placed inside a long solenoid with 10 rounds per cm, normal to the solenoid’s axis. What is the induced emf in the loop during a steady change in the solenoid carrying current from 2.0 A to 4.0 A lasting for 0.5 seconds?

Solution:

The magnetic field produced inside the solenoid,

B = μ0nI

Suppose the area of the loop placed inside the solenoid is A then the magnetic flux linked with the loop,

φ = BA
= μ0nIA

e is the induced e.m.f. produced due to change in current through the solenoid, then,

e = dφ / dt = d/dt [μ0nIA]

Or, e = -μ0 × n × A × dI/dt

Number of turns per unit length of the solenoid (n) = 10 turns cm-1 = 1000 turns m-1

A = 4cm2 = 4 × 10-4m2

dI/dt = 4-2/0.5 = 4 As-1

Therefore, e = -4π × 10-7 × 1000 × 4 × 10-4 × 4

= -2 × 10-6V.

Problem 2: A rectangular wire loop of sides 10 cm and 5 cm with a small cut is moving out of a region of a uniform magnetic field of magnitude 0.2 T directed normally to the loop. What is the emf developed across the cut if the velocity of the loop is 5 cm s-1 in a direction normal to the (i) longer side, and (ii) shorter side of the loop? In each case, determine the time period of the induced voltage.

Solution:

Given:

Length of the wired loop (I) = 10 cm = 0.1 m

Width of the wired loop (b) = 5 cm = 0.05 m

Area of loop, A = l×b = 0.1 × 0.05 = 5 × 10-3m2

Strength of magnetic field (B) = 0.2T

Velocity of the loop (v) = 5 cm/s = 0.05 m/s

(i) EMF developed in the loop (e) = Blv

= 0.2 x 0.1 x 0.05 = 1 × 10-3 V

Time required to travel along the width (t) = Distance travelled / Velocity  =  b/v

= 0.05/0.05 = 1s

Hence, the induced voltage is 1 × 10-3 V which lasts for 1 s.

(ii) EMF developed (e) = Bbv

= 0.2 × 0.05 × 0.05 = 5 × 10-4 V

Time required to travel along the length, t’ = l/v

= 0.1 / 0.05 = 2 s.

Therefore, the induced voltage is 5 x 10-4 V which lasts for 2 s.

Problem 3: When the 350 rad s-1 of angular frequency is applied to the rotation of a metallic rod that is 3.0 m long along an axis that passes through one end of the rod. The rod’s other end is in touch with a metal ring that has a circular shape. Everywhere there is a 0.2 T steady and uniform magnetic field parallel to the axis. Determine the emf that was created between the center and the ring.

Solution:

Given:

Length of the rod (l) = 3m

Angular frequency (ω) = 350 rad/s

Magnetic field strength, B = 0.2 T

Average linear velocity (v) = (Iω + 0)/2 = Iω/2

e = Blv = Bl (Iω / 2)  =  (Bl2ω) / 2

= (0.2 × 32 × 350) / 2

= 315 V

Problem 4: A circular coil of radius 10 cm and 50 turns is rotated about its vertical diameter with an angular speed of 20 rad s-1 in a uniform horizontal magnetic field of magnitude 5.0 × 10-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 20 Ω calculates the maximum value of current in the coil. Calculate the average power loss due to joule heating. Where does this power come from?

Solution:

Given:

Max emf induced = 0.603 V

Average emf induced = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power which is coming from external rotor)

Circular coil radius (r) = 10 cm = 0.1 m

Area of the coil (A) = πr2 = π × (0.1)2 m2

Number of turns on the coil (N) = 50

Angular speed (ω) = 20 rad/s

Strength of magnet (B) = 5 × 10-2 T

Total resistance produced by the loop (R) = 20 Ω

Maximum induced emf is given as:

e = N ω A B

= 50 × 20 × π × (0.1)2 × 5 × 10-2

= 1.57 V

The maximum emf induced in the coil is 1.57 V.

Maximum current is given as :

I = e/R

= 1.57 / 20  =  0.0785 A

Average power loss due to Joule heating:

P = el / 2

= (1.57 x 0.0785) / 2  =  0.06162 W

Problem 5: A 50 m long horizontal straight wire that runs from east to west is falling at a speed of 1.0 m s-1, at right angles to the horizontal component of the earth’s magnetic field, 0.60 x 10-4 Wb m-2.

(a) What is the instantaneous value of the electro-motive force induced in the wire?

(b) What is the direction of the electro-motive force?

(c) Determine which end of the wire has a higher electrical potential?

Solution:

Given:

Wire’s Length (l) = 50 m

Speed of the wire with which it is falling (v)= 1.0 m/s

Strength of magnetic field (B) = 0.6 × 10-4 Wbm-2

(a) EMF induced in the wire (e) = Blv

= 0.6 × 10-4 × 1 × 50

= 3 × 10-3 V

(b) Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf if from West to East.

(c) The west end of the wire is at a higher potential.

Problem 6: Current in a circuit falls from 7.0 A to 0.0 A in 0.3 s. If an average emf of 500 V is induced, give an estimate of the self-inductance of the circuit.

Solution:

Given:

Initial current (I1) : 7 A

Final current (I2) : 0 A

Change in current (dl) = I1 – I2 = 7A

Total time (t) = 0.3 s

Average EMF (e) = 500 V

For self inductance (L) of the coil, we have the relation for average emf as:

e = L (di/dt)

L = e / (di/dt)

L = 500 / (7/0.3)

L = 21.4286 H

Thus, the self inductance of the coil is 21.4286 H.

Problem 7: A pair of adjacent coils has a mutual inductance of 3 H. If the current in one coil changes from 0 to 50 A in 0.8 s, What is the change of flux linkage with the other coil?

Solution:

Given:

Mutual inductance (μ) = 3 H

Initial current (I1) = 0 A

Final current (I2) = 50 A

Therefore, change in current is (dl)= I2 – I1 = 50 – 0 = 50 A

Time taken (t) = 0.8 seconds

The mutual conductance of the circuit can be given by,

e = μ (dI/dt)

μ = e/(dI/dt)

we know, e = dφ/dt

dφ/dt  =  μ (dI/dt)

dφ  =  3 × (50)

dφ  =  150 Wb

Problem 8: A airplane is traveling west at a speed of 1500 km/h. What is the voltage difference developed between the ends of the wing having a span of 32 m, If the Earth’s magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 60o?

Solution:

Given:

Speed of plane (v) = 1500 km/h = 1500 × (5 / 18) = 416.67 m/s

Distance between the ends of the wings (l) = 32m

Magnetic field strength by earth (B) = 5 × 10-4 T

Dip angle (δ) = 60o

Vertical component of Earth’s magnetic field,

Bv = B sin δ

=5 × 10-4 sin 60o

= 4.33 × 10-4 T

motional emf (e) = (Bv) × l × v

= 4.33 × 10-4 × 32 × 416.67

e = 5.773 V

Problem 9: Suppose the loop in problem 2 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.5 T at the rate of 0.06 T s-1. If the cut is joined and the loop has a resistance of 2.3 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

Solution:

Given:

Rectangular loop with sides 10 cm and 5 cm.

Area of loop (A) = L × B = 10 × 5

= 50 cm2

= 5 × 10-3 m2

Magnetic field’s initial value (B) = 0.5T

Magnetic field’s rate of decrease (dB/dt) = 0.06 T/s

Emf induced in the loop is :

e = dφ/dt

dφ = change in flux in the loop area = AB

Therefore, e = d(AB) / dt = AdB / dt

= 5 × 10-3 × 0.06 = 0.3 × 10-3 V

Resistance in the loop (R) = 2.3 Ω

The current developed in the loop will be:

i = e/R

= 0.3 × 10-3 / 2.3  =  1.3 × 10-4 A

Power dissipated in the loop in the form of heat is given as:

P = i2R

= (1.3 × 10-4)2 × 2.3

= 3.887 × 10-8 W

Problem 10: In a setup with a magnetic field in the positive z-direction, a square loop with a side of 15 cm and sides that are parallel to the X and Y axes is moved with a velocity of 6 cm s-1 in the positive x-direction. The field is neither constant in time nor uniform in space. It increases by 10-3 T cm-1 as one move in the negative x-direction and decreases in time at a rate of 10-3 Ts-1. The gradient along the negative x-direction is 10-3 T cm-1. If the loop’s resistance is 3.50 mΩ, determine the magnitude and direction of the induced current in the loop.

Solution:

Given:

Side of the Square loop (s) = 15cm = 0.15 m

Area of the loop, A = s × s = 0.15 × 0.15 = 0.0225 m2

Velocity of the loop (v) = 6 cm s-1 = 0.06 m s-1

Gradient of the magnetic field along negative x-direction,

dB/dx = 10-1m-1

And, the rate of decrease of the magnetic field,

dB/dt = 10-3Ts-1

Resistance (R) = 3.50 mΩ = 3.5 × 10-3 Ω

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as :

dB/dt = A × dB/dx × v

= 225 × 10-4m2 × 10-1 × 0.06

= 13.5× 10-5Tm2s-1

Rate of change of the flux due to explicit time variation in field B is given as:

dφ/dt = A × dB/dt

= 225 × 10-4 × 10-3

= 2.25 × 10-5Tm2s-1

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

e = 2.25 × 10-5 + 13.5 × 10-5

= 15.75 × 10-5V

Therefore, Induced current (i) = e/R = 15.75 × 10-5 / 3.5 × 10-3

i = 4.5 × 10-2 A

Problem 11: A strong loudspeaker magnet has a magnetic field, the strength between its poles needs to be measured. A 4 cm2 flat search coil with 35 densely wound turns is immediately snatched out of the field area after being placed normal to the field direction. Alternatively, one can quickly rotate it 90 degrees such that its plane is parallel to the field direction. A ballistic galvanometer connected to the coil recorded 7.5 mC as the complete charge flowed in the coil. The coil and galvanometer’s combined resistance is 0.70 Ω. Calculate the magnet’s field strength.

Solution:

Given:

Coil’s Area (A) = 4 cm2 = 4 × 10-4 m2

Number of turns on the coil (N) = 35

Total Change in the coil (Q) = 6.5 mC = 6.5 × 10-3 C

Total Resistance produced (R) = 0.7 Ω

Current generated in the coil,

I = e / R                     …(1)

EMF induced is shown as :

e = -N (dφ/dt)                                …(2)

Where,

Form equation (1) and (2), we have

I = -N (dφ/dt)  /  R

I dt  =  (-N/R) dφ

∫I dt = -N/R ∫ dφ

Total Charge, Q = ∫ I dt

Therefore, Q  =  -N/R (φf – φi)  =  -N/R (-φi)  =  +Nφi / R

Q = NBA / R

Therefore, B = QR / NA

= (6.5 × 10-3 × 0.7) / (35 × 4 × 10-4)= 0.325 T

Problem 12: An air-cored solenoid with a length of 40cm, an area of cross-section of 45cm2, and a number of turns 650 carry a current of 3.6 A. In just 10-3 s, the current is abruptly cut off. What is the average back emf induced in the circuit across the open switch’s ends? Ignore the variation in the magnetic field near the ends of the solenoid.

Solution:

Given,

Length of solenoid (l) = 40 cm = 0.4 m

Area of solenoid (A) = 45 cm2 = 45 × 10-4 m2

Number of turns on the solenoid (N) = 650

Current in the solenoid (I) = 3.6 A

Time duration (t) = 10-3 s

B = Strength of the magnetic field = μ0 (NI/l)

Where,

μ0 = Permeability of free space = 4π × 10-7 T m A-1

e = μ0N2IA / lt

= 4π × 10-7 × (650)2 × 3.6 × 45 × 10-4 / 0.4 × 10-3

= 21.5 V

## FAQs on  Electromagnetic Induction

Question 1: Define Electromagnetic Induction.

Electromagnetic or magnetic induction is the production of an electromagnetic force around the electrical conductor in a changing magnetic field.

Question 2: What are the applications of electromagnetic induction?

Electromagnetic induction is essentially used for production of electricity. Electricity is generated by subjecting a metal to a changing magnetic field. This induces a current in the metal due to a force on the free electrons.

• Dams use the energy of flowing water to drive coils of wire in a magnetic field to generate electricity.
• Electromagnetic induction is also used for braking purpose in magnetic trains.
• Nuclear reactors generate heat, which is used to convert water to steam, that is used to drive turbines in a magnetic field.

Question 3: What is the relationship between magnetic induction and current?

We know that electric current can produce magnetic fields also support that magnetic field could produce electric currents. The production of electro motive force and currents by the changing magnetic field through a conducting loop is called magnetic induction.

Question 4: State the difference between a magnetic field and an Electric field.