NCERT CLASS 12 PHYSICS

CHAPTER 6 – ELECTROMAGNETIC INDUCTION

IN-TEXT QUESTIONS, EXERCISE QUESTIONS AND IMPORTANT QUESTIONS WITH ANSWERS

PART A – IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is electromagnetic induction?

Electromagnetic induction is the phenomenon in which an electromotive force (EMF) is produced in a conductor or circuit whenever the magnetic flux linked with it changes.

If the circuit is closed, an induced current flows through it.

2. Who discovered electromagnetic induction?

Electromagnetic induction was discovered independently by Michael Faraday and Joseph Henry around 1831.

3. What is induced electromotive force?

The electromotive force produced in a circuit due to a change in magnetic flux linked with the circuit is called induced electromotive force or induced EMF.

4. What is induced current?

The current produced in a closed conducting circuit due to electromagnetic induction is called induced current.

5. What is a galvanometer?

A galvanometer is a sensitive instrument used to detect and measure small electric currents in a circuit.

6. What happens when a bar magnet is moved towards a coil connected to a galvanometer?

When a bar magnet is moved towards a coil connected to a galvanometer, the galvanometer shows a momentary deflection. This indicates that an induced current is produced in the coil.

7. What happens when the bar magnet is moved away from the coil?

When the bar magnet is moved away from the coil, the galvanometer deflects in the opposite direction. This shows that the direction of induced current is reversed.

8. What happens when the bar magnet is kept stationary near the coil?

When the bar magnet is kept stationary near the coil, there is no change in magnetic flux through the coil. Therefore, no induced EMF or current is produced.

9. How does the speed of a moving magnet affect induced current?

The faster the magnet moves towards or away from the coil, the greater is the rate of change of magnetic flux. Therefore, the magnitude of induced EMF and current increases.

10. Is relative motion between a magnet and a coil necessary for electromagnetic induction?

No. Relative motion is not always necessary. Electromagnetic induction can also occur when the magnetic field changes with time, even if the magnet and coil remain stationary.

11. What is magnetic flux?

Magnetic flux is a measure of the total magnetic field passing through a given surface.

For a uniform magnetic field and a plane surface,

ΦB = BA cos θ

where,
B = Magnetic field
A = Area of the surface
θ = Angle between the magnetic field and the normal to the surface.

12. What is the SI unit of magnetic flux?

The SI unit of magnetic flux is weber (Wb).

1 Wb = 1 tesla × 1 m²

13. When is magnetic flux maximum?

Magnetic flux is maximum when the magnetic field is perpendicular to the plane of the surface.

In this case, the angle between the magnetic field and the normal to the surface is 0°.

ΦB = BA cos 0°

ΦB = BA

14. When is magnetic flux zero?

Magnetic flux is zero when the magnetic field is parallel to the plane of the surface.

In this case, the angle between the magnetic field and the normal to the surface is 90°.

ΦB = BA cos 90° = 0

15. State Faraday's first law of electromagnetic induction.

Whenever the magnetic flux linked with a circuit changes, an electromotive force is induced in the circuit.

If the circuit is closed, an induced current also flows through it.

16. State Faraday's second law of electromagnetic induction.

The magnitude of induced EMF is directly proportional to the rate of change of magnetic flux linked with the circuit.

Mathematically,

ε = −dΦB/dt

For a coil having N turns,

ε = −N(dΦB/dt)

17. What is the significance of the negative sign in Faraday's law?

The negative sign indicates that the induced EMF acts in a direction that opposes the change in magnetic flux responsible for producing it.

This is explained by Lenz's law.

18. State Lenz's law.

Lenz's law states that the direction of induced current is such that the magnetic field produced by it opposes the change in magnetic flux that produces the induced current.

19. How does Lenz's law follow the law of conservation of energy?

According to Lenz's law, the induced current always opposes the change in magnetic flux that produces it.

Therefore, external work must be done to change the magnetic flux. This work is converted into electrical energy.

Thus, Lenz's law is consistent with the law of conservation of energy.

20. Why does a magnet experience an opposing force when it is moved towards a coil?

When a magnet is moved towards a coil, the magnetic flux linked with the coil increases. According to Lenz's law, the induced current produces a magnetic field that opposes this increase.

As a result, the coil develops a magnetic pole that repels the approaching pole of the magnet.

21. What is motional EMF?

The EMF induced across the ends of a conductor due to its motion through a magnetic field is called motional EMF.

For a straight conductor moving perpendicular to a uniform magnetic field,

ε = Blv

where,
B = Magnetic field
l = Length of conductor
v = Velocity of conductor.

22. What is the condition for producing motional EMF?

Motional EMF is produced when a conductor moves in a magnetic field in such a way that the magnetic flux linked with the circuit changes.

For the formula ε = Blv, the conductor, its velocity and the magnetic field must have the required mutually perpendicular arrangement.

23. What is eddy current?

Eddy currents are circulating currents induced inside a bulk conductor when the magnetic flux linked with different parts of the conductor changes.

They flow in closed paths within the conductor.

24. Give two applications of eddy currents.

Eddy currents are used in electromagnetic braking systems.
They are used in induction furnaces for heating metals.

25. How can eddy currents be reduced?

Eddy currents can be reduced by using laminated cores. The core is divided into thin insulated sheets, which increases the resistance of the paths available for circulating currents.

26. What is self-induction?

Self-induction is the phenomenon in which a change in current flowing through a coil produces an induced EMF in the same coil.

27. What is self-inductance?

Self-inductance is the property of a coil by which it opposes any change in the current flowing through it.

It is represented by L.

28. What is the SI unit of self-inductance?

The SI unit of self-inductance is henry (H).

29. Define one henry.

A coil has a self-inductance of one henry if an EMF of one volt is induced in it when the current changes at the rate of one ampere per second.

30. What is mutual induction?

Mutual induction is the phenomenon in which a change in current in one coil produces an induced EMF in a nearby coil.

31. What is mutual inductance?

Mutual inductance is the property by which a change in current in one coil induces an EMF in another nearby coil.

It is represented by M.

32. What is the SI unit of mutual inductance?

The SI unit of mutual inductance is henry (H).

33. What is the difference between self-induction and mutual induction?

Self-induction occurs when a changing current in a coil induces EMF in the same coil.

Mutual induction occurs when a changing current in one coil induces EMF in another nearby coil.

34. Why is a soft iron core used in some induction experiments?

A soft iron core increases the magnetic flux linked with the coils because it provides a path of high magnetic permeability. Therefore, the magnitude of induced EMF can increase.

35. What is an AC generator?

An AC generator is a device that converts mechanical energy into electrical energy in the form of alternating current by using electromagnetic induction.

36. What is the principle of an AC generator?

An AC generator works on the principle of electromagnetic induction. When a coil rotates in a magnetic field, the magnetic flux linked with it changes continuously and an alternating EMF is induced.

37. What is the function of slip rings in an AC generator?

Slip rings provide continuous electrical contact between the rotating coil and the external circuit. They help transfer the induced alternating current to the external circuit.

38. What is the function of carbon brushes in an AC generator?

Carbon brushes maintain electrical contact with the rotating slip rings and transfer current between the rotating part and the external circuit.

39. What is the energy stored in an inductor?

The energy stored in an inductor carrying current I is given by:

U = ½LI²

where L is the self-inductance of the inductor.

40. Why does an inductor oppose a change in current?

When current through an inductor changes, the magnetic flux linked with it also changes. This produces a self-induced EMF that opposes the change in current according to Lenz's law.

PART B – NCERT EXERCISE QUESTIONS AND ANSWERS

QUESTION 6.1

Question: Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).

Answer:

The direction of induced current can be determined using Lenz's law.

According to Lenz's law, the induced current always opposes the change in magnetic flux responsible for producing it.

The answers for the six situations are:

(a) The induced current flows along qrpq.

(b) The induced current flows along prqp in one coil and yzxy in the other coil.

(d) The induced current flows along zyxz.

(e) The induced current flows along xryx.

(f) No induced current is produced because the magnetic flux through the loop does not change.

Final Answer:

(a) qrpq

(b) prqp and yzxy

(d) zyxz

(e) xryx

(f) No induced current.

QUESTION 6.2

Question: Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

Answer:

According to Lenz's law, the induced current flows in such a direction that it opposes the change in magnetic flux.

(a) When a wire of irregular shape is turned into a circular shape:

For a given length of wire, a circle encloses the maximum possible area.

Therefore, when the irregular wire is changed into a circular shape, the area enclosed by the wire increases.

As the area increases, the magnetic flux through the loop also increases.

According to Lenz's law, the induced current produces a magnetic field that opposes this increase in magnetic flux.

Therefore, the induced current flows in the anticlockwise direction along adcb.

(b) When a circular loop is deformed into a narrow straight wire:

When the circular loop is changed into a narrow straight wire, the area enclosed by the loop decreases.

Therefore, the magnetic flux through the loop decreases.

According to Lenz's law, the induced current produces a magnetic field that opposes this decrease in magnetic flux.

Therefore, the induced current flows in the anticlockwise direction along a′d′c′b′.

Final Answer:

(a) Anticlockwise direction along adcb.

(b) Anticlockwise direction along a′d′c′b′.

QUESTION 6.3

Question: A long solenoid has 15 turns per cm and a small loop of area 2.0 cm² is placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Answer:

Given:

Number of turns per unit length,

n = 15 turns/cm

n = 15 × 100 = 1500 turns/m

Area of the loop,

A = 2.0 cm²

A = 2.0 × 10⁻⁴ m²

Initial current,

I₁ = 2.0 A

Final current,

I₂ = 4.0 A

Time interval,

Δt = 0.1 s

Permeability of free space,

μ₀ = 4π × 10⁻⁷ T m/A

Step 1: Calculate the rate of change of current.

dI/dt = (I₂ − I₁)/Δt

dI/dt = (4.0 − 2.0)/0.1

dI/dt = 2/0.1

dI/dt = 20 A/s

Step 2: Calculate the magnetic field inside the solenoid.

Magnetic field inside a long solenoid is:

B = μ₀nI

Step 3: Calculate the induced EMF.

The magnetic flux through the loop is:

ΦB = BA

ΦB = μ₀nIA

According to Faraday's law,

ε = μ₀nA(dI/dt)

Substituting the values,

ε = 4π × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × 20

ε = 7.54 × 10⁻⁶ V

Final Answer:

Induced EMF = 7.54 × 10⁻⁶ V

Approximately,

ε = 7.5 μV

QUESTION 6.4

Question: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s⁻¹ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Answer:

Given:

Length of longer side = 8 cm = 0.08 m

Length of shorter side = 2 cm = 0.02 m

Magnetic field,

B = 0.3 T

Velocity,

v = 1 cm/s = 0.01 m/s

The formula for motional EMF is:

ε = Blv

where l is the length of the side cutting across the magnetic field boundary.

(a) When the loop moves in a direction normal to its longer side:

Here,

l = 0.08 m

Using,

ε = Blv

ε = 0.3 × 0.08 × 0.01

ε = 2.4 × 10⁻⁴ V

The induced voltage lasts until the entire loop comes out of the magnetic field.

The distance travelled during this time is equal to the shorter side, 0.02 m.

Time = Distance/Velocity

t = 0.02/0.01

t = 2 s

(b) When the loop moves in a direction normal to its shorter side:

Here,

l = 0.02 m

Using,

ε = Blv

ε = 0.3 × 0.02 × 0.01

ε = 0.6 × 10⁻⁴ V

The distance travelled until the entire loop comes out of the field is equal to the longer side, 0.08 m.

Time = Distance/Velocity

t = 0.08/0.01

t = 8 s

Final Answer:

(a) Induced EMF = 2.4 × 10⁻⁴ V

Time = 2 s

(b) Induced EMF = 0.6 × 10⁻⁴ V

Time = 8 s

QUESTION 6.5

Question: A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Answer:

Given:

Length of the rod,

l = 1.0 m

Angular frequency,

ω = 400 rad/s

Magnetic field,

B = 0.5 T

The formula for EMF induced in a rotating rod is:

ε = ½Bωl²

Substituting the values,

ε = ½ × 0.5 × 400 × (1.0)²

ε = 0.25 × 400

ε = 100 V

Final Answer:

Induced EMF = 100 V

QUESTION 6.6

Question: A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the earth's magnetic field, 0.30 × 10⁻⁴ Wb m⁻². (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?

Answer:

Given:

Length of wire,

l = 10 m

Velocity,

v = 5.0 m/s

Horizontal component of Earth's magnetic field,

B = 0.30 × 10⁻⁴ T

(a) Calculate the induced EMF.

The formula for motional EMF is:

ε = Blv

Substituting the values,

ε = 0.30 × 10⁻⁴ × 10 × 5.0

ε = 15 × 10⁻⁴ V

ε = 1.5 × 10⁻³ V

(b) What is the direction of the EMF?

The wire is falling vertically downwards, while the horizontal component of Earth's magnetic field acts towards the north.

Using Fleming's right-hand rule or the magnetic force direction on positive charges, the induced EMF acts from west to east.

(c) Which end is at a higher electrical potential?

Positive charges are pushed towards the eastern end of the wire.

Therefore, the eastern end becomes positively charged and is at a higher electrical potential.

Final Answer:

(a) Induced EMF = 1.5 × 10⁻³ V or 1.5 mV.

(b) Direction of EMF = From west to east.

QUESTION 6.7

Question: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Answer:

Given:

Initial current,

I₁ = 5.0 A

Final current,

I₂ = 0.0 A

Time interval,

Δt = 0.1 s

Average induced EMF,

ε = 200 V

The formula for self-induced EMF is:

ε = L |ΔI/Δt|

Step 1: Calculate the rate of change of current.

|ΔI/Δt| = |I₂ − I₁|/Δt

= |0 − 5.0|/0.1

= 5.0/0.1

= 50 A/s

Step 2: Calculate self-inductance.

ε = L |ΔI/Δt|

L = ε/(|ΔI/Δt|)

L = 200/50

L = 4 H

Final Answer:

Self-inductance of the circuit = 4 H

QUESTION 6.8

Question: A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Answer:

Given:

Mutual inductance,

M = 1.5 H

Initial current,

I₁ = 0 A

Final current,

I₂ = 20 A

Change in current,

ΔI = 20 − 0 = 20 A

Time interval,

Δt = 0.5 s

The relation between mutual inductance and flux linkage is:

NΦ = MI

Therefore,

Δ(NΦ) = M × ΔI

Substituting the values,

Δ(NΦ) = 1.5 × 20

Δ(NΦ) = 30 Wb

Final Answer:

Change in flux linkage = 30 Wb

Note: The given time interval is not required to calculate the change in flux linkage. It can be used to calculate the average induced EMF:

ε = M(ΔI/Δt)

ε = 1.5 × (20/0.5)

ε = 60 V

PART C – IMPORTANT DERIVATIONS

1. Derive Faraday's law of electromagnetic induction.

According to Faraday's law, the magnitude of induced EMF in a circuit is equal to the rate of change of magnetic flux linked with the circuit.

For a coil having N turns,

ε ∝ dΦB/dt

Therefore,

ε = −N(dΦB/dt)

The negative sign represents Lenz's law.

Hence,

ε = −N(dΦB/dt)

This is the mathematical expression of Faraday's law of electromagnetic induction.

2. Derive the expression for motional EMF induced in a straight conductor.

Consider a straight conducting rod of length l moving with uniform velocity v in a uniform magnetic field B.

Assume that the rod moves perpendicular to the magnetic field and its length is also perpendicular to the direction of motion.

Let the rod move through a distance x in time t.

The area swept by the rod is:

A = lx

Magnetic flux through the area is:

ΦB = BA

ΦB = Blx

According to Faraday's law,

ε = |dΦB/dt|

ε = d(Blx)/dt

Since B and l are constant,

ε = Bl(dx/dt)

But,

dx/dt = v

Therefore,

ε = Blv

Hence, the induced EMF in a moving straight conductor is:

ε = Blv

3. Derive the expression for EMF induced in a rotating rod.

Consider a conducting rod of length l rotating with angular velocity ω about one of its ends in a uniform magnetic field B.

The magnetic field is parallel to the axis of rotation.

Consider a small element of the rod of length dr at a distance r from the axis.

The linear velocity of this element is:

v = ωr

The small EMF induced across this element is:

dε = Bv dr

Substituting v = ωr,

dε = Bωr dr

The total EMF induced between the two ends of the rod is:

ε = ∫₀ˡ Bωr dr

Taking B and ω outside the integral,

ε = Bω ∫₀ˡ r dr

ε = Bω [r²/2]₀ˡ

ε = Bωl²/2

Therefore,

ε = ½Bωl²

This is the expression for EMF induced in a rotating rod.

4. Derive the expression for self-inductance of a long solenoid.

Consider a long solenoid having:

N = Number of turns

l = Length of solenoid

A = Cross-sectional area

I = Current flowing through the solenoid

The number of turns per unit length is:

n = N/l

The magnetic field inside a long solenoid is:

B = μ₀nI

B = μ₀NI/l

Magnetic flux through each turn is:

ΦB = BA

ΦB = μ₀NIA/l

Total flux linkage is:

NΦB = μ₀N²IA/l

By definition of self-inductance,

NΦB = LI

Therefore,

LI = μ₀N²IA/l

Cancelling I,

L = μ₀N²A/l

Since n = N/l,

L = μ₀n²Al

This is the expression for self-inductance of a long solenoid in air or vacuum.

5. Derive the expression for energy stored in an inductor.

Consider an inductor of inductance L carrying current I.

The induced EMF is:

ε = L(dI/dt)

The small amount of work done in moving charge dq against this induced EMF is:

dW = ε dq

Since,

dq = I dt

Therefore,

dW = L(dI/dt) × I dt

dW = LI dI

The total work done in increasing the current from 0 to I is:

W = ∫₀ᴵ LI dI

Taking L outside the integral,

W = L ∫₀ᴵ I dI

W = L[I²/2]₀ᴵ

W = ½LI²

Therefore, the energy stored in an inductor is:

U = ½LI²

6. Derive the expression for EMF induced in an AC generator.

Consider a rectangular coil of N turns and area A rotating with angular velocity ω in a uniform magnetic field B.

At any instant t, the angle between the magnetic field and the normal to the coil is:

θ = ωt

Magnetic flux through one turn is:

ΦB = BA cos θ

ΦB = BA cos ωt

Total flux linkage is:

NΦB = NBA cos ωt

According to Faraday's law,

ε = −d(NΦB)/dt

ε = −d(NBA cos ωt)/dt

ε = NBAω sin ωt

Therefore,

ε = ε₀ sin ωt

where,

ε₀ = NBAω

ε₀ is the maximum value of induced EMF.

Hence, the EMF produced by an AC generator is:

ε = NBAω sin ωt

This shows that the induced EMF changes periodically with time.

PART D – ADDITIONAL IMPORTANT QUESTIONS WITH ANSWERS

1. Why is induced EMF produced when a magnet is moved towards a coil?

When a magnet moves towards a coil, the magnetic flux linked with the coil changes. According to Faraday's law, this change in magnetic flux produces an induced EMF in the coil.

2. Why is no EMF induced when a magnet and coil are stationary relative to each other?

When both the magnet and coil remain stationary, the magnetic flux linked with the coil remains constant. Since there is no change in magnetic flux, no EMF is induced.

3. Why does induced current oppose the cause producing it?

According to Lenz's law, induced current always flows in such a direction that the magnetic field produced by it opposes the change in magnetic flux responsible for producing it.

4. Why is a laminated iron core used in electrical devices?

A laminated iron core reduces eddy current losses. The thin insulated layers increase the resistance of the paths in which eddy currents could flow.

5. What is the difference between magnetic flux and magnetic field?

Magnetic field describes the magnetic influence in a region of space.

Magnetic flux measures the magnetic field passing through a particular surface.

Magnetic field is measured in tesla (T), whereas magnetic flux is measured in weber (Wb).

6. What factors affect the magnitude of induced EMF?

The magnitude of induced EMF depends on:

The number of turns in the coil.
The strength of the magnetic field.
The area of the coil.
The orientation of the coil with respect to the magnetic field.
The rate at which magnetic flux changes.

7. What happens to the induced EMF if the number of turns in a coil is doubled?

According to Faraday's law,

ε = −N(dΦB/dt)

If the rate of change of magnetic flux remains constant, doubling the number of turns doubles the induced EMF.

8. What happens to the induced EMF if the rate of change of magnetic flux is doubled?

According to Faraday's law, induced EMF is directly proportional to the rate of change of magnetic flux.

Therefore, if the rate of change of magnetic flux is doubled, the magnitude of induced EMF also doubles.

9. Why is an AC generator called an energy conversion device?

An AC generator converts mechanical energy into electrical energy through electromagnetic induction.

10. What is the difference between an AC generator and a DC generator?

An AC generator produces alternating current, whereas a DC generator produces unidirectional current.

An AC generator generally uses slip rings, while a simple DC generator uses a split-ring commutator.

PART E – IMPORTANT NUMERICAL QUESTIONS

1. A coil of 100 turns experiences a change in magnetic flux from 0.02 Wb to 0.08 Wb in 0.2 seconds. Calculate the average induced EMF.

Given:

N = 100

Initial flux = 0.02 Wb

Final flux = 0.08 Wb

Time = 0.2 s

Formula:

ε = N × |ΔΦB/Δt|

ε = 100 × (0.08 − 0.02)/0.2

ε = 100 × 0.06/0.2

ε = 30 V

Answer: Average induced EMF = 30 V.

2. A conducting rod of length 0.5 m moves with a velocity of 4 m/s perpendicular to a magnetic field of 0.2 T. Calculate the induced EMF.

Given:

B = 0.2 T

l = 0.5 m

v = 4 m/s

Formula:

ε = Blv

ε = 0.2 × 0.5 × 4

ε = 0.4 V

Answer: Induced EMF = 0.4 V.

3. A coil has 200 turns and a magnetic flux of 5 × 10⁻⁴ Wb per turn. Calculate the total flux linkage.

Given:

N = 200

ΦB = 5 × 10⁻⁴ Wb

Formula:

Flux linkage = NΦB

= 200 × 5 × 10⁻⁴

= 0.1 Wb

Answer: Total flux linkage = 0.1 Wb.

4. An inductor of 2 H carries a current of 3 A. Calculate the energy stored in it.

Given:

L = 2 H

I = 3 A

Formula:

U = ½LI²

U = ½ × 2 × 3²

U = 9 J

Answer: Energy stored = 9 J.

5. A coil has a self-inductance of 4 H. If the current changes at the rate of 5 A/s, calculate the induced EMF.

Given:

L = 4 H

dI/dt = 5 A/s

Formula:

ε = L(dI/dt)

ε = 4 × 5

ε = 20 V

Answer: Magnitude of induced EMF = 20 V.

6. Two coils have a mutual inductance of 0.5 H. If the current in the first coil changes at the rate of 10 A/s, calculate the induced EMF in the second coil.

Given:

M = 0.5 H

dI/dt = 10 A/s

Formula:

ε = M(dI/dt)

ε = 0.5 × 10

ε = 5 V

Answer: Magnitude of induced EMF = 5 V.

PART F – IMPORTANT FORMULAE

Magnetic flux:

ΦB = BA cos θ

Faraday's law:

ε = −N(dΦB/dt)

Average induced EMF:

ε = −N(ΔΦB/Δt)

Motional EMF:

ε = Blv

EMF induced in a rotating rod:

ε = ½Bωl²

Magnetic field inside a long solenoid:

B = μ₀nI

Self-inductance:

L = NΦB/I

Self-induced EMF:

ε = −L(dI/dt)

Mutual inductance:

M = N₂Φ₂₁/I₁

Mutual induced EMF:

ε₂ = −M(dI₁/dt)

Self-inductance of a long solenoid:

L = μ₀N²A/l

Mutual inductance of two coils:

M = N₂Φ₂₁/I₁

Energy stored in an inductor:

U = ½LI²

EMF produced by an AC generator:

ε = NBAω sin ωt

Maximum EMF of an AC generator:

ε₀ = NBAω

Angular velocity:

ω = 2πf

PART G – IMPORTANT SI UNITS

Magnetic field – tesla (T)
Magnetic flux – weber (Wb)
Induced EMF – volt (V)
Electric current – ampere (A)
Resistance – ohm (Ω)
Self-inductance – henry (H)
Mutual inductance – henry (H)
Area – square metre (m²)
Length – metre (m)
Velocity – metre per second (m/s)
Angular velocity – radian per second (rad/s)
Energy – joule (J)

PART H – QUICK REVISION

Electromagnetic induction is the production of EMF due to a change in magnetic flux.
Faraday's law gives the magnitude of induced EMF.
Lenz's law determines the direction of induced current.
The SI unit of magnetic flux is weber.
The SI unit of self-inductance and mutual inductance is henry.
Motional EMF is given by ε = Blv.
Self-induction occurs in the same coil.
Mutual induction occurs between two nearby coils.
Eddy currents are circulating currents induced in bulk conductors.
Laminated cores are used to reduce eddy current losses.
Energy stored in an inductor is U = ½LI².
An AC generator works on the principle of electromagnetic induction.
An AC generator converts mechanical energy into electrical energy.