Electromagnetic Induction

SPEC 3.7.5.3 - 3.7.5.4 • MAGNETIC FIELDS

Magnetic flux, flux linkage, Faraday's law, Lenz's law, and rotating coils

Introduction

In 1831, Michael Faraday discovered one of the most important principles in physics: a changing magnetic field can induce an electric current. This principle underpins the generation of almost all electrical power we use today.

Key Insight

Electromagnetic induction is the process by which an emf is induced in a conductor when it experiences a changing magnetic flux. This is the basis for generators, transformers, and many other electrical devices.

In this lesson, we will explore:

Magnetic flux (Φ) - the "amount" of magnetic field through a surface

Flux linkage (NΦ) - the total flux through a coil of N turns

Faraday's law - relationship between changing flux and induced emf

Lenz's law - why induced current always opposes the change

Rotating coils - how generators produce AC electricity

Magnetic Flux

Magnetic flux (Φ) is a measure of the total magnetic field passing through a given area. Think of it as counting how many magnetic field lines pass through a surface.

Definition

Magnetic flux (Φ) is the product of the magnetic flux density (B) and the area (A) perpendicular to the magnetic field.

Φ = B A

SymbolQuantityUnit

ΦMagnetic fluxWeber (Wb)

BMagnetic flux densityTesla (T)

AArea perpendicular to fieldm²

Definition of Weber

One weber is the magnetic flux that, linking a circuit of one turn, produces in the circuit an emf of 1 V when the flux is reduced to zero in 1 second.

Unit Relationship

1 Weber = 1 Tesla × 1 m², so 1 Wb = 1 T m². Also: 1 Wb = 1 V s (volt-second).

Flux at an Angle

When the magnetic field is not perpendicular to the surface, we must consider the angle between the field and the normal (perpendicular) to the surface.

General Formula

Φ = B A cos θ

where θ is the angle between the magnetic field direction and the normal to the surface.

Key Points

θ = 0° → cos θ = 1 → Φ = BA (maximum flux, field perpendicular to surface)

θ = 90° → cos θ = 0 → Φ = 0 (zero flux, field parallel to surface)

Only the perpendicular component of B contributes to flux

Worked Example01

A rectangular coil has dimensions 8.0 cm × 5.0 cm. It is placed in a uniform magnetic field of flux density 0.25 T with the plane of the coil perpendicular to the field. Calculate the magnetic flux through the coil.

1

Identify: B = 0.25 T, dimensions = 8.0 cm × 5.0 cm

2

Calculate area: A = 0.080 × 0.050 = 4.0 × 10⁻³ m²

3

Plane perpendicular to field → normal parallel to field (θ = 0°)

4

Apply Φ = BA cos(0°) = 0.25 × 4.0 × 10⁻³ × 1 = 1.0 × 10⁻³ Wb

Answer: Φ = 1.0 × 10⁻³ Wb = 1.0 mWb

Worked Example02

A circular coil of radius 6.0 cm is placed in a uniform magnetic field of flux density 0.40 T. The normal to the plane of the coil makes an angle of 60° with the magnetic field direction. Calculate the magnetic flux through the coil.

1

Identify: B = 0.40 T, r = 0.060 m, θ = 60°

2

Calculate area: A = πr² = π × (0.060)² = 1.131 × 10⁻² m²

3

Apply Φ = BA cos θ: Φ = 0.40 × 1.131 × 10⁻² × cos(60°)

4

Φ = 0.40 × 1.131 × 10⁻² × 0.5 = 2.26 × 10⁻³ Wb

Answer: Φ = 2.3 × 10⁻³ Wb = 2.3 mWb (2 s.f.)

Flux Linkage

When a coil has multiple turns, each turn experiences the same magnetic flux. The total effect is multiplied by the number of turns.

Definition

Magnetic flux linkage (NΦ) is the magnetic flux multiplied by the number of turns N in a coil:

N Φ = B A N

For a coil at angle θ to the field:

N Φ = B A N cos θ

SymbolQuantityUnit

NΦFlux linkageWeber (Wb) or Weber-turns

NNumber of turnsDimensionless

Key Concept

A coil of 100 turns with flux 0.01 Wb through each turn has flux linkage of 100 × 0.01 = 1 Wb. It is the flux linkage (not flux alone) that determines the induced emf in Faraday's law.

Worked Example03

A rectangular coil of 200 turns has dimensions 4.0 cm × 3.0 cm. It is placed in a uniform magnetic field of 0.15 T with the plane of the coil perpendicular to the field. Calculate the flux linkage through the coil.

1

Identify: N = 200, B = 0.15 T, dimensions = 4.0 cm × 3.0 cm

2

Calculate area: A = 0.040 × 0.030 = 1.2 × 10⁻³ m²

3

Plane perpendicular to field → normal parallel to field (θ = 0°)

4

Apply NΦ = BAN: NΦ = 0.15 × 1.2 × 10⁻³ × 200 = 0.036 Wb

Answer: NΦ = 3.6 × 10⁻² Wb = 36 mWb

Rotating Coil in a Magnetic Field

When a coil rotates in a uniform magnetic field, the angle θ changes continuously. If the coil rotates with angular velocity ω, then θ = ωt. The flux linkage varies depending on the angular speed of the coil.

Rotating Coil Flux Linkage

N Φ = B A N cos (ω t)

where ω is the angular velocity (rad s⁻¹) and t is time.

How the Formula Arises

As the coil rotates, the angle θ between the normal to the coil and the field direction changes continuously. Since θ = ωt for constant angular velocity, and Φ = BA cos θ, we get NΦ = BAN cos(ωt).

Understanding the Positions

θ = 0° (coil perpendicular to B): Maximum flux linkage (NΦ = BAN), zero rate of change, so zero EMF

θ = 90° (coil parallel to B): Zero flux linkage, maximum rate of change, so maximum EMF

Important

Flux linkage is maximum when the coil plane is perpendicular to the field (θ = 0°). Flux linkage is zero when the coil plane is parallel to the field (θ = 90°).

Faraday's Law

Faraday's law quantifies the relationship between changing magnetic flux and induced emf.

Faraday's Law

The magnitude of the induced emf is equal to the rate of change of magnetic flux linkage through the circuit.

ε = N \frac{ΔΦ}{Δ t}

Understanding the Formula

ε = magnitude of induced emf (V)

N = number of turns

ΔΦ/Δt = rate of change of flux (Wb s⁻¹)

Key Insight

The induced emf depends on how quickly the flux changes, not the total flux. A large, slow change produces less emf than a small, rapid change. No change in flux = no induced emf.

Lenz's Law

Lenz's law explains the direction of the induced current and is a consequence of conservation of energy.

Lenz's Law

The induced EMF opposes the change in magnetic flux that produces it.

ε = - N \frac{ΔΦ}{Δ t}

The negative sign indicates opposition to the change.

Why Opposition?

If the induced current aided the change, it would create more change, inducing more current—this would violate conservation of energy. The opposition ensures work must be done to change the flux, which is converted into electrical energy.

Applying Lenz's Law

The direction of induced current is such that the magnetic field it creates opposes the change in flux. If flux is increasing, the induced field opposes the original field. If flux is decreasing, the induced field reinforces it.

Lenz's Law: Falling Magnet

A classic demonstration: a magnet falling through a coil takes longer to fall than through air alone because the induced current opposes its motion.

The Physics Explained

As the N pole approaches the coil, the flux through the coil increases. By Lenz's law, the induced current must create a magnetic field that opposes this increase—so the top of the coil becomes a N pole, repelling the approaching magnet and slowing it down.

Understanding Each Stage

Approaching: Flux through coil increases → induced current creates N pole at top → repels incoming N pole → magnet slows

Through centre: Magnet symmetrically inside coil → flux momentarily constant → no induced EMF → no force

Leaving: Flux through coil decreases → induced current creates S pole at top → attracts departing N pole → magnet slows

Exam Application

When explaining Lenz's law: (1) Identify whether flux is increasing or decreasing, (2) State that induced current creates a field that opposes this change, (3) Explain the resulting force (repulsion or attraction).

Fleming's Right-Hand Rule

Fleming's right-hand rule gives the direction of induced emf and current when a conductor moves through a magnetic field (generator action).

How to Apply

Hold your right hand with thumb, first finger, and second finger all at right angles. Point the first finger in the direction of the magnetic field and the thumb in the direction of motion. The second finger then gives the direction of the induced current.

Remember

Right-hand rule is for generators (motion → current). Left-hand rule is for motors (current → motion).

Worked Example04

A coil of 400 turns experiences a change in magnetic flux from 9.0 × 10⁻⁶ Wb to zero in a time of 6.0 ms. Calculate the magnitude of the induced emf.

1

Identify: N = 400, Φ₁ = 9.0 × 10⁻⁶ Wb, Φ₂ = 0, Δt = 6.0 × 10⁻³ s

2

Calculate |ΔΦ|: |ΔΦ| = |0 - 9.0 × 10⁻⁶| = 9.0 × 10⁻⁶ Wb

3

Apply Faraday's law: |ε| = N|ΔΦ/Δt|

4

|ε| = 400 × (9.0 × 10⁻⁶)/(6.0 × 10⁻³) = 400 × 1.5 × 10⁻³ = 0.60 V

Answer: |ε| = 0.60 V

Worked Example05

A coil of 500 turns and area 2.0 × 10⁻³ m² is placed perpendicular to a magnetic field. The field strength increases uniformly from 0 to 0.80 T in 0.25 s. Calculate the induced emf.

1

Given: N = 500, A = 2.0 × 10⁻³ m², ΔB = 0.80 T, Δt = 0.25 s

2

Initial flux linkage: NΦ₁ = 0 × 2.0 × 10⁻³ × 500 = 0 Wb

3

Final flux linkage: NΦ₂ = 0.80 × 2.0 × 10⁻³ × 500 = 0.80 Wb

4

|ε| = |Δ(NΦ)/Δt| = |0.80 - 0|/0.25 = 3.2 V

Answer: |ε| = 3.2 V

EMF in a Moving Conductor

When a straight conductor moves through a magnetic field, cutting field lines, an emf is induced across its ends.

Moving Conductor EMF

ε = B Lv

where L is the length of conductor in the field and v is the velocity perpendicular to both the field and the conductor.

If the conductor moves at an angle θ to the field:

ε = B Lv sin θ

Maximum emf occurs when θ = 90° (conductor moves perpendicular to field).

Derivation

In time Δt, the wire sweeps out area A = L × vΔt. Change in flux: ΔΦ = BA = BLvΔt. Therefore: ε = ΔΦ/Δt = BLv

Worked Example06

An aircraft with a wingspan of 35 m flies horizontally at 250 m s⁻¹. The vertical component of Earth's magnetic field is 4.0 × 10⁻⁵ T. Calculate the emf induced between the wingtips.

1

Identify: L = 35 m (wingspan), v = 250 m s⁻¹, B = 4.0 × 10⁻⁵ T

2

The wings cut through the vertical component as the aircraft moves horizontally

3

Apply ε = BLv: ε = 4.0 × 10⁻⁵ × 35 × 250

4

ε = 0.35 V

Answer: ε = 0.35 V = 350 mV

EMF in a Rotating Coil

When a coil rotates at constant angular velocity ω in a uniform magnetic field, the emf varies sinusoidally with time. The emf is found by differentiating the flux linkage.

Derivation by Differentiation

Flux linkage:

Φ = B A N cos (ω t)

EMF is the negative rate of change of flux linkage:

ε = - \frac{d Φ}{d t} = - \frac{d}{d t} [B A N cos (ω t)] = - B A N \times (- ω sin (ω t)) = B A N ω sin (ω t)

Since

\frac{d}{d t} [cos (ω t)] = - ω sin (ω t)

, the EMF becomes a sine function.

Rotating Coil EMF Formulas

Instantaneous emf:

ε = B A N ω sin (ω t) = ε_{0} sin (ω t)

Peak (maximum) emf:

ε_{0} = B A N ω

RMS emf: The root mean square value is the equivalent steady DC voltage that would deliver the same average power. For a sinusoidal waveform:

ε_{rms} = \frac{ε _{0}}{2} = \frac{B A N ω}{2}

Critical Point

The emf is maximum when the coil plane is parallel to the field (wires cutting through flux at right angles). This is when flux linkage is zero but changing most rapidly.

Exam Tip

Angular velocity ω must be in rad s⁻¹. To convert: 1 rev s⁻¹ = 2π rad s⁻¹. Frequency f in Hz gives ω = 2πf.

Flux Linkage and EMF: Phase Relationship

For a rotating coil, flux linkage and emf are 90° out of phase. When flux linkage is maximum, emf is zero, and vice versa.

The Mathematics

If NΦ = BAN cos(ωt), then by Faraday's law: ε = -d(NΦ)/dt = BANω sin(ωt) = ε₀ sin(ωt)

The peak EMF is ε₀ = BANω, occurring when the coil plane is parallel to the field.

Key Relationship

EMF = -d(NΦ)/dt. The emf is the negative gradient of the flux linkage graph. Maximum emf at zero flux linkage (steepest gradient). Zero emf at maximum flux linkage (zero gradient).

Reading the Graphs

The small coil diagrams show the orientation at each point. When the coil is edge-on to the field (horizontal in diagram), the flux is changing most rapidly and EMF is maximum. When the coil face is perpendicular to B (vertical in diagram), flux is maximum but momentarily not changing, so EMF = 0.

Worked Example07

A circular coil of 100 turns, each of radius 10 cm, is rotated at 10 rev s⁻¹ about an axis at right angles to a field of flux density 0.10 T. Find the position of the coil when the emf is maximum, and calculate this emf.

1

The position when emf is maximum is when the plane of the coil is parallel to the field

2

Identify: N = 100, r = 0.10 m, f = 10 rev s⁻¹, B = 0.10 T

3

Calculate area: A = πr² = π × (0.10)² = π × 0.01 m²

4

Convert to angular velocity: ω = 2πf = 2π × 10 = 20π rad s⁻¹

5

Apply ε₀ = BANω: ε₀ = 0.10 × (π × 0.01) × 100 × 20π

6

ε₀ = 0.10 × 0.01 × π × 100 × 20π = 2π² = 19.7 V

Answer: Coil plane parallel to field; ε₀ = 19.7 V ≈ 20 V

Worked Example08

A coil of 350 turns and area 3.5 × 10⁻⁵ m² rotates in a magnetic field of 0.070 T. The coil starts with its plane perpendicular to the field and rotates through 40° in 0.18 s. Calculate: (a) the initial flux linkage, (b) the flux linkage after rotation, (c) the average induced emf.

1

Given: N = 350, A = 3.5 × 10⁻⁵ m², B = 0.070 T, Δt = 0.18 s

2

(a) Initially θ = 0° (plane perpendicular to field → normal parallel to B)

3

NΦ₁ = BAN cos(0°) = 0.070 × 3.5 × 10⁻⁵ × 350 × 1 = 8.58 × 10⁻⁴ Wb

4

(b) After rotation: θ = 40°

5

NΦ₂ = 0.070 × 3.5 × 10⁻⁵ × 350 × cos(40°) = 8.58 × 10⁻⁴ × 0.766 = 6.57 × 10⁻⁴ Wb

6

(c) |ε| = |Δ(NΦ)|/Δt = |6.57 - 8.58| × 10⁻⁴/0.18 = 1.1 × 10⁻² V

Answer: (a) 8.6 × 10⁻⁴ Wb, (b) 6.6 × 10⁻⁴ Wb, (c) 11 mV

Practice

Question 1: A square coil of 50 turns has sides of length 4.0 cm. It is placed in a uniform magnetic field of 0.20 T with the plane of the coil at 30° to the field direction. Calculate the flux linkage through the coil.

1

If the plane is at 30° to the field, the normal is at 60° to the field (θ = 60°)

2

Area: A = (0.040)² = 1.6 × 10⁻³ m²

3

NΦ = BAN cos θ = 0.20 × 1.6 × 10⁻³ × 50 × cos(60°)

4

NΦ = 0.20 × 1.6 × 10⁻³ × 50 × 0.5 = 8.0 × 10⁻³ Wb

Answer: NΦ = 8.0 mWb

Practice

Question 2: A coil of 800 turns is removed from a magnetic field in 0.40 s. The average emf induced is 2.0 V. Calculate the original magnetic flux through each turn of the coil.

1

Given: N = 800, |ε| = 2.0 V, Δt = 0.40 s

2

Flux changes from Φ to 0 when removed from field

3

From Faraday's law: |ε| = N|ΔΦ|/Δt = NΦ/Δt

4

Rearrange: Φ = |ε|Δt/N = 2.0 × 0.40 / 800 = 1.0 × 10⁻³ Wb

Answer: Φ = 1.0 mWb

Practice

Question 3: A coil of 250 turns and area 4.0 × 10⁻³ m² rotates at 50 Hz in a magnetic field of flux density 0.12 T. Calculate: (a) the peak emf, (b) the rms emf.

1

Given: N = 250, A = 4.0 × 10⁻³ m², f = 50 Hz, B = 0.12 T

2

Calculate ω: ω = 2πf = 2π × 50 = 100π rad s⁻¹

3

(a) Peak emf: ε₀ = BANω = 0.12 × 4.0 × 10⁻³ × 250 × 100π = 37.7 V

4

(b) RMS emf: ε_rms = ε₀/√2 = 37.7/√2 = 26.7 V

Answer: (a) ε₀ = 38 V, (b) ε_rms = 27 V

Eddy Currents

Eddy currents are induced currents in bulk metal objects (larger than pieces of wire) when the magnetic flux through them changes. Because metals have low resistance, these currents can be large.

Key Properties

Flow in circular paths within the metal

Create their own magnetic field that opposes the change (Lenz's law)

Cause heating of the metal (I²R losses)

Applications

Electromagnetic brakes: Eddy currents oppose motion, providing braking force (used in trains, rollercoasters). Induction heating: Rapidly changing field induces large eddy currents for heating.

Unwanted Effects

Transformer cores: Eddy currents cause energy loss. Cores are laminated (thin insulated sheets) to reduce eddy current paths and minimise losses.

Summary

Key Formulas

ConceptFormulaUnits

Magnetic flux

Φ = B A cos θ

Wb

Flux linkage

N Φ = B A N cos θ

Wb

Faraday's law

ε = - N \frac{ΔΦ}{Δ t}

V

Moving conductor

ε = B Lv

or

ε = B Lv sin θ

V

Rotating coil (peak)

ε_{0} = B A N ω

V

Rotating coil (inst.)

ε = B A N ω sin (ω t)

V

RMS emf

ε_{rms} = \frac{ε _{0}}{2}

V

Key Exam Skills

Distinguish between θ to the surface vs θ to the normal

Apply Lenz's law to determine direction of induced current

Use Fleming's right-hand rule for generator situations

EMF is max when coil plane is parallel to field (flux linkage = 0)

Convert frequency to ω using ω = 2πf (1 rev s⁻¹ = 2π rad s⁻¹)

Eddy currents: laminated cores reduce energy losses