You are viewing a free preview of this lesson.
Subscribe to unlock all 8 lessons in this course and every other course on LearningBro.
Electromagnetic induction (AQA specification 3.7.5) describes how a changing magnetic field produces an EMF (and hence a current in a closed circuit). This is the basis of generators, transformers, and much of modern electrical technology.
Spec mapping: AQA 7408 A-Level Physics, Section 3.7.5 (Electromagnetic induction). This lesson covers magnetic flux Φ = BA cos θ, flux linkage NΦ for a coil of N turns, Faraday's law as the rate of change of flux linkage, Lenz's law as the consequence of energy conservation, and the derivation ε = BLv for a moving rod. The sinusoidal EMF of a rotating coil ε = NBAω sin(ωt) is also developed. Faraday's and Lenz's laws as named statements with worked examples, and transformer operation, are extended in the dedicated lessons that follow. Refer to the official AQA 7408 specification document for the authoritative wording.
Synoptic links: (i) Capacitance and time-dependent circuits (3.7.4) — exponential charge/discharge formulas Q = Q₀e^(−t/RC) and induced-EMF time-dependence have the same mathematical character, and questions sometimes splice an RC discharge with an induced EMF transient. (ii) Mechanics — work-energy (3.4) — Lenz's law is fundamentally a re-statement of conservation of energy: the induced current opposes its cause because otherwise mechanical work and electrical energy would be created from nothing. (iii) Generators and transformers (engineering applications, 3.7.5) — the rotating-coil EMF derived here is the basis of every alternator on the National Grid; transformer voltage step-up/step-down relies on the same Faraday's law applied to two coils sharing flux through an iron core.
Key Definition: Magnetic flux (Φ) through a surface is the product of the magnetic flux density B and the area A of the surface, when the field is perpendicular to the surface.
Φ = BA cos θ
where Φ is the magnetic flux (Wb, webers), B is the magnetic flux density (T), A is the area (m²), and θ is the angle between the field direction and the normal to the surface.
When B is perpendicular to the surface (θ = 0°): Φ = BA
When B is parallel to the surface (θ = 90°): Φ = 0
The unit of magnetic flux is the weber (Wb), where 1 Wb = 1 T m² = 1 V s.
For a coil of N turns, the flux linkage is:
NΦ = NBA cos θ
Flux linkage has units of Wb turns (or simply Wb, since N is dimensionless).
If the coil rotates in a uniform field at angular velocity ω:
θ = ωt
NΦ = NBA cos(ωt)
This sinusoidal variation in flux linkage is the basis of the AC generator.
Key Definition: Faraday's law states that the magnitude of the induced EMF is equal to the rate of change of flux linkage through the circuit.
ε = −N(dΦ/dt) = −d(NΦ)/dt
The negative sign indicates that the induced EMF opposes the change producing it (Lenz's law).
In practice, for calculations:
|ε| = N × |ΔΦ/Δt|
Consider a straight conductor of length L moving at velocity v perpendicular to a uniform magnetic field B.
In time Δt, the conductor sweeps out an area ΔA = LvΔt.
Change in flux: ΔΦ = BΔA = BLvΔt
Rate of change of flux: ΔΦ/Δt = BLv
Therefore: ε = BLv (for a single conductor moving perpendicular to the field)
Question: A metal rod of length 0.40 m moves at 5.0 m s⁻¹ perpendicular to a uniform magnetic field of 0.25 T. Calculate the EMF induced across the rod.
Solution:
ε = BLv = 0.25 × 0.40 × 5.0 = 0.50 V
Question: A coil of 200 turns and area 8.0 × 10⁻³ m² is placed in a uniform magnetic field. The field increases uniformly from 0 to 0.50 T in 0.10 s. Calculate the magnitude of the induced EMF.
Solution:
ΔΦ = BΔA? No — the area is constant, the field changes.
ΔΦ = ΔB × A = (0.50 − 0) × 8.0 × 10⁻³ = 4.0 × 10⁻³ Wb
|ε| = N × |ΔΦ/Δt| = 200 × (4.0 × 10⁻³ / 0.10) = 200 × 0.040 = 8.0 V
Key Definition: Lenz's law states that the direction of the induced EMF (and hence the induced current) is such as to oppose the change in flux that produces it.
This is a consequence of conservation of energy. If the induced current enhanced the flux change, it would create a runaway process that would generate energy from nothing — violating the first law of thermodynamics.
Example: A bar magnet is pushed into a solenoid (north pole first). The increasing flux through the solenoid induces an EMF that drives a current. By Lenz's law, this current must create a magnetic field that opposes the increasing flux — so the solenoid's near end becomes a north pole (to repel the incoming magnet). You must do work pushing the magnet in, and this work is the source of the electrical energy.
Example: If the magnet is pulled away, the flux decreases, and the induced current creates a field to try to maintain the flux — the near end becomes a south pole (to attract the retreating magnet).
A coil of N turns and area A rotating at angular velocity ω in a uniform field B has flux linkage:
NΦ = NBA cos(ωt)
By Faraday's law:
ε = −d(NΦ)/dt = NBAω sin(ωt)
ε = ε₀ sin(ωt) where ε₀ = NBAω
This is the peak EMF. The EMF varies sinusoidally with time — this is how an AC generator works.
Question: A rectangular coil of 150 turns, each of area 0.020 m², rotates at 50 revolutions per second in a uniform magnetic field of 0.40 T. Calculate: (a) the peak EMF; (b) the EMF at time t = 3.0 ms.
Solution:
(a) ω = 2πf = 2π × 50 = 100π = 314.2 rad s⁻¹
ε₀ = NBAω = 150 × 0.40 × 0.020 × 314.2 = 150 × 0.40 × 6.283 = 377 V
(b) ε = ε₀ sin(ωt) = 377 × sin(314.2 × 3.0 × 10⁻³) = 377 × sin(0.9425) sin(0.9425 rad) = 0.8106 ε = 377 × 0.8106 = 306 V
Question: A circular copper disc of radius 0.10 m rotates at 100 rad s⁻¹ in a region of its rim where a localised magnetic field B = 0.50 T is applied perpendicular to the disc, over a small area of 1.0 × 10⁻⁴ m². The disc has resistance 5.0 × 10⁻³ Ω in the relevant current loop. Estimate the magnitude of the induced EMF and the resulting braking power dissipated.
Solution:
The rim of the disc moves through the field region at velocity v = ωr = 100 × 0.10 = 10 m s⁻¹.
The disc passes through the field region in a time of order Δt = √A / v = √(1.0 × 10⁻⁴) / 10 = 1.0 × 10⁻³ s.
The change of flux through the local current loop is ΔΦ ≈ B × A = 0.50 × 1.0 × 10⁻⁴ = 5.0 × 10⁻⁵ Wb.
|ε| ≈ |ΔΦ/Δt| = 5.0 × 10⁻⁵ / 1.0 × 10⁻³ = 5.0 × 10⁻² V = 50 mV
Induced current: I = ε/R = 0.050 / 5.0 × 10⁻³ = 10 A
Braking power dissipated: P = I²R = (10)² × 5.0 × 10⁻³ = 0.50 W
This is the basis of eddy-current braking in trains and lorries — the braking power scales as v² (because both ε and I scale linearly with v for given B and geometry), so the brake force grows with speed but vanishes at rest, making this technique self-limiting.
A transformer transfers electrical energy from one circuit to another using electromagnetic induction. It consists of two coils (primary and secondary) wound on a shared iron core.
For an ideal transformer (no energy losses):
Ns/Np = Vs/Vp
where Np and Ns are the numbers of turns in the primary and secondary coils, and Vp and Vs are the primary and secondary voltages.
For an ideal transformer (100% efficient):
VpIp = VsIs
Power in = Power out. If the voltage increases, the current decreases by the same factor.
Question: A transformer has 400 turns on the primary coil and 2000 turns on the secondary coil. The primary voltage is 230 V and the primary current is 5.0 A. Assuming the transformer is ideal, calculate: (a) the secondary voltage; (b) the secondary current.
Solution:
(a) Vs/Vp = Ns/Np → Vs = Vp × Ns/Np = 230 × 2000/400 = 230 × 5.0 = 1150 V
(b) Using power conservation: VpIp = VsIs Is = VpIp/Vs = 230 × 5.0 / 1150 = 1150 / 1150 = 1.0 A
Check: Ns/Np = 5, so voltage increases by 5× and current decreases by 5× ✓
Real transformers are not 100% efficient. Energy losses occur due to:
Efficiency = (output power / input power) × 100% = (VsIs / VpIp) × 100%
Exam Tip: You should be able to explain each type of transformer loss and how it is minimised. The laminated core (eddy current reduction) is the most commonly asked about. The laminations are thin sheets of iron, each insulated from the next, oriented parallel to the flux direction — this prevents large eddy current loops from forming.
Specimen question modelled on the AQA A-Level Physics 7408 Paper 2 format. Not from any published paper.
Q. (a) State Faraday's law of electromagnetic induction. (2 marks) (b) A metal rod of length L = 0.25 m slides along two parallel horizontal rails that complete a circuit through a stationary resistor. The rod moves at constant speed v = 0.80 m s⁻¹ perpendicular to a uniform vertical magnetic field of flux density B = 0.45 T. Calculate the EMF induced across the rod, and the current that flows if the resistor has resistance R = 1.5 Ω. (3 marks) (c) State Lenz's law, and explain in terms of energy conservation why the induced current must oppose the change of flux. (2 marks) (d) A flat coil of 320 turns and cross-sectional area 2.4 × 10⁻³ m² is rotated at constant angular speed of 25 rev s⁻¹ inside a uniform magnetic field of flux density 0.18 T, with the rotation axis perpendicular to the field. Derive an expression for the EMF as a function of time and calculate the peak EMF. By considering the rate of change of flux linkage, explain qualitatively why the induced EMF is greatest when the plane of the coil is parallel to the field and zero when it is perpendicular. (9 marks)
(Total: 16 marks)
Subscribe to continue reading
Get full access to this lesson and all 8 lessons in this course.