Lesson 23: Electromagnetic Induction (HL) HL ONLY

Lesson 11 established Faraday's and Lenz's laws; this HL lesson (Theme D.4) completes induction: generators and alternating current, rms values, transformers, and rectification — the physics of the entire electricity supply system, and a reliable source of multi-step HL problems.

The generator and sinusoidal emf

A coil of N turns and area A rotating at angular frequency ω in a uniform field B has flux linkage NΦ = NBA cos(ωt) (taking the angle from the normal), so the induced emf is ε = NBAω sin(ωt) with peak value ε₀ = NBAω. Doubling the rotation rate doubles both the frequency and the peak emf — a scaling law Paper 1 loves. The output is inherently alternating: one full cycle per revolution.

RMS values exist because AC power calculations need an equivalent DC value: for a sinusoid, Vrms = V₀/√2 and Irms = I₀/√2, with average power P̄ = IrmsVrms = ½I₀V₀ for a resistive load. Mains "230 V" is an rms value; the peak is 325 V — which is why insulation and component ratings exceed the nominal voltage.

Worked problem 1 — generator output

A generator coil has N = 120 turns, area 0.015 m², rotating at 50 Hz in a field of 0.42 T. Find the peak emf and the rms emf.

ω = 2πf = 314 rad s⁻¹
ε₀ = NBAω = 120 × 0.42 × 0.015 × 314 = 237 V
εrms = 237/√2 = 168 V

Check the scaling: at 25 Hz the same machine gives ε₀ = 119 V and frequency 25 Hz — both halves matter.

Transformers

A transformer works only on AC: the alternating primary current creates an alternating flux in the iron core, inducing emf in the secondary. For an ideal transformer:

Vs/Vp = Ns/Np   and   VpIp = VsIs (power conserved)

Stepping voltage up steps current down, which is the whole point of grid transmission: at 400 kV a given power needs 1/1000th the current of a 400 V supply, cutting resistive losses (P = I²R) by a factor of 10⁶. Real transformers lose energy to eddy currents in the core (reduced by lamination), hysteresis (soft iron), and coil resistance — expect efficiency calculations of 90–98%.

Worked problem 2 — transformer and transmission

A transformer with 1800 primary turns is connected to a 240 V rms supply and delivers 12 V rms to a 60 W lamp. (a) How many secondary turns? (b) Find the secondary and primary currents (ideal transformer).

(a) Ns = Np × Vs/Vp = 1800 × 12/240 = 90 turns
(b) Is = P/Vs = 60/12 = 5.0 A ; Ip = P/Vp = 60/240 = 0.25 A

The 20:1 step-down in voltage is a 1:20 step-up in current — if the lamp cable were fed at primary current values it would need to be twenty times thinner... which is precisely backwards: the *secondary* (low voltage) side carries the heavy current and needs the thick cable, as anyone who has seen battery jumper leads knows.

Motional emf and Lenz in machines

A conductor of length L moving at speed v perpendicular to field B generates ε = BLv (the flux-sweeping argument of Lesson 11). Rotating machinery fights its own generation: a motor's spinning coil generates a back emf opposing the supply, so the net driving voltage is V_supply − ε_back; at start-up (no back emf yet) the current surges — the reason motors need starter circuits and lights dip when large motors start.

Practical worked example — reading an oscilloscope trace

An oscilloscope with timebase 2.0 ms per division and gain 5.0 V per division shows a sine wave 6.4 divisions peak-to-peak vertically and 8.0 divisions per cycle horizontally. Extract the signal's parameters, with uncertainties of ±0.2 division on each reading:

V₀ = (6.4/2) × 5.0 = 16 V (±3%)
Period T = 8.0 × 2.0 ms = 16 ms → f = 62.5 Hz (±2.5%)
Vrms = 16/√2 = 11.3 V → report 11.3 ± 0.4 V (combining % uncertainties)

Graticule reading skill — peak-to-peak divided by two, and the trap of reading the period between non-corresponding points — is exactly what Paper 2's oscilloscope items assess.

Rectification and smoothing (awareness)

A single diode gives half-wave rectification (current in pulses, mean output low); a bridge of four diodes gives full-wave; a capacitor in parallel with the load smooths the ripple by discharging between peaks. The physics is the capacitor's exponential (Lesson 12) harnessed: choose RC much larger than the ripple period and the output stays near the peak value.

Nature of Science

The "war of the currents" (Edison's DC versus Tesla/Westinghouse's AC) was settled by transformer physics: AC could be stepped up for transmission and down for use, DC (in the 1880s) could not. The technical argument, demonstrated at Niagara in 1896, decided a continent's infrastructure — evidence and engineering economics, not celebrity, won.

Real-world application

Every phone charger is this lesson in a box: a transformer (now usually a high-frequency switched design, smaller because ε ∝ rate of change of flux) steps mains down, diodes rectify, a capacitor smooths, and a regulator holds 5 V. The same chain at gigawatt scale — generator, step-up transformer, grid, step-down transformers — delivers power from power station to street.

Common mistakes (examiner style)

• Using peak values in P = VI for AC mains calculations (use rms throughout for power).
• Transformer equation inverted (Ns/Np upside down) — check plausibility: fewer secondary turns must mean lower voltage.
• Assuming an ideal transformer's currents scale like its voltages (they scale inversely).
• Forgetting that ε₀ = NBAω needs ω in rad s⁻¹, not f in Hz.
• Believing a transformer can step up energy, not just voltage.

IB Exam Experience

Paper 1-style questions. 1. The peak value of an AC supply is 170 V. Its rms value is about: A. 85 V B. 120 V C. 170 V D. 240 V. Key: B — 170/√2 = 120 V. 2. An ideal transformer has Np = 500, Ns = 50, Vp = 240 V, Ip = 0.20 A. The secondary current is: A. 0.020 A B. 0.20 A C. 2.0 A D. 20 A. Key: C — Vs = 24 V; power 48 W conserved → Is = 48/24 = 2.0 A. 3. A generator's rotation speed is tripled. The peak emf: A. stays the same B. triples C. increases ninefold D. falls to a third. Key: B — ε₀ ∝ ω.

Paper 2-style structured question (8 marks). A small wind turbine generator produces an rms emf of 26 V at a rotation frequency of 8.0 Hz in a field of 0.55 T; its coil has 400 turns. (a) Calculate the peak emf. [2] (b) Show that the coil's area is about 3.3 × 10⁻³ m². [3] (c) The output is stepped up by a transformer of turns ratio 1:9 for transmission. Calculate the rms transmission voltage and explain why stepping up reduces energy loss in the cables. [3] *Marking notes:* (a) ε₀ = 26 × √2 = 37 V. (b) ε₀ = NBAω with ω = 2π × 8.0 = 50.3 rad s⁻¹: A = 37/(400 × 0.55 × 50.3) = 37/11 066 = 3.3 × 10⁻³ m² ✓. Full credit requires the ω conversion to be visible; using f instead of ω is the standard error (giving 2.1 × 10⁻² m²). (c) V = 26 × 9 = 234 V; loss explanation: for the same transmitted power, higher V means lower I, and cable loss P = I²R falls as the square of the current reduction (here 81× less for the same power) — both reasoning steps required for full marks.

IA idea: Investigate the efficiency of a small transformer as a function of load current, identifying which loss mechanism dominates at low and at high load from the shape of the efficiency curve.

AC power in practice

A kettle rated "230 V, 2.0 kW" draws rms current I = P/V = 2000/230 = 8.7 A; its element resistance (at operating temperature) is R = V²/P = 26 Ω; the peak current is 8.7 × √2 = 12.3 A and the peak voltage 325 V — numbers the fuse (13 A) and insulation are chosen around. Average power over a cycle for a resistive load is IrmsVrms; instantaneously the power oscillates at twice the supply frequency between 0 and I₀V₀ = 4.0 kW. Motors and other inductive loads draw current out of phase with voltage, so their delivered power is IrmsVrms cos φ — the power factor — a refinement worth mentioning qualitatively in discussion answers.

Eddy-current braking, quantified in principle

A conducting disc spinning through a magnet's field has emfs induced in the metal; the resulting currents and their own fields oppose the motion, converting kinetic energy to heat at a rate proportional to B² and to the speed. Because the braking force grows with speed and vanishes at rest, eddy brakes are superb at smoothing high-speed stops (trains, rollercoasters, exercise bikes) but cannot hold a vehicle stationary — friction brakes take the final metres. This speed-dependent force profile is precisely Lenz's law plus P = ε²/R reasoning, and writing that chain explicitly is what HL explanation marks reward.

Why RMS and not mean?

The arithmetic mean of a sinusoidal voltage over a full cycle is zero, yet the supply clearly delivers power. The mean of the *rectified* voltage is 0.637V₀, but power depends on v², whose mean is V₀²/2 — hence the root-mean-square: square it, mean it, root it. Any averaging that answers a *power* question must pass through the square. Analogously, gas molecules' speeds enter kinetic theory as a mean square (Lesson 8's ⟨c²⟩) for the same reason: energy is quadratic in the underlying variable. Recognising "the average of the square is not the square of the average" as one idea in two costumes is synoptic thinking at its best.