The big idea: A generator is a motor run backwards: instead of using current to make a coil spin, you spin a coil in a magnetic field and it makes a current.
As the coil turns, the magnetic flux through it changes smoothly, so by Faraday's law it induces an emf (a driving voltage). Because the flux rises and falls as the coil rotates, the emf comes out as a sine wave — this is alternating current (AC).
Why a sine wave?: When the coil is flat to the field the flux is largest but momentarily unchanging, so the emf is zero. When the coil is edge-on to the field the flux is changing fastest, so the emf is at its peak. One full turn gives one full cycle of the sine.
The output is described by the rotational frequency of the coil: ω is the angular frequency (how fast it spins, in rad s⁻¹). The faster you spin it, the bigger and the more frequent the peaks.
For a coil of N turns and area A rotating with angular frequency ω in a magnetic field B, the induced emf follows a sine curve. Its highest value — the peak emf ε₀ — is reached each time the coil is edge-on to the field.
- instantaneous induced emf (V)
- peak (maximum) emf (V)
- magnetic flux density (T)
- area of the coil (m²)
- number of turns on the coil
- angular frequency of rotation (rad s⁻¹)
Four ways to get a bigger emf: ε₀ = BANω, so the peak voltage grows if you increase any of:
- B — a stronger magnet - A — a larger coil - N — more turns of wire - ω — spin it faster
IB-style question — peak emf of a generator
A flat coil of 200 turns, each of area 0.015 m², spins at an angular frequency of 50 rad s⁻¹ in a uniform magnetic field of flux density 0.40 T. Determine the peak emf of the generator.
Solution
- Write the given peak-emf formula first:
- Substitute the values:
- Work it out — keep the unit:
Final answer
ε₀ = 60 V. (Spin it twice as fast and the peak emf doubles to 120 V.)
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Why we need rms: An AC voltage is constantly changing — its average over a cycle is zero, which is useless for describing how much power it delivers. So we quote the root-mean-square (rms) value instead.
The rms value is the steady DC value that would deliver the same average power (the same heating effect in a resistor). When the mains is called "230 V", that 230 V is the rms voltage.
- rms current (A) / rms voltage (V)
- peak current (A) / peak voltage (V)
- ≈ 1.41 (the sine-wave factor)
Worked example — rms of the mains
A mains supply has a peak voltage of V₀ = 325 V. Determine the rms voltage of the supply.
Solution
- Write the given rms formula first:
- Substitute the peak voltage:
- Work it out — keep the unit:
Final answer
Vrms = 230 V — the figure quoted for the household mains.
What a transformer does: A transformer changes an AC voltage up or down. Two coils share an iron core: AC in the primary (Np turns) makes a changing flux that induces an emf in the secondary (Ns turns).
The voltages are in the same ratio as the turns. More turns on the secondary ⇒ step-up (higher voltage); fewer turns ⇒ step-down (lower voltage).
- primary / secondary voltage (V)
- number of turns on primary / secondary
- primary / secondary current (A)
An ideal transformer conserves power: An ideal (100% efficient) transformer wastes no energy, so the power in equals the power out: ε_p I_p = ε_s I_s.
That is why the current ratio is upside down: if a step-down transformer halves the voltage, it doubles the current — and vice versa.
Worked example — a step-down transformer
A step-down transformer has Np = 1000 turns on the primary and Ns = 50 turns on the secondary. The primary voltage is Vp = 230 V. Determine the secondary voltage.
Solution
- Write the given transformer ratio first:
- Substitute the turns and primary voltage:
- Work it out — keep the unit:
Final answer
Vs = 11.5 V. Fewer turns on the secondary ⇒ the voltage steps down.
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Where it shows up: AC generators, rms and transformers are HL only (D.4):
- Paper 1A — a one-step 'find the rms value', 'is this step-up or step-down?', or 'what does an ideal transformer conserve?'. - Paper 2 — determine a peak emf with ε₀ = BANω, or use the transformer ratio (and power conservation) to find a secondary voltage or current.
Three easy marks: (1) rms = peak ÷ √2 (and peak = rms × √2 going the other way). (2) Voltage ratio = turns ratio; current ratio is inverted. (3) "Ideal transformer" ⇒ power in = power out (εp Ip = εs Is).
IB-style question — current in a charger transformer
An ideal transformer in a phone charger steps 230 V (rms) down to 5.0 V (rms). The output current to the phone is 2.0 A. Determine the current drawn from the mains by the primary coil.
Solution
- An ideal transformer conserves power, so write power in = power out:
- Make the primary current the subject and substitute:
- Work it out — keep the unit:
Final answer
Ip ≈ 0.043 A (43 mA). Stepping the voltage down by ~46× steps the current up by ~46×, so the primary draws far less current than the secondary delivers.