Why a turning coil makes a sine wave
A coil of N turns and area A sits in a magnetic field of flux density B. As it turns through angle θ, the flux linking it is N·B·A·cos θ: largest when the coil faces the poles, zero when it lies along the field.
Faraday’s law says the induced EMF is the rate of change of that flux linkage. Differentiate cos θ at a steady angular speed ω and you get a sine:
λ = N·B·A·cos(ωt) e = −dλ/dt = N·B·A·ω·sin(ωt) Emax = N·B·A·ω, Erms = Emax / √2
When the EMF is zero, and when it peaks
The EMF depends on how fast the flux changes, not on how much flux there is. Facing the poles (θ = 0°) the coil holds the most flux, but for that instant it isn’t changing, so e = 0. A quarter turn later the coil holds no flux at all, yet the flux is changing fastest, so e is at its peak.
In the simulator the coil sides are marked ⊙ (current out of the page) and ⊗ (into the page). Watch them fade out and swap each half turn: that swap is the alternation in alternating current.
Frequency: poles and speed
A machine with P poles passes P/2 north–south pairs for every mechanical turn, so each turn of the shaft gives P/2 electrical cycles. With the speed n in revolutions per minute:
That is why grid generators run at fixed speeds: a 2-pole steam turbine at 3000 rpm and a 4-pole machine at 1500 rpm both give 50 Hz, and a slow hydro turbine needs many poles to reach the same frequency.
ωe = (P/2)·ωm f = P·n / 120
From EMF to terminal voltage and power
Slip rings and brushes carry the coil’s current out to the load. The coil itself has resistance Ra, so some voltage is lost inside the machine: the terminal voltage is a little below the EMF, and the gap widens as the load draws more current.
I = E / (R + Ra) V = I·R P = V·I, torque = P_gen / ωm
Worked example (the simulator’s default values)
- N = 100 turns, B = 0.50 T, A = 0.010 m², 2 poles at 3000 rpm, load R = 100 Ω, Ra = 1 Ω.
- ωm = 2π·3000/60 = 314.2 rad/s; with 2 poles ωe = ωm, so f = 2·3000/120 = 50 Hz and T = 20 ms.
- N·B·A = 100·0.50·0.010 = 0.50 Wb-turns, so Emax = 0.50·314.2 = 157.1 V and Erms = 111.1 V.
- I = 111.1 / (100 + 1) = 1.10 A rms; terminal V = 1.10·100 = 110.0 V; load power = 110.0·1.10 = 120.9 W.
Common mistakes
- Thinking the EMF is largest when the flux is largest. It is largest when the flux is changing fastest, a quarter turn later.
- Thinking more turns or a stronger magnet raises the frequency. They raise the voltage; only speed and poles set the frequency.
- Using the 2-pole formula for every machine: with P poles the electrical angle runs P/2 times faster than the shaft.
Common questions
What is the formula for the EMF of an AC generator?
e = N·B·A·ω·sin(ωt), so the peak EMF is Emax = N·B·A·ω and the RMS EMF is Emax/√2, with ω the electrical angular speed in rad/s.
How do you calculate generator frequency from speed?
f = P·n/120, with P the number of poles and n the speed in rpm. A 2-pole machine at 3000 rpm gives 50 Hz; at 3600 rpm it gives 60 Hz.
What do slip rings do in an AC generator?
They connect the rotating coil to the stationary circuit: each end of the coil goes to its own ring, and carbon brushes pressing on the rings carry the alternating current out to the load.