Rotating Coils & Generators

Hold the field still and turn the coil instead. Let the rotation axis lie in the plane of the coil and perpendicular to B, and let θ be the angle from B to the coil's normal. The flux through one turn is Φ = BA cos θ, and with N turns wound on the same former the flux linkage is NΦ = NBA cos θ. Turning at a steady angular speed ω makes θ = ωt + θ₀, so NΦ = NBA cos(ωt + θ₀). Every symbol except the cosine is a constant of the machine: B is fixed by the magnets, A by the winding, N by the wire. The whole time dependence has been pushed into the geometry. Take N = 150 turns, A = 8.0 × 10⁻³ m² and B = 0.40 T: the flux linkage swings between +0.48 Wb and −0.48 Wb once per revolution, no matter how fast or slowly the shaft is turned.

Faraday's law needs the rate of change, so differentiate: ε = −N dΦ/dt = −NBA d(cos ωt)/dt = NBAω sin ωt. Three facts fall out at once. The output is a sinusoid, because the derivative of a cosine is a sine. Its frequency is the rotation frequency, since nothing in the derivative changed how fast θ advances. And its amplitude carries an extra factor ω that the flux never had. The phase deserves a second look. At θ = 0 the coil is face-on to the field, the flux linkage is at its maximum, and the emf is zero, because a cosine has zero slope at its peak. A quarter turn later the coil plane contains B, the flux linkage is zero, and the emf is at its peak: that is where the long sides of the coil cut straight across the field lines. The emf peak therefore trails the flux peak by a quarter cycle rather than leading it. With the coil above spun at 30.0 Hz, ω = 188.5 rad s⁻¹ and εₘₐₓ = 0.48 Wb × 188.5 rad s⁻¹ = 90.5 V.

Write the amplitude as εₘₐₓ = 2πf NBA and the design choices separate cleanly. More turns and a larger area raise the output in direct proportion and cost only winding space; a stronger field does the same until the iron saturates. Frequency is different, because raising f raises the amplitude and the frequency together — you cannot buy more volts this way without changing the waveform the load receives. For a simple two-pole machine the electrical frequency equals the mechanical rotation rate, which is why a 50 Hz alternator turns at 3000 revolutions per minute. Since loads respond to the heating value rather than the peak, quote εᵣₘₛ = εₘₐₓ/√2. A coil giving εₘₐₓ = 90.5 V delivers 64.0 V rms, and mains quoted as 230 V rms actually peaks at 325 V — the figure that matters when you choose the insulation.

The coil turns while the external circuit does not, so the connection has to be made through sliding contacts. Two continuous slip rings, one per coil end, keep each end permanently attached to its own terminal: whatever the coil produces reaches the load unaltered, so the output is the full sinusoid and the machine is an alternator. Replace them with a single ring split into two halves — a commutator — and the brushes change which half they touch at every half turn, precisely when the emf crosses zero. The negative half of the sinusoid is flipped over, giving |εₘₐₓ sin ωt|, an output that never reverses but pulses from 0 to εₘₐₓ twice per revolution. Its mean value is (2/π)εₘₐₓ ≈ 0.637εₘₐₓ, rather than zero. Real DC machines stack several coils at different angles onto one commutator, so the brushes always sit on whichever coil is near its peak and the ripple almost disappears.

An open-circuited generator is easy to turn. Close the circuit through a load R, with the coil's own resistance r, and the current is I(t) = ε(t)/(R + r) = [NBAω/(R + r)] sin ωt. That current flows in a coil sitting in the same field B, so the coil acquires a magnetic moment NIA and the field exerts a torque on it, τ = NIAB sin ωt. By Lenz's law it opposes the rotation, and it is largest exactly where the emf is largest. Multiply by ω and the mechanical power appears: τω = I (NBAω sin ωt) = εI, the instantaneous electrical power, instant by instant. Averaging over a turn, with the mean of sin²ωt equal to ½, gives a total P̄ = εₘₐₓ²/[2(R + r)] and a mean torque (NBA)²ω/[2(R + r)]. Halve the total resistance R + r and both the current and the torque double: a generator resists in proportion to what is taken from it.

The model is worth more when it is run backwards. Spin the coil at several rates, record the peak emf on an oscilloscope, and plot εₘₐₓ against f. The model predicts a straight line through the origin of gradient 2πNBA, so one graph measures the product NBA — and if N and A are known from the winding, it measures B inside the machine, where no probe fits. A curved plot or a non-zero intercept is telling you an assumption has failed. The period on the same trace gives f directly, a cross-check on the drive. Watch the waveform shape as well: a flat-topped trace means the field is not uniform over the coil's sweep, which is deliberate in machines with shaped pole pieces, while a load-voltage peak arriving later than the rotor angle predicts betrays self-inductance, the point at which sin ωt bookkeeping must give way to AC circuit theory.