How does a spinning coil make the 50 Hz power in your wall socket?

How to use this simulation

1. Press play. The coil starts in the neutral plane (θ = 0), where it faces the field squarely. The graph on the right begins at zero and the lamp is dark.
2. Watch the coil turn towards 90°. The emf climbs along the sine curve and the lamp brightens. At 90° the coil is parallel to the field, the flux is zero, yet the emf is at its peak of 314.2 V.
3. Look at the arrows on the two vertical sides of the coil. They point in opposite directions, so the current runs round the loop. After 180° they swap, and that swap is what makes the current alternating.
4. Drag the turns slider N. Double it and E₀ doubles, because every turn adds its own emf. Do the same with the area A and the field B and see the same straight-line response.
5. Raise the rotation speed. The wave squeezes together (higher frequency) and gets taller (higher peak), because E₀ = NBAω has ω in it. Notice that speed changes both things at once.
6. Set the speed to 3,000 rpm. The frequency readout shows 50 Hz, the mains frequency used in Bangladesh and India. Compare the two dashed lines on the graph: the peak and the rms level.

From rotation to electricity

Every power station in the country, whether it burns gas, runs on coal, spins in a hydro dam or turns in the wind, ends in the same machine: a generator. Steam or water pushes a turbine, the turbine spins a shaft, and the shaft spins a coil in a magnetic field. That is the whole trick. Mechanical energy goes in, electrical energy comes out.

Why does spinning a coil produce a voltage? Because of Faraday’s law. The emf in a coil equals the rate at which magnetic flux through it changes. As the coil turns, the angle between its face and the field keeps changing, so the flux keeps changing, so there is always an emf. If you want the full story of that law, the electromagnetic induction simulation on this site walks through it with a moving magnet and a coil.

In this simulation you control the four things that matter: the number of turns, the area of the coil, the strength of the field and how fast it spins. The rest of this page shows how they combine into one short formula.

The parts of the machine

A simple AC generator has five parts, and the simulation draws all of them.

• The field magnet: the N and S poles that supply a nearly uniform field B between them. Big machines use electromagnets, but the physics is identical.
• The armature coil: a rectangular loop of N turns of insulated wire that rotates on a shaft. Its area is A.
• The axle and prime mover: whatever turns the coil. In a power station it is a steam or water turbine; in a bicycle dynamo it is the wheel.
• Slip rings: two metal rings fixed to the shaft, one joined to each end of the coil. They turn with the coil.
• Brushes: two carbon contacts that press on the rings and stay still. They connect the spinning coil to the stationary outside circuit, here a lamp.

Why the emf is a sine wave

Let θ be the angle between the normal to the coil and the magnetic field. The flux through one turn is the field times the area facing the field, which is B·A·cos θ. At θ = 0 the coil faces the field head-on and the flux is at its maximum. At θ = 90° the coil is edge-on and the flux is zero.

Now let the coil turn steadily with angular speed ω, so θ = ωt. The emf is minus the rate of change of the total flux linkage N·Φ. Differentiating cos ωt gives −ω sin ωt, and the two minus signs cancel.

What each slider does

Because E₀ = NBAω is a plain product, every factor scales the peak in a straight line. Start from the default coil: N = 100 turns, A = 200 cm² (which is 0.020 m²), B = 0.50 T and 3,000 rpm.

First convert the speed. 3,000 revolutions per minute is 50 revolutions per second, so ω = 2π × 50 = 314.2 rad/s. Then E₀ = 100 × 0.50 × 0.020 × 314.2 = 314.2 V. That is the number the readings panel shows when the page loads.

Double N and the peak becomes 628.3 V. Halve B and it falls to 157.1 V. Notice that rpm is the only slider that also changes the frequency; the other three change only the height of the wave.

Reading the graph

The horizontal axis is the rotation angle θ, drawn over two full turns. The bold curve is what the generator has produced so far and the orange dot is the present position. The faint curve ahead of it is what is still to come.

Here is the same default coil at a few angles. Notice how the flux column is a cosine and the emf column is a sine, shifted by a quarter of a turn.

Frequency and the 50 Hz supply

One full turn of the coil gives one full cycle of the emf, so the frequency in hertz equals the revolutions per second. At 3,000 rpm that is 50 revolutions each second, so f = 50 Hz and the period is T = 20 ms. The emf reaches its first peak a quarter of a period after the start, which is 5.0 ms.

Bangladesh and India both supply mains at 50 Hz, which means the voltage reverses direction 100 times every second. A big power-station alternator has several pairs of poles, so it does not need to spin at the full 3,000 rpm. With two pairs of poles, f = (rpm × pairs) ÷ 60 needs only 1,500 rpm. Some countries, including the United States, use 60 Hz instead, which needs 3,600 rpm for one pair.

That is why the lamp in the simulation is shown in slow motion. Real mains lights flicker 100 times a second, far too fast to see, so the picture runs at only 4% of the real speed (2.0 turns per second at 3,000 rpm) while the readings keep the real values.

Peak value and rms value

A voltage that swings between +314.2 V and −314.2 V has an average of zero over a cycle, which says nothing about how much work it does. We need a different average, one that counts positive and negative halves alike. The answer is the root-mean-square, or rms, value: square the emf, take its average over a cycle, then take the square root. For a sine wave the answer is the peak divided by √2.

The rms value of an AC voltage is the DC voltage that would heat a resistor at the same rate. That is why “220 V mains” always means 220 V rms. Its peak is 220 × √2 = 311 V, which is what the insulation in your wall has to survive. For our default coil, E₀ = 314.2 V gives E_rms = 222.1 V. Watch the dashed lines on the graph: they sit at ±E_rms, about 70.7% of the way up to the peak.

Worked example 1: the default coil

A coil has 100 turns, area 200 cm², and spins at 3,000 rpm in a field of 0.50 T. Find the peak emf, the rms emf, the frequency and the emf at 30°.

Worked example 2: how many turns for 220 V?

You want a generator whose output is 220 V rms, using the same area, field and speed as the default. The peak must be 220 × √2 = 311 V.

Rearrange E₀ = NBAω for N: N = E₀ ÷ (BAω) = 311 ÷ (0.50 × 0.020 × 314.2) = 99.0. So about 100 turns will do. That is why the default of 100 turns gives a figure so close to mains: 222.1 V rms.

Worked example 3: a small dynamo

A bicycle-style dynamo coil has 50 turns of area 30 cm² in a 0.2 T field and turns at 600 rpm. Then f = 10 Hz and E₀ = 50 × 0.2 × (30 × 10⁻⁴) × ω gives 1.885 V, about 1,885 mV. The rms value is 1.333 V.

The lesson: a household-sized voltage needs many turns, a bigger field or higher speed. Real machines get there with hundreds of turns and strong electromagnets.

Worked example 4: the emf at a given instant

For the default coil, what is the emf 2.5 ms after the coil is in the neutral plane? The period is 20 ms, so 2.5 ms is the fraction 2.5/20 of a turn, which is θ = 45°. Then e = E₀ sin θ = 222.1 V, which is exactly E_rms. The emf equals the rms value twice every cycle, at 45° and 135°, and again with a minus sign at 225° and 315°.

Where energy comes from

When current flows in the lamp, the coil becomes an electromagnet and the field pushes back on it, like a brake. By Lenz’s law the induced current always opposes the turning. The turbine has to keep pushing against this magnetic drag, and that work is the electrical energy delivered. Open the circuit and the drag disappears: the coil spins freely and you get no power. A generator creates nothing from nowhere.

Common mistakes

These four lose marks every year.

• Putting the maximum emf at θ = 0. With θ measured from the normal to the field, the emf is zero at θ = 0 and maximum at 90°. The opposite statement is true when the angle is measured from the plane of the coil.
• Using centimetres squared in E₀ = NBAω. Convert the area to m² and the speed to rad/s first.
• Using rpm as ω. Multiply the revolutions per second by 2π.
• Quoting the peak value when the question gives “220 V”. The household value is always rms, so the peak is 311 V.

Exam corner

Typical questions: explain the principle and working of an AC generator with a labelled diagram, derive e = NBAω sin ωt, find the peak or rms value, or find the frequency from the speed. In a derivation show the flux, differentiate once and state the sign by Lenz’s law. In a numerical, write the conversions explicitly and check that your answer for the default case matches the readings panel.

A classic follow-up asks what happens to E₀ and f if the speed is doubled. Both double. If the number of turns is doubled, only E₀ doubles.

Revision list

Check yourself without looking.

• Principle: electromagnetic induction in a rotating coil.
• e = E₀ sin ωt with E₀ = NBAω.
• The emf is maximum when the flux is zero (coil parallel to the field).
• E_rms = E₀/√2 and 220 V mains has a peak of about 311 V.
• f = rpm ÷ 60 for one pair of poles; mains in Bangladesh and India is 50 Hz.
• Slip rings give AC; a split-ring commutator gives DC.
• Mechanical work against the magnetic drag becomes the electrical energy.

Frequently asked questions