Physics · 4 Electricity and magnetism

4.5.2 The a.c. generator

Before you start

An e.m.f. is induced in a conductor that cuts across magnetic field lines, or when the magnetic field linking a coil changes (4.5.1).
The induced e.m.f. is larger for a faster movement, a stronger field and more turns of wire (4.5.1).
Reversing the direction a wire moves through a field reverses the induced e.m.f. (4.5.1).
Between the flat N and S poles of a magnet, the field lines run straight across from N to S (4.1).
An alternating current keeps reversing its direction; a direct current flows one way only (4.2.2).
Frequency is the number of complete cycles per second, measured in hertz (Hz).

Syllabus objectives covered (2)

Describe a simple form of a.c. generator (rotating coil or rotating magnet) and the use of slip rings and brushes where needed
Sketch and interpret graphs of e.m.f. against time for simple a.c. generators and relate the position of the generator coil to the peaks, troughs and zeros of the e.m.f.

Part 1 of 7

What a generator does

From 4.5.1 you know that an e.m.f. is induced in a wire when it cuts across magnetic field lines. A generator does this without stopping: it keeps a coil and a magnetic field moving relative to each other by turning one of them.
There are two ways to arrange that. Either the coil turns between fixed magnet poles (a rotating-coil generator), or a magnet turns while the coil stays still (a rotating-magnet generator).
Both give an alternating e.m.f.: one that rises, falls to zero, reverses direction, and repeats. That is why this kind of generator is called an a.c. generator.
When the generator supplies a current, the energy delivered to the circuit comes from whatever turns the axle — an engine, a turbine or a person pedalling.

Part 2 of 7

The rotating-coil generator

Start with the parts. A rectangular coil of wire, with many turns, is mounted on an axle between the N and S poles of a magnet. The field in the gap runs straight across from N to S.
The axle is turned at a steady speed. As the coil turns, its two sides that run parallel to the axle move through the field and cut field lines, so an e.m.f. is induced in each of them.
apparatus diagram — not drawn yet
A rectangular coil of wire on a horizontal axle between a magnet's N pole (left) and S pole (right), with a curved arrow showing the direction of rotation; the two ends of the coil run along the axle to two separate complete metal rings fixed on the axle; a stationary carbon block presses on each ring, and wires from the two blocks go to a lamp. Label: coil, N, S, axle, slip rings, brushes.
The coil and the slip rings turn together. The brushes stay still and rub on the rings, so the lamp stays connected without its wires twisting.
Notice that the two sides move in opposite directions — one goes up while the other comes down. Because of that, their e.m.f.s push charge the same way round the coil, so the two e.m.f.s add together.
More turns on the coil means more wire cutting the field at once, which is why a real generator coil has many turns rather than one loop.

Part 3 of 7

Slip rings and brushes

Here is the problem slip rings solve. The coil turns, but the lamp it supplies does not. If the ends of the coil were wired straight to the lamp, the wires would wind round each other once every turn and soon twist and break.
Slip rings. Each end of the coil is joined to its own complete metal ring. Both rings are fixed on the axle, so they turn with the coil.
Brushes. A brush — usually a block of carbon — is pressed against each ring. The brushes do not turn. The rings slide round underneath them, keeping contact all the time, and the external circuit is connected to the brushes.
Because each brush always touches the same ring, each side of the external circuit stays connected to the same end of the coil for the whole turn. Whatever e.m.f. the coil produces is passed to the circuit unchanged.
The trap here: slip rings do not make the e.m.f. alternate. The coil produces an alternating e.m.f. by itself, and the slip rings only pass it on. A d.c. motor uses a different part, a split-ring commutator; do not mix the two up.

Part 4 of 7

The rotating-magnet generator

The second design keeps the coil still and turns a magnet beside it or inside it.
As the magnet turns, the magnetic field linking the coil keeps changing, and every half turn of the magnet it reverses direction. That changing field induces an alternating e.m.f. in the fixed coil.
apparatus diagram — not drawn yet
A bar magnet on a vertical spindle with a curved arrow showing it spinning, inside a fixed coil wound round it, with the two ends of the coil going straight to a lamp by ordinary wires; no rings or brushes.
Here the coil never moves, so its ends are wired straight to the lamp. Nothing twists, so no slip rings or brushes are needed.
Because the coil never moves, its ends can be wired straight to the circuit. There is nothing to twist, so no slip rings or brushes are needed.
That is what the words 'where needed' in the objective point at. Slip rings and brushes are needed when the coil rotates, and not when the magnet rotates.
One example is a bicycle dynamo, in which the turning wheel spins a small magnet inside a fixed coil.

Part 5 of 7

Which coil position gives which e.m.f.

Before you can draw the graph you need one idea: the size of the e.m.f. depends on how fast the sides of the coil are cutting field lines.
The coil turns at a steady speed, so its sides always move at the same speed. What changes as it turns is the direction they move in compared with the field.
Describe the coil by its plane — the flat face you would get if you stretched a sheet across it. Picture looking along the axle, with N on the left and S on the right, so the field lines run horizontally across.
Plane perpendicular to the field (coil vertical). The two sides are at the top and the bottom, moving horizontally — along the field lines, not across them. No lines are being cut, so at this instant the e.m.f. is zero.
apparatus diagram — not drawn yet
Two end-on views side by side, each looking along the axle with horizontal field lines running from an N pole on the left to an S pole on the right: on the left the coil is a vertical line with small velocity arrows at its top and bottom ends pointing horizontally, labelled 'plane perpendicular to field: e.m.f. zero'; on the right the coil is a horizontal line with a velocity arrow pointing up at one end and down at the other, labelled 'plane parallel to field: e.m.f. maximum'.
Vertical coil: the sides move along the field lines and cut none. Horizontal coil: the sides move straight across the lines and cut them fastest.
Plane parallel to the field (coil horizontal). The sides are at the left and the right, one moving straight up and the other straight down. They move at right angles to the field lines, cutting them at the greatest rate, so the e.m.f. is at its maximum.
In between those two positions the sides move partly across the lines, so the e.m.f. has an in-between value. As the coil turns from vertical to horizontal, the e.m.f. rises smoothly from zero to its maximum.
The awkward case: when the coil is vertical, the most field passes through it, which makes it look like the best position. It gives zero. At that instant the field through the coil is not changing and no side is cutting lines, and it is the cutting (the change) that induces an e.m.f., not the amount of field.

Part 6 of 7

One turn on the graph

Now follow the coil a quarter turn at a time, starting the clock with the coil vertical. Call the time for one complete turn $T$ .
At $t = 0$ the coil is vertical, so the e.m.f. is zero.
At $t = T / 4$ the coil has turned a quarter turn and is horizontal, so the e.m.f. is at its maximum. Draw this as a peak.
At $t = T / 2$ the coil is vertical again, so the e.m.f. is back to zero.
At $t = 3 T / 4$ the coil is horizontal again, but the side that was moving up is now moving down, and the other side is now moving up. Reversing the direction of motion reverses the induced e.m.f., so this maximum is in the opposite direction: a trough, as deep as the peak was high.
At $t = T$ the coil is back where it started and the e.m.f. is zero. One turn of the coil gives one complete cycle of the e.m.f.
graph diagram — not drawn yet
A graph of e.m.f. (y-axis, positive and negative) against time (x-axis) showing one complete smooth sine-shaped cycle starting at zero, with ticks at T/4, T/2, 3T/4 and T; above the curve at each of t = 0, T/4, T/2, 3T/4 and T a small end-on sketch of the coil between N (left) and S (right) poles: vertical at the three zeros, horizontal at the peak and at the trough, with one side of the coil marked by a dot so you can see it has swapped sides at the trough.
Vertical coil at every zero, horizontal coil at the peak and the trough. The dotted side is on the opposite side at the trough, which is why the e.m.f. has reversed.
So in every turn the e.m.f. is zero twice (each time the coil is vertical) and at its maximum size twice (each time the coil is horizontal) — once as a peak and once as a trough. The e.m.f. changes direction every half turn, each time the coil passes through the vertical.
To sketch the graph: label the axes e.m.f. and time; draw a smooth curve that is rounded at the peaks and troughs; make every peak and every trough the same height; and space the zeros evenly, half a period apart.
If the coil starts horizontal instead, the graph starts at a peak (or a trough) instead of at zero. It is the same curve moved along by a quarter of a period. Read the starting position from the question before you draw anything.
Whether the first maximum is a peak or a trough depends on the direction of rotation and on which way round the output is connected. Turning the coil the other way turns the whole graph upside down. If a question gives you part of the graph, carry on in the same pattern.

Part 7 of 7

Speed, period and frequency

The period $T$ is the time for one complete cycle, which for this simple generator is the time for one turn of the coil. The frequency $f$ is the number of cycles per second, in hertz.
Because one turn gives one cycle, the frequency equals the number of turns per second. And if one cycle takes $T$ seconds, the number of cycles in one second is $1 \div T$ :
$f = \frac{1}{T}$
For example, a coil turning $50$ times per second gives $50$ Hz, and its period is $T = 1 / 50 = 0.02$ s, which is $20$ ms.
Turning the coil faster changes two things at once. The sides cut field lines faster, so the peak e.m.f. is larger: doubling the speed doubles it. Each turn also takes less time, so the period is shorter: doubling the speed halves it.
graph diagram — not drawn yet
One pair of e.m.f.–time axes with two sine curves both starting at zero: a solid curve showing exactly one cycle across the time axis, and a dashed curve with twice the peak height showing exactly two cycles over the same time.
Double the speed: the peaks are twice as high and twice as many cycles fit into the same time.
A stronger magnet, more turns on the coil or a coil of larger area also make the peak e.m.f. larger, but they do not change how long a turn takes. They change the height of the graph and leave the time axis alone.
The trap here: speed is the only one of these changes that alters the timing. A sketch for a faster coil must change both the height and the spacing of the peaks.

Worked examples

Example 1. Describe how a simple rotating-coil a.c. generator produces an alternating e.m.f. in an external circuit, including the use of slip rings and brushes.

Show the working

Start with the parts and what moves: a coil of wire on an axle, between the N and S poles of a magnet, turned at a steady speed.
Give the physics next. As the coil turns, its sides cut the magnetic field lines, so an e.m.f. is induced in the coil.
Then say why it alternates. Every half turn, each side of the coil starts moving the opposite way through the field, so the direction of the induced e.m.f. reverses.
Now the connection. Each end of the coil is joined to a slip ring; the slip rings turn with the coil. Name both parts — slip rings and brushes — because the objective asks for both.
Brushes, which stay still, press against the slip rings and are connected to the external circuit. This keeps the coil connected to the circuit as it turns, without the wires twisting.

Answer: A coil on an axle is turned between the poles of a magnet. Its sides cut the field lines, so an e.m.f. is induced, and because each side moves the opposite way through the field every half turn, the e.m.f. reverses direction. The ends of the coil are joined to two slip rings that turn with the coil; fixed brushes press on the rings and connect them to the external circuit, so the coil stays connected without the wires twisting.

Example 2. The coil of an a.c. generator turns at $25$ revolutions per second. At time $t = 0$ the plane of the coil is perpendicular to the magnetic field. Determine (a) the frequency and period of the e.m.f., and (b) the times during the first revolution at which the e.m.f. is zero, at a peak and at a trough.

Show the working

One revolution gives one cycle, so the frequency equals the revolutions per second: $f = 25$ Hz.
The period is the time for one revolution.
$T = \frac{1}{f} = \frac{1}{25} = 0.04 s = 40 ms$
At $t = 0$ the plane is perpendicular to the field (coil vertical), so the sides are moving along the field lines. The e.m.f. starts at zero.
Each quarter turn takes $T / 4 = 40 \div 4 = 10$ ms. After one quarter turn the coil is parallel to the field, so the first maximum — the peak — is at $10$ ms.
After half a turn ( $20$ ms) the coil is vertical again: zero. After three quarters ( $30$ ms) it is parallel again with its sides swapped: the trough. After a full turn ( $40$ ms) it is back to zero.

Answer: (a) $f = 25$ Hz and $T = 0.04$ s ( $40$ ms). (b) Zero at $0$ , $20$ ms and $40$ ms; peak at $10$ ms; trough at $30$ ms.

Example 3. A graph of e.m.f. against time for an a.c. generator starts at a maximum of $3.6$ V at $t = 0$ . The e.m.f. falls to zero at $5$ ms, reaches $- 3.6$ V at $10$ ms, is zero again at $15$ ms and is back at $3.6$ V at $20$ ms. (a) State the position of the coil at $t = 0$ . (b) Determine the number of revolutions per second. (c) The coil is now turned at half the speed. Describe the new graph.

Show the working

(a) The e.m.f. is at a maximum, so the sides must be cutting the field lines at the greatest rate. That happens when the plane of the coil is parallel to the field.
(b) Find one full cycle. The graph goes from a peak at $0$ to the next peak at $20$ ms, so the period is $20$ ms. Do not take $10$ ms: peak to trough is only half a cycle.
Convert to seconds before dividing: $20$ ms $= 0.020$ s.
$f = \frac{1}{T} = \frac{1}{0.020} = 50 Hz$
One cycle is one revolution, so the coil makes $50$ revolutions per second.
(c) Halving the speed halves the rate of cutting, so the peak e.m.f. halves: $3.6 \div 2 = 1.8$ V. Each turn takes twice as long, so the period doubles: $2 \times 20 = 40$ ms.
The coil still starts in the same position, so the new graph still starts at its maximum at $t = 0$ .

Answer: (a) The plane of the coil is parallel to the magnetic field. (b) Period $20$ ms, so $50$ revolutions per second. (c) The same shape, still starting at a maximum, but with peaks of $1.8$ V and troughs of $- 1.8$ V, and a period of $40$ ms.

Where people go wrong

Mistake: Saying the e.m.f. is largest when the coil is vertical, facing the poles, because the most field passes through it.

Instead: The e.m.f. is zero when the plane of the coil is perpendicular to the field, and largest when it is parallel to the field.

An e.m.f. is induced by cutting field lines (by change), not by the amount of field present. In the vertical position the sides move along the field lines and cut none.

Mistake: Saying the slip rings make the current alternate, or calling them a commutator.

Instead: The coil produces the alternating e.m.f.; the slip rings only keep it connected to the circuit. A split-ring commutator is a different part, used in a d.c. motor.

Each slip ring stays joined to the same end of the coil all the time, so it cannot reverse anything. The reversal comes from each side of the coil changing direction through the field every half turn.

Mistake: Showing a faster coil with taller peaks but the same spacing, or with closer peaks but the same height.

Instead: Faster rotation changes both: the peak e.m.f. increases and the period decreases.

Faster sides cut field lines at a greater rate (a bigger e.m.f.), and each turn takes less time (a shorter period, a higher frequency).

Mistake: Saying a rotating-magnet generator needs slip rings and brushes too.

Instead: Only a rotating coil needs them. A fixed coil is wired straight to the circuit.

Slip rings exist to stop the connecting wires twisting. If the coil does not move, there is nothing to twist.

Mistake: Taking the period as the time from a peak to the next trough.

Instead: The period is one full cycle: peak to the next peak, or zero to the zero after next.

Peak to trough is half a turn of the coil, so using it gives a period half the true value and a frequency twice the true value.

Check yourself

What would happen if the ends of a rotating coil were wired straight to the external circuit, with no slip rings?
Answer
The connecting wires would twist round each other once every turn and would soon tangle and break.
State what the brushes are and what they do.
Answer
They are fixed contacts (usually carbon blocks) pressed against the slip rings. They do not turn, and they connect the turning slip rings to the external circuit by sliding contact.
Why does a rotating-magnet generator not need slip rings or brushes?
Answer
Its coil is fixed, so the ends of the coil can be wired straight to the circuit with nothing to twist.
Explain why the e.m.f. is zero when the plane of the coil is perpendicular to the magnetic field.
Answer
The sides of the coil are moving parallel to the field lines, so they are not cutting any lines and no e.m.f. is induced at that instant.
In one complete revolution of the coil, how many times is the e.m.f. zero, and how many times is it at its maximum size?
Answer
Zero twice (each time the coil is perpendicular to the field) and at maximum size twice (each time it is parallel to the field), once as a peak and once as a trough.
A coil turns at $10$ revolutions per second, starting perpendicular to the field. Determine the period, and the time from $t = 0$ to the first peak.
Answer
$T = 1 / 10 = 0.1$ s. The first peak is a quarter of a turn later: $0.1 \div 4 = 0.025$ s.
On an e.m.f.–time graph, one complete cycle takes $25$ ms. How many revolutions per second is the coil making?
Answer
$25$ ms $= 0.025$ s, so $f = 1 / 0.025 = 40$ Hz: $40$ revolutions per second.
State three changes that would increase the peak e.m.f. of a rotating-coil generator, and say which one also changes the frequency.
Answer
Turning the coil faster, using a stronger magnet, using more turns on the coil (or a larger coil). Only turning faster also changes the frequency.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A rectangular coil is turned at a constant speed between the poles of a magnet. The ends of the coil are connected to an external circuit through slip rings and brushes.

(a) Describe how the slip rings and brushes connect the turning coil to the external circuit. 3 marks No calculator

Mark scheme

B1 each end of the coil is joined to a (separate) slip ring
B1 the slip rings turn with the coil while the brushes stay still and press / slide on the rings
B1 so the coil stays connected to the circuit without the wires twisting / tangling, owtte

(b) Explain why the e.m.f. is zero at the instant the plane of the coil is perpendicular to the magnetic field. 2 marks No calculator

Mark scheme

B1 the sides of the coil move parallel to the field lines (at that instant)
B1 so no field lines are cut / the field through the coil is not changing, so no e.m.f. is induced

(c) Explain why the direction of the e.m.f. reverses every half turn of the coil. 2 marks No calculator

Mark scheme

B1 after half a turn each side of the coil moves in the opposite direction through the field, owtte
B1 reversing the direction of motion (of a wire through the field) reverses the direction of the induced e.m.f. / the sides swap places, so the side moving up is now moving down, owtte

Question 2. The coil of a simple a.c. generator turns at $20$ revolutions per second. At $t = 0$ the plane of the coil is perpendicular to the magnetic field.

(a) Calculate the period of the e.m.f. 2 marks

Mark scheme

C1 T = 1 / f or 1 / 20 seen
A1 0.05 s (50 ms); unit required

(b) Sketch a graph of e.m.f. against time for the first two revolutions of the coil. Mark the time at the end of the second revolution. 3 marks No calculator

Mark scheme

B1 smooth curve alternating above and below the time axis, starting at zero at t = 0; the first half-cycle may be above or below the axis
B1 two complete cycles ending at 0.10 s; ecf from (a)
B1 all peaks and troughs the same height, with zeros evenly spaced

(c) Determine the two times during the first revolution at which the plane of the coil is parallel to the magnetic field. 1 mark

Mark scheme

B1 0.0125 s and 0.0375 s / 12.5 ms and 37.5 ms (both needed; unit needed); ecf from (a)

(d) The coil is now turned at $40$ revolutions per second. Describe how the graph changes. 2 marks No calculator

Mark scheme

B1 peaks (and troughs) are higher / twice the height
B1 period halves to 0.025 s / frequency doubles to 40 Hz / twice as many cycles in the same time; ecf from (a)

Question 3. A bicycle dynamo contains a magnet that is spun by the wheel inside a fixed coil. The coil is connected to the bicycle lamp.

(a) Explain why this generator does not need slip rings or brushes. 2 marks No calculator

Mark scheme

B1 the coil does not move / is fixed (it is the magnet that turns)
B1 so the coil can be connected directly to the lamp with wires that cannot twist

(b) Suggest two changes to the dynamo that would increase the peak e.m.f. 2 marks No calculator

Mark scheme

B1 any one from: spin the magnet faster (ride faster) / use a stronger magnet / use more turns on the coil / use a coil of larger area, owtte
B1 a second, different change from the same list

How this is examined

Sketching the e.m.f.–time graph, and redrawing it for a faster coil: a Sketch on given axes, possibly with a stated starting position for the coil and a second line to add for a different speed A smooth curve alternating above and below the time axis; all peaks and troughs the same height; zeros evenly spaced; the correct starting value (zero for a coil starting perpendicular to the field, a maximum for one starting parallel). For double speed: peaks twice as high and the period halved.
What the slip rings and brushes are and why the rotating coil needs them: a Describe beside a labelled diagram of a rotating-coil generator, or a State asking for the purpose of one labelled part Each end of the coil is joined to its own slip ring; the slip rings turn with the coil; fixed brushes press on the rings and connect to the external circuit; so the coil stays connected without the wires twisting. Each of these is a separate point.
Why the e.m.f. is zero at one coil position and largest at another: an Explain about a diagram of the coil drawn at one position, or a question asking which points on a graph match a vertical or a horizontal coil For zero: 'the sides move parallel to the field lines, so they do not cut them'. For a maximum: 'the sides move at right angles to the field lines, so they cut them at the greatest rate'. Naming the position without the cutting is a description, not an explanation.
Getting the period and the turning rate from the graph: a Determine or Calculate using the time axis of an e.m.f.–time graph The period read over one full cycle (peak to next peak, not peak to trough); the period converted to seconds; $f = 1 / T$ ; the answer with its unit, Hz, or as revolutions per second.
Why a rotating-magnet generator has no slip rings: a short Explain or Suggest about a generator in which the magnet turns 'The coil does not move', followed by 'so it can be connected directly to the circuit and the wires cannot twist'.

Words to know

a.c. generator: A device in which a coil and a magnetic field are turned relative to each other, inducing an e.m.f. that keeps reversing direction.
Slip rings: Two complete metal rings on the axle of a rotating coil, one joined to each end of the coil, turning with it.
Brushes: Fixed contacts, usually carbon, pressed against the slip rings to connect the turning coil to the external circuit.
Plane of the coil: The flat face of the coil. Perpendicular to the field gives zero e.m.f.; parallel to the field gives the maximum e.m.f.
Peak e.m.f.: The largest value the e.m.f. reaches in a cycle, at the moment the plane of the coil is parallel to the field.
Period: The time for one complete cycle of the e.m.f., which for a simple generator is the time for one turn of the coil.
Frequency: The number of complete cycles per second, in hertz; for a simple generator it equals the number of turns per second, and $f = 1 / T$ .

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