Full chapter
Electromagnetic induction
All 5 topics and the revision summary on one page.
01
Producing an induced voltage
A changing magnetic field through a coil can induce an e.m.f. across its ends. A strong field that remains unchanged through the coil gives no sustained induced e.m.f.
E.m.f. is energy supplied per unit charge, measured in volts. Current is the rate of flow of charge, measured in amperes. They describe different quantities.
Move a magnet relative to a coil
Connect a coil to a suitable sensitive voltmeter and move a bar magnet along the coil's axis. As a pole approaches, the field through the coil changes and the meter can register a voltage. Stop and hold the magnet still: after the transient settles, the reading returns to zero. Withdraw the same pole and the induced polarity reverses.
One winding, fixed meter connections
The sensitive voltmeter reads VA - VB: its positive lead stays at A. The display shows polarity, with no numerical calibration. Blue arrows show magnet motion.
1. The north pole approaches
The magnetic field through the turns is changing. This winding and these connections give a positive reading during the approach.
2. The magnet is held still
Once the motion has stopped and the transient has passed, this unchanged arrangement gives no sustained induced e.m.f.
3. The north pole withdraws
Withdrawal reverses the magnetic change and the meter polarity. Swapping the meter leads would reverse both signs.
| Change made | Voltage observation |
|---|---|
| Bring N towards the coil | A reading of one polarity; use this as the reference. |
| Hold N stationary | Zero after the transient settles. |
| Withdraw the same N pole | The opposite polarity to the reference approach. |
| Bring S towards the coil instead | The opposite polarity to the N-pole approach. |
| Make the same approach faster | A greater peak magnitude over a shorter event, if the meter can resolve it. |
The labels positive and negative depend on the winding and which meter lead is connected to each coil end. Swapping the meter leads reverses the displayed sign; it does not change the physical cause of induction.
An e.m.f. does not guarantee a current
An induced e.m.f. can exist across the ends of an open coil. A continuing induced current also needs a closed conducting path. A high-resistance voltmeter draws little current and measures the terminal voltage with little loading.
A sensitive galvanometer in a closed coil circuit shows induced current instead. Its deflection is evidence of induction, but its current scale is not a voltage scale. The current depends on the induced e.m.f. and the resistance of the complete circuit.
Identify what changes through the loop
- Move the coil: a coil moving relative to a stationary magnet can experience a changing field through its loops.
- Change a nearby current: varying the current in an electromagnet changes its field, even when the coil and electromagnet stay still.
- Rotate the coil: changing its orientation in a steady external field changes the field passing through its loops.
Moving the magnet and coil together while keeping their relative arrangement unchanged does not by itself produce induction. Ask what changes through the coil, rather than only whether something moves.
Compare the magnitude fairly
- Faster change
- Move the same magnet along the same path more quickly. The more rapid field change gives a greater induced e.m.f.
- A larger field change
- A stronger magnet can give a greater induced e.m.f. when it produces a larger field change over the same path and time.
- More linked turns
- More turns experiencing the same changing field increase the total induced e.m.f. Keep the turns' position and orientation comparable.
More turns can also increase circuit resistance. A change in galvanometer current alone does not establish the same ratio of induced e.m.f.s. For a voltage comparison, use suitable voltage measurements and account for loading.
Investigate induction with controlled observations
- Secure the coil and connect a voltage sensor or meter able to show both polarities. Select a voltage range that includes the expected peak with useful resolution.
- Mark the magnet's path and endpoints. Record which pole faces the coil and keep the winding and meter connections fixed.
- Record an approach, a stationary interval and a withdrawal. Compare the polarities and identify when the magnetic situation is changing.
- Change one factor at a time. For a speed comparison, use the same magnet, coil and path. For different magnets or turn counts, keep the motion and geometry comparable.
- Repeat the comparison and inspect the recorded peaks and timing for variation.
A brief peak needs a sufficiently fast recording method: a slow display may miss its maximum. Faster sampling does not correct inconsistent magnet motion. Use the same marked travel and comparable duration when motion is meant to be controlled, and keep the magnet aligned without striking the coil.
Optional check A coil and a permanent magnet are kept still. Current in a nearby electromagnet is then increased, changing the magnetic field through the coil. Which statement is correct?
02
The direction opposes the change
The induced e.m.f. has a polarity that, in a closed conducting circuit, drives a current whose magnetic effect opposes the change producing it.
Like magnetic poles repel and unlike poles attract. Viewed from one face of a current-carrying coil, anticlockwise current makes that face N; clockwise current makes it S.
North approaches: oppose the approach
As a north pole approaches a closed coil, the field through the coil changes. The induced current makes the near face north, so it repels the approaching north pole. Viewed from the magnet end, that current is anticlockwise.
North withdraws: oppose the separation
When the same north pole withdraws, the change is reversed. The coil's near face becomes south, attracting the magnet and opposing its withdrawal. Viewed from the same magnet end, the current is clockwise.
The induced effect opposes the change
Here the conducting loop is closed, so an induced current can flow. In each lower view, look from the magnet towards the loop. Brown shows current, teal shows the induced field, blue shows magnet motion and purple shows magnetic force on the magnet.
North approaches: near face becomes north
Repulsion opposes the approach. The induced field points out of the near face.
North withdraws: near face becomes south
Attraction opposes the separation. The induced field points into the near face.
South approaches: near face becomes south
Repulsion opposes the approach. The induced field points into the near face.
South withdraws: near face becomes north
Attraction opposes the separation. The induced field points out of the near face.
| Magnet motion | Near coil face | Current in that face view |
|---|---|---|
| N approaches | N: repels | Anticlockwise |
| N withdraws | S: attracts | Clockwise |
| S approaches | S: repels | Clockwise |
| S withdraws | N: attracts | Anticlockwise |
Predict a new case
A south pole moves away
First identify the change: the magnet and coil are separating. To oppose that separation, the near coil face must attract the south pole, so it becomes north.
A north face requires anticlockwise current when viewed from the magnet end. Work from the change to the pole, then from the pole to the current direction.
The induced field does not always oppose the magnet's original field. During withdrawal, it acts to maintain a field that is decreasing. The rule concerns opposition to the change.
Where does the electrical energy come from?
When the induced current supplies a load, work is needed to keep moving the magnet or coil against the opposing magnetic effect. That external work provides the electrical output and any heating losses.
The opposing effect need not prevent the motion: an external force can maintain it while doing work. The magnet is not steadily consumed as fuel. In a generator, the mechanical energy source must keep turning the rotor.
Optional check A north pole is withdrawn from a closed conducting coil. Viewed from the magnet end, which induced pole and current direction oppose that withdrawal?
03
A simple a.c. generator
A generator uses mechanical motion to keep changing the magnetic field through a coil. In the rotating-coil model, the induced voltage reverses every half-turn, producing alternating output.
Induction depends on change through the coil. A closed external circuit allows the induced e.m.f. to drive current and transfer energy to a load. The mechanical drive supplies this energy.
Keep electrical contact while the coil rotates
- Magnetic poles and rotating coil
- The poles provide an external field. A mechanical drive turns the coil on an axle, changing its orientation relative to that field.
- Two complete slip rings
- Each coil end is permanently connected to its own ring. The rings rotate with the coil and remain insulated from one another.
- Stationary brushes and load
- Each brush maintains sliding contact with one ring and connects it to the external circuit. The coil ends remain distinct while the coil rotates.
A complete circuit at the positive peak
This oblique view shows the coil at one quarter-turn. A and B mark its near ends; the far ends join at the top. Brown arrows show conventional current through the connected load. The teal arrow shows field direction.
Two full rings, two separate connections. Ring A always connects to physical end A; Ring B always connects to B. Each stationary brush stays in contact with its own ring.
The dashed grey line is the insulated axle, not a wire. Its projected crossings make no electrical connection. The coil leads are insulated from the axle and from each other.
At this instant the output is A positive relative to B. Half a turn later the same connections carry the opposite current; the slip rings do not rectify it.
Slip rings maintain contact; they do not swap the coil ends or make the output one-way. The split-ring commutator in a d.c. motor has a different role: its segments exchange brushes every half-turn.
Connect the orientation to the graph
Use a uniform field directed right, and look along the axle from the near end. The coil turns clockwise. A and B name the same physical sides throughout.
Define the output voltage as the potential of near terminal A relative to near terminal B, written VA - VB. Positive output means A is at the higher potential. This stated connection and rotation fix the graph's signs.
Follow the same A and B through one turn
Look along the axle from the near end. Rotation is clockwise and the field points right in every stage. The circles locate the active sides. The dashed diameter shows the coil-plane orientation in this end view; it is not another wire joining the near ends.
The output reference stays fixed: voltage at A minus voltage at B. Blue arrows show mechanical motion.
1. Start: 0 s
0 VPlane perpendicular to the field
A is above the axle and B below. The field through the coil is at an extreme, but is momentarily not changing: the output is zero.
2. One quarter-turn: 0.010 s
+6.0 VPlane parallel to the field
A is on the right moving down; B is on the left moving up. Near A is positive relative to near B, giving the first positive peak.
3. Half a turn: 0.020 s
0 VPlane perpendicular to the field
The physical sides have exchanged positions: A is below the axle and B above. The output is momentarily zero again.
4. Three quarters of a turn: 0.030 s
-6.0 VPlane parallel to the field
A is on the left moving up; B is on the right moving down. Near A is now negative relative to near B, giving the negative peak.
5. One complete turn: 0.040 s
0 VPlane perpendicular to the field
A is above and B below again. The original orientation has returned after one period, and the next cycle begins.
The same five instants on a smooth graph
Supplied model: uniform rotation in a uniform field, peak output 6.0 V, and one complete turn every 0.040 s. The sign refers to A relative to B throughout.
At zero output, the magnetic field itself has not disappeared. At maximum output magnitude, it is the rate of change through the coil that is greatest. Keep the coil's plane distinct from a line drawn perpendicular to that plane.
For the stated uniform field and uniform rotation, sketch a smooth sinusoidal curve through the cycle, with output voltage on the vertical axis and time on the horizontal axis. Straight segments between the five marked points would describe a different waveform.
The period is T = 0.040 s, so the frequency is f = 1/T = 25 Hz. Constant rotation speed does not mean constant output voltage. Swapping the output leads would reverse every voltage sign while leaving the period unchanged.
Change the output under matched conditions
- Rotate faster: the field through the coil changes faster, so peak voltage increases and the period becomes shorter.
- Use a stronger field: peak voltage increases for the same coil and rotation speed.
- Use more turns linked by the changing field: peak voltage increases with the field, coil geometry and rotation speed otherwise comparable.
Changing field strength or turn count does not by itself change the frequency when rotation speed stays fixed.
A rotating magnet is another arrangement
A rotating magnet can change the field through stationary coils and produce alternating output. The stationary coil ends can connect directly to the external circuit, so that arrangement does not need slip rings on those output connections. Slip rings are needed where the chosen design must maintain contact with rotating coil ends.
Optional check A coil rotates uniformly in a uniform magnetic field. At t = 0 its plane is perpendicular to the field. Why is the instantaneous output zero at that position in the stated generator model?
04
How transformers change voltage
A transformer uses a changing magnetic field to transfer energy between two separate windings. The turns ratio sets the voltage ratio in the ideal model.
A current in a coil creates a magnetic field. A changing field through another coil can induce an e.m.f. Power is the rate of energy transfer.
Two windings, one linked field
The primary winding connects to the a.c. input. Its changing current creates a changing magnetic field in an iron core. The core strengthens and channels the field so that it links the secondary winding, inducing an e.m.f. there.
Separate circuits linked by a changing core field
Each winding is continuous and insulated from the iron core. The primary supply circuit and secondary load circuit have no ordinary conducting connection.
The drawing illustrates fewer secondary turns. Use the stated Np and Ns in a calculation. Continuing a.c. changes the core field and can sustain a secondary output; steady d.c. cannot after switching transients.
The windings are insulated from the core and from one another. Primary electrons do not travel through the core into the secondary circuit. The coupling is magnetic, while charge flows in each winding's own complete circuit.
A steady primary current gives a steady field and no sustained induced secondary output in this model. Switching on or off can produce a brief transient; continuing transformer action needs a changing field, as provided by an a.c. input.
Use the ideal-transformer relationships
Let N be the number of turns, V the voltage across a winding and I its current. Subscript p means primary and s means secondary.
More secondary turns give a step-up voltage transformer; fewer give a step-down voltage transformer. A useful rearrangement is Vs = Vp x Ns / Np.
The examples use the stated operating a.c. values and resistive loads. At the same ideal transferred power, a higher voltage comes with a lower current. The turns ratio alone does not determine current for an unspecified load.
Step down from 240 V to 12 V
Voltage ratio, then power
1000 primary turns and 50 secondary turns
The ideal transformer has Np = 1000, Ns = 50 and Vp = 240 V.
Vs = 240 x 50/1000 = 12.0 V.
The connected load takes Is = 2.0 A, so output power = 12.0 x 2.0 = 24 W.
For the ideal transformer, input power is also 24 W. Therefore Ip = 24/240 = 0.100 A.
The lower-voltage winding carries the larger current. The primary and secondary currents are not equal, because these are separate circuits linked by energy transfer.
Step up from 12 V to 60 V
Keep transferred power fixed
200 primary turns and 1000 secondary turns
Vp = 12 V, so Vs = 12 x 1000/200 = 60 V.
The specified ideal transferred power is 30 W. Primary current Ip = 30/12 = 2.5 A; secondary current Is = 30/60 = 0.50 A.
The voltage rises fivefold and the current falls fivefold at this same power. The transformer has not created additional electrical energy.
Recognise what the ideal model leaves out
Real transformers have losses, including heating in the windings and core. Their useful output power is less than their input power; energy is still conserved because the difference is transferred elsewhere.
An iron core made of insulated laminations reduces circulating induced currents and associated heating in the core. This improves efficiency, but it does not make every practical transformer loss-free.
Optional check An ideal transformer has 1000 primary turns and 50 secondary turns. Its primary voltage is 240 V, and a connected resistive load takes 2.0 A from the secondary. What are the secondary voltage and primary current?
05
Reducing transmission losses
Transmission cables have resistance and warm when current flows. For the same delivered power, using a higher transmission voltage reduces current and therefore reduces cable heating.
Electrical power is P = VI for the stated operating conditions. For a resistance R carrying current I, the voltage drop across that resistance is V = IR.
Use the voltage across the cable
Let R be the total resistance of the outgoing and return cables. Their combined voltage drop is Vdrop = IR.
The receiving-end voltage is across the equipment receiving the useful power. It is not the same as the drop along the cables. Multiplying cable current by the full receiving-end voltage gives the delivered power in this model, not the cable loss.
A grid overview and a separate cable power model
Overview: change voltage before and after transmission
This overview has no numerical voltages. The outgoing and return conductors are both shown. Brown arrows show one instant of alternating current.
The receiving-end transmission voltage is across the step-down transformer's input. The final load receives the lower output voltage.
Direct cable model: 1000 V received
This is a direct cable-and-load model, separate from the overview. Its receiving voltage is across this load. The sending supply must provide the useful 10.0 kW plus the cable heating.
Direct cable model: 10000 V received
This is a direct cable-and-load model, separate from the overview. Its receiving voltage is across this load. The sending supply must provide the useful 10.0 kW plus the cable heating.
Compare two systems delivering 10.0 kW
Use an idealised power account with receiving-end power 10.0 kW = 10 000 W and total cable resistance 2.0 ohm. The receiving equipment is arranged to take that same useful power at either stated voltage; this is not a comparison of different voltages applied to an unchanged resistor.
Case A
1000 V at the receiving end
Cable current I = Preceived/Vreceived = 10 000/1000 = 10.0 A.
Cable drop Vdrop = IR = 10.0 x 2.0 = 20.0 V.
Cable heating Ploss = I2R = 10.02 x 2.0 = 200 W.
The sending-end voltage must be 1000 + 20.0 = 1020 V. Its input power is 1020 x 10.0 = 10 200 W: 10 000 W delivered plus 200 W lost in the cables.
Case B
10 000 V at the receiving end
Cable current I = 10 000/10 000 = 1.00 A.
Cable drop Vdrop = 1.00 x 2.0 = 2.00 V.
Cable heating Ploss = 1.002 x 2.0 = 2.00 W.
The sending-end voltage must be 10 000 + 2.00 = 10 002 V. Its input power is 10 002 x 1.00 = 10 002 W: 10 000 W delivered plus 2.00 W lost in the cables.
The receiving-end voltage is ten times larger in B, so the current is one tenth as large. Because heating depends on current squared, the cable loss is one hundredth as large: 2 W instead of 200 W.
Keep the full power account
A step-up transformer can raise the voltage before a long transmission cable. A step-down transformer reduces it afterwards to the voltage required by the receiving equipment. The transformers change the voltage/current combination; the cable resistance still causes a finite loss.
The advantage of higher voltage follows from keeping delivered power and cable resistance fixed. Increasing voltage does not universally reduce heating: across the same unchanged resistance, a higher voltage would increase current and heating. Name the conditions before making the comparison.
Real transformers and other equipment also have losses. Include them if supplied; the calculated 200 W and 2 W are the cable losses under these stated conditions.
Optional check Two systems each deliver 10.0 kW at the receiving end through cables with total resistance 2.0 ohm. The receiving-end voltage rises from 1000 V to 10000 V. How does the cable heating loss change?
Revision summary
Produce and measure an induced e.m.f.
A changing magnetic field through a coil can induce an e.m.f., measured in volts. The change may come from relative magnet/coil motion, a changing nearby current or a rotating coil. A steady magnetic arrangement gives no sustained induced e.m.f.
An open coil can have induced e.m.f.; induced current needs a closed conducting path and depends on its resistance. A high-resistance voltmeter measures terminal voltage with little loading. A galvanometer shows current, whose scale must not be relabelled as voltage.
- Approach and withdrawal of the same pole give opposite polarities with the meter leads fixed.
- Reversing which pole approaches reverses the polarity. Holding it still gives zero after transients settle.
- Faster change, a larger field change in the same time, or more turns linked by that change increase induced e.m.f. under comparable conditions.
- Control path, orientation and motion when comparing. Use a suitable voltage range and fast enough recording to resolve a brief peak. Repetition does not fix inconsistent motion or changed circuit resistance.
Oppose the change
The induced polarity drives a current, when the circuit is closed, whose magnetic effect opposes the change producing it. Viewed from the magnet end:
- N approaches: near face N, anticlockwise current.
- N withdraws: near face S, clockwise current.
- S approaches: near face S, clockwise current.
- S withdraws: near face N, anticlockwise current.
Approach is opposed by repulsion and withdrawal by attraction. The induced field can support a decreasing original field; it does not always oppose the original field itself. External work maintains the changing arrangement and supplies electrical output and losses.
Read the generator cycle
Each rotating coil end stays connected to its own complete slip ring and stationary brush. The rings maintain contact without swapping ends. A rotating magnet with stationary output coils is another arrangement.
For uniform rotation in a uniform field, the output-time curve is sinusoidal. With the stated clockwise rotation, rightward field and output VA - VB, the cycle is:
- 0 and 0.040 s
- A above, B below; coil plane perpendicular to field; output 0 V.
- 0.010 s
- A right moving down, B left moving up; plane parallel to field; near A positive; output +6.0 V.
- 0.020 s
- A below, B above; plane perpendicular to field; output 0 V.
- 0.030 s
- A left moving up, B right moving down; plane parallel to field; near A negative; output -6.0 V.
Zero output means zero instantaneous rate of change through the coil, not an absent field. The period is 0.040 s and frequency 25 Hz. Faster rotation increases frequency and peak magnitude; stronger field or more linked turns increases magnitude at fixed rotation speed.
Use the ideal transformer
VpIp = VsIsp = primary; s = secondary. Use matching voltage units. Ideal input power equals output power.
Changing primary current creates a changing field in the iron core, inducing secondary e.m.f. The windings are insulated, separate circuits. Steady d.c. gives no sustained secondary output after switching transients.
1000:50 turns with 240 V input gives 12.0 V output. A 2.0 A secondary load receives 24 W and requires 0.100 A ideal primary current. Conversely, 200:1000 turns with 12 V input gives 60 V output; at 30 W, the primary/secondary currents are 2.5 A and 0.50 A.
Step-up voltage means more secondary turns. At the same ideal power the secondary current is smaller. Real losses make useful output power lower than input power.
Calculate cable loss with cable quantities
At 10.0 kW received and R = 2.0 ohm, a receiving voltage of 1000 V needs 10.0 A and loses 200 W in the cables. The sending end supplies 1020 V and 10 200 W. At 10 000 V received, current is 1.00 A, cable loss 2.00 W, and the sending end supplies 10 002 V and 10 002 W.
A tenfold voltage increase reduces current tenfold and cable loss one hundredfold at fixed delivered power and cable resistance. Do not substitute load voltage for cable drop, or ignore the difference between sending and receiving power.
Back to induced voltage