Electromagnetic Induction

How a changing magnetic flux produces an emf, why the induced current always opposes the change, and how transformers use this to move energy across the grid with little loss.

60 min · Module 6: Electromagnetism

Assumes you have read The Motor Effect.

The full teaching. Start here.

Syllabus requirementsModule 6: Electromagnetism

describe how magnetic flux can change, with reference to the relationship $Φ = B_{∥} A = B A cos θ$ (ACSPH083, ACSPH107, ACSPH109)ACSPH083 ACSPH107 ACSPH109
analyse qualitatively and quantitatively, with reference to energy transfers and transformations, examples of Faraday’s Law and Lenz’s Law $ε = - N \frac{ΔΦ}{Δ t}$ , including but not limited to: (ACSPH081, ACSPH110)ACSPH081 ACSPH110
the generation of an electromotive force (emf) and evidence for Lenz’s Law produced by the relative movement between a magnet, straight conductors, metal plates and solenoids
the generation of an emf produced by the relative movement or changes in current in one solenoid in the vicinity of another solenoid
analyse quantitatively the operation of ideal transformers through the application of: (ACSPH110)ACSPH110
$\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}$
$V_{p} I_{p} = V_{s} I_{s}$
evaluate qualitatively the limitations of the ideal transformer model and the strategies used to improve transformer efficiency, including but not limited to:
incomplete flux linkage
resistive heat production and eddy currents
analyse applications of step-up and step-down transformers, including but not limited to:
the distribution of energy using high-voltage transmission lines

What this actually asks for

Four formulas, all on the data sheet, and one idea behind them: an emf appears only while the magnetic flux through a circuit is changing.

The Faraday dot point says with reference to energy transfers and transformations, so every Lenz's law answer should say where the energy comes from and where it goes. The transformer dot points split into a quantitative one (analyse quantitatively) and a qualitative one (evaluate qualitatively), so expect both a calculation and a judgement about how good the ideal model is. The last dot point is the reason transformers matter: the grid.

The short version

Magnetic flux measures how much field passes through an area:

$Φ = B A cos θ$

with $θ$ between the field and the normal to the area. It changes if the field, the area or the angle changes.

A changing flux induces an emf (Faraday's law):

$ε = - N \frac{ΔΦ}{Δ t}$

The minus sign is Lenz's law: the induced current flows so that its own field opposes the change. If it helped the change instead, energy would come from nothing.

An ideal transformer on AC obeys

$\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}, V_{p} I_{p} = V_{s} I_{s}$

Real transformers lose energy through flux that misses the secondary, heating in the coils and eddy currents in the core. A closed soft iron core, thick copper windings and a laminated core reduce these losses.

The grid steps the voltage up for transmission because, for fixed power, a higher voltage means a smaller current, and the line loss $I^{2} R$ falls with the square of the current.

Magnetic flux

Counting field lines through a loop

AnswersWhy does tilting a loop change the flux even though the field is unchanged?

Picture the field as lines, packed more densely where it is stronger. Flux counts how many lines pass through a loop. A bigger loop catches more of them, a stronger field packs more into the same space, and tilting the loop lets lines slip past its edge.

The tilt is measured by the angle between the field and the normal, the line sticking straight out of the loop. Face on, $θ = 0^{\circ}$ and every line through that area passes through. Edge on, $θ = 9 0^{\circ}$ and none do. Only the field component along the normal, $B_{∥} = B cos θ$ , counts.

Magnetic flux through an areaon the NESA formulae sheet

Φ = B_{∥} A = B A cos θ

Symbols: $Φ$ magnetic flux through the areaWb
$B$ magnetic field strengthT
$B_{∥}$ component of the field along the normal to the areaT
$A$ area the field passes throughm²
$θ$ angle between the field and the normal to the area°
Valid when: The field is uniform over a flat area. θ is measured from the normal, so a loop facing the field has θ = 0 and the largest flux.
Not valid when: The field varies across the area, the surface is curved, or θ is measured from the plane of the loop rather than from its normal.
Rearranged: for B: $B = \frac{Φ}{A c o s θ}$ for A: $A = \frac{Φ}{B c o s θ}$ for \theta: $cos θ = \frac{Φ}{B A}$
Where it turns up: Finding the change in flux when a coil turns, moves or its field changes
Setting up a Faraday's law calculation
Explaining why a loop edge on to a field has no flux through it
Where marks go missing: Measuring θ from the plane of the loop, which swaps cosine for sine
Leaving the area in cm² instead of converting to m²
Multiplying by the number of turns, which belongs in Faraday's law, not in the flux

Where this comes from

Φ = BA cos θ

The angle is measured from the normal, the line at right angles to the loop. Face on to the field, θ is 0° and the flux is largest; edge on, θ is 90° and no field lines pass through.

Field strength0.20 T

Loop area50 cm²

Angle between field and normal60 °

Flux through the loop: 5.00 × 10⁻⁴ Wb
Field component along the normal, B cos θ: 0.100 T
Largest flux, face on: 1.00 × 10⁻³ Wb

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Past $9 0^{\circ}$ the flux is negative: the field now enters the loop from the other face. This matters in generators, where a coil turns through a full circle and the flux swings between $+ B A$ and $- B A$ .

CheckpointAnswer before reading on.

A square loop of wire with sides of $5.0$ cm sits in a uniform magnetic field of $0.40$ T. The normal to the loop makes an angle of $6 0^{\circ}$ with the field.

Calculate the magnetic flux through the loop.

Your answerWb

Give it to 2 significant figures.

Hint 1Convert the side length to metres before finding the area.

Hint 2 $θ$ in $Φ = B A cos θ$ is measured from the normal, which is what the question gives.

Hint 3 $Φ = 0.40 \times (0.050)^{2} \times cos 6 0^{\circ}$

CheckpointAnswer before reading on.

A flat coil of area $A$ lies in a uniform field $B$ . The plane of the coil makes an angle of $3 0^{\circ}$ with the field lines.

Which expression gives the flux through the coil?

A $B A cos 6 0^{\circ}$
B $B A cos 3 0^{\circ}$
C $B A$
D $0$

Hint 1Draw the normal to the coil. What angle does it make with the field?

Hint 2The normal is at right angles to the plane.

Hint 3If the plane is at $3 0^{\circ}$ to the field, the normal is at $6 0^{\circ}$ to it.

Exam question

Straightforward · about 4 min

3 marks

With reference to $Φ = B A cos θ$ , describe how the magnetic flux through a coil can be changed. Give a practical example of each way.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Look at each quantity on the right hand side of the formula.

Hint 2Which of them could a magnet moving towards a coil change?

Hint 3What changes when a coil spins in a fixed field?

Where students lose marks on this one

Mentioning only moving a magnet.
Why it happens: It is the demonstration most students have seen.
Work through the formula term by term so no way is missed.

Written for this site.

Faraday's law

Faraday found that a coil connected to a galvanometer registers a current only while something is changing: a magnet moving, a coil turning, a nearby current switching on or off. A strong magnet held still inside the coil does nothing. What matters is the rate at which the flux changes, and each turn of a coil adds its own emf in series:

Faraday's law of induction, with Lenz's lawon the NESA formulae sheet

ε = - N \frac{ΔΦ}{Δ t}

Symbols: $ε$ average induced emfV
$N$ number of turns in the coil1
$ΔΦ$ change in flux through one turnWb
$Δ t$ time taken for the changes
Valid when: Any change in flux through a circuit: a changing field, a changing area or a changing angle. It gives the average emf over Δt; the minus sign is Lenz's law, saying the induced current opposes the change that caused it.
Not valid when: Used for the emf at an instant when the flux is changing at a varying rate, where it gives only the average, or used with the total flux through all N turns and then multiplied by N again.
Rearranged: for \Delta t: $Δ t = \frac{N ∣ΔΦ∣}{∣ ε ∣}$ for N: $N = \frac{∣ ε ∣ Δ t}{∣ΔΦ∣}$ for \varepsilon: $ε = B l v (rod moving across a field)$
Where it turns up: Average emf when a magnet is pushed into a coil or a coil is rotated
The emf across a rod moving along rails, ε = Blv
Explaining which way an induced current flows
Where marks go missing: Using the final flux instead of the change in flux
Forgetting the number of turns
Treating the minus sign as a numerical sign to carry through, instead of as the direction rule

Where this comes from

$ΔΦ$ is the change in flux through one turn. The formula gives the average emf over $Δ t$ ; if the flux changes at a steady rate, that is also the emf at every instant.

Worked example3 marks

A magnet pushed towards a coil

A coil of $100$ turns, each of area $3.0 \times 1 0^{- 3}$ m $^{2}$ , faces a bar magnet. As the magnet is pushed closer, the field through the coil rises from $0.020$ T to $0.12$ T in $0.25$ s. Find the average emf.

Change in flux through one turn

The coil faces the field, so cos θ = 1 throughout and only B changes.

$ΔΦ = Δ B A = (0.12 - 0.020) \times 3.0 \times 1 0^{- 3} = 3.0 \times 1 0^{- 4} Wb$

Apply Faraday's law

The turns are in series, so their emfs add.

$∣ ε ∣ = N \frac{ΔΦ}{Δ t} = 100 \times \frac{3.0 \times 1 0 ^{- 4}}{0.25} = 0.12 V$

Answer

The average emf is $0.12$ V while the magnet moves. Once it stops, the emf drops to zero.

Is that answer sensible?

Pushing twice as fast halves $Δ t$ and doubles the emf; the flux change is the same because it depends only on where the magnet starts and ends.

CheckpointAnswer before reading on.

A coil of $200$ turns, each of area $4.0 \times 1 0^{- 3}$ m $^{2}$ , sits at right angles to a magnetic field. The field increases steadily from $0$ to $0.50$ T in $0.20$ s.

Calculate the size of the average emf induced in the coil.

Your answerV

Give it to 2 significant figures.

Hint 1Find the change in flux through one turn first.

Hint 2At right angles, $cos θ = 1$ , so $ΔΦ = Δ B A$ .

Hint 3 $ε = N \frac{ΔΦ}{Δ t} = 200 \times \frac{2.0 \times 1 0 ^{- 3}}{0.20}$

CheckpointAnswer before reading on.

A coil of $50$ turns and area $0.020$ m $^{2}$ starts facing a uniform $0.30$ T field, with its normal along the field. It is turned through $9 0^{\circ}$ in $0.050$ s, so that it ends edge on to the field.

Calculate the size of the average emf induced.

Your answerV

Give it to 2 significant figures.

Hint 1What is the flux at the start, with $θ = 0^{\circ}$ ?

Hint 2What is the flux at the end, with $θ = 9 0^{\circ}$ ?

Hint 3Use the change: $ΔΦ = 0 - B A$ .

Lenz's law and energy

Why the induced current always pushes back

AnswersWhat would go wrong if the induced current helped the change instead?

Push the north pole of a magnet into a coil. Suppose the induced current made the near end of the coil a south pole. It would attract the magnet, which would speed up, change the flux faster, induce a bigger current, attract harder still, and keep going. Kinetic energy and electrical energy would both grow with nothing supplying them.

That cannot happen, so the induced current must do the opposite: it makes the near end a north pole and repels the magnet. You have to do work to push the magnet in, and that work is what the circuit turns into electrical energy and then heat. Lenz's law is conservation of energy written as a direction rule.

CheckpointAnswer before reading on.

The north pole of a bar magnet is pushed into one end of a coil connected to a galvanometer.

Which statement describes the induced current?

AIt makes the near end of the coil a north pole, repelling the magnet.
BIt makes the near end of the coil a south pole, attracting the magnet.
CIt flows only while the magnet is inside the coil and at rest.
DIt is the same whether the magnet is pushed in or pulled out.

Hint 1Lenz's law: the induced current opposes the change that produced it.

Hint 2The change is the magnet approaching. What would oppose an approaching north pole?

Hint 3Like poles repel.

The syllabus lists magnets, straight conductors, metal plates and solenoids. The same reasoning applies to each:

A straight conductor moving across a field sweeps out area, so it cuts flux at a rate $B l v$ . The simulation below shows this.
A metal plate or pipe has no coil, but it is full of closed paths. A changing flux drives eddy currents around them, and their fields oppose the motion. This is magnetic braking, and it is why a magnet falls slowly through a copper pipe.
A solenoid near another solenoid links them through its field. Changing the current in one changes the flux through the other, which is how a transformer works.

CheckpointAnswer before reading on.

A strong magnet dropped down a vertical copper pipe falls much more slowly than an identical unmagnetised steel block. Copper is not attracted to magnets.

Which explanation is correct?

ACurrents induced in the pipe produce fields that oppose the motion of the magnet, and its energy becomes heat in the copper.
BThe copper is weakly attracted to the magnet, which holds it back.
CInduced currents pull the magnet down, but air trapped in the pipe slows it more.
DThe magnet repels the pipe equally at every point, so the net force on it is upwards even when it is held still.

Hint 1The walls of the pipe are a conductor, and the flux through each ring of the pipe changes as the magnet passes.

Hint 2What does a changing flux do in a conductor?

Hint 3Which way must the resulting force point, by Lenz's law?

A rod on rails

A rod moving at speed $v$ along rails a distance $l$ apart adds area $l v Δ t$ to the circuit in each interval $Δ t$ . The flux grows at $B l v$ per second, so the emf is

$ε = B l v$

The induced current in the rod sits in the magnetic field, so it feels a motor effect force $B I l$ , and Lenz's law says that force points backwards. Give the rod a push in the simulation and watch it slow down. The two energy readouts come from separate calculations: the kinetic energy lost, and the heat worked out from $I^{2} R$ over time. They agree.

Speed

Starting speed2.0 m s⁻¹

Field strength0.50 T

Resistance0.50 Ω

Rod mass100 g

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

Dots show the field coming out of the screen. The induced current runs down the rod, and the motor effect force on it points backwards. Every joule of kinetic energy the rod loses appears as heat in the resistor.

Set the field to zero and the rod slides on at a constant speed, because with no flux there is no current and no braking. Lower the resistance and it stops sooner, since a larger current means a larger force.

Worked example4 marks

Keeping a rod moving

A rod on rails $0.50$ m apart is pulled at a steady $2.0$ m s $^{- 1}$ through a $0.50$ T field. The rails are joined by a $0.50 Ω$ resistor. Find the current, the force needed to keep the rod moving, and show that the power supplied equals the heating in the resistor.

The emf

The rod, the field and the velocity are mutually perpendicular.

$ε = B l v = 0.50 \times 0.50 \times 2.0 = 0.50 V$

The current

The rod and rails have negligible resistance, so all of it is in the resistor.

$I = \frac{ε}{R} = \frac{0.50}{0.50} = 1.0 A$

The braking force

The current in the rod is in the field, so it feels a motor effect force. Lenz's law makes it oppose the motion, so the puller must supply an equal forward force.

$F = B I l = 0.50 \times 1.0 \times 0.50 = 0.25 N$

Compare powers

The dot point asks for energy transfers, and this comparison is the evidence that energy is conserved.

$P_{pull} = F v = 0.25 \times 2.0 = 0.50 W, P_{heat} = I^{2} R = 1. 0^{2} \times 0.50 = 0.50 W$

Answer

The current is $1.0$ A and the rod needs a steady $0.25$ N pull. The $0.50$ W of mechanical work done by the puller becomes $0.50$ W of heat in the resistor.

Is that answer sensible?

In symbols, $F v = B (\frac{B l v}{R}) l v = \frac{( B l v ) ^{2}}{R} = \frac{ε ^{2}}{R}$ , the electrical power. The equality holds for any values, which is what Lenz's law guarantees.

CheckpointAnswer before reading on.

A straight metal rod $0.50$ m long moves at $3.0$ m s $^{- 1}$ through a uniform $0.20$ T magnetic field, with the rod, its velocity and the field all mutually perpendicular.

Calculate the emf between the ends of the rod.

Your answerV

Give it to 2 significant figures.

Hint 1In a time $Δ t$ the rod sweeps out an area $l v Δ t$ .

Hint 2So the flux it cuts per second is $B l v$ .

Hint 3 $ε = B l v = 0.20 \times 0.50 \times 3.0$

Exam question

Harder · about 5 min

3 marks

A metal rod slides along two horizontal conducting rails joined by a resistor, in a vertical magnetic field. A student must keep pushing to keep the rod moving at a constant speed, even with no friction.

Explain why, with reference to energy transfers and Lenz's law.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1The moving rod changes the flux through the circuit. What flows as a result?

Hint 2A current in a magnetic field feels a force. Which way must that force point?

Hint 3Where does the energy the student supplies end up?

Where students lose marks on this one

Saying the rod needs a push to overcome its own inertia.
Why it happens: Carrying over the idea that motion needs a force.
A rod at constant speed with no net force keeps moving. The push is needed only because the induced current creates a backwards force.

Written for this site.

The graph below shows a coil driven at a steady speed into, through and out of a strip of field. There is an emf only at the edges, while the flux is changing, and its sign flips between entering and leaving.

An emf appears only while the flux is changing: as the coil enters and as it leaves. Inside the field the flux is steady, so the emf is zero. Leaving gives the opposite sign to entering, as Lenz’s law requires.

Speed0.50 m s⁻¹

Coil side length10 cm

Field strength0.40 T

Turns20

Size of the emf while an edge crosses, NBlv: 0.400 V
Flux through one turn when fully inside: 4.00 × 10⁻³ Wb
Time to enter: 0.200 s
Fully inside the field: for 0.400 s, emf zero

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Two solenoids

Switch on a current in one coil and the field it makes passes through a second coil nearby. While the current rises, the flux through the second coil rises, and an emf is induced in it. Once the current is steady, nothing is changing and the emf is zero. Switching off gives a pulse the other way. Moving one solenoid towards or away from the other has the same effect as changing its current.

To get an emf that never stops, keep the current changing: supply the first coil with alternating current. That is a transformer.

CheckpointAnswer before reading on.

Coil A is connected to a battery and a switch. Coil B, placed end to end with coil A, is connected only to a sensitive galvanometer.

The switch is closed and left closed. What does the galvanometer show?

AA brief deflection when the switch closes, then zero.
BA steady deflection for as long as the switch is closed.
CNo deflection at any time, because the coils are not connected.
DA deflection that grows steadily while the switch is closed.

Hint 1An emf is induced in B only while the flux through B is changing.

Hint 2When does the current in A, and so its field, change?

Hint 3Once the current in A is steady, is the flux through B changing?

Ideal transformers

A step-down transformer

1AC supply
The supply must be alternating. A steady DC current would give a steady flux, and a steady flux induces nothing in the secondary, whatever the turns ratio.
2Primary coil
The alternating current in the primary makes an alternating field. The changing flux induces a back emf in each primary turn, and in an ideal transformer the total back emf matches the supply voltage $V_{p}$ .
3Flux in the core
The core guides almost all the field lines around the loop, so each turn of the secondary carries the same changing flux as each turn of the primary. That shared $ΔΦ/Δ t$ is what makes the voltage ratio equal the turns ratio.
4Laminated iron core
Soft iron concentrates the flux far more than air would. The core is built from thin insulated sheets, drawn here as lines, to block eddy currents, which would otherwise circulate in the iron and heat it.
5Secondary coil
This coil has fewer turns than the primary, so its emf is smaller: a step-down transformer. With more turns than the primary it would be a step-up transformer.
6Load
The secondary current flows through the load. Power drawn here must come from the supply, so a larger load current makes the primary draw more current too.

An alternating current in the primary coil sets up an alternating flux in the iron core. The core carries that flux through the secondary coil, where it induces an alternating emf.

Each turn of either coil has the same emf, $ΔΦ/Δ t$ , because the core carries the same changing flux through both. The voltage across a coil is that emf times its number of turns, which gives the turns ratio:

Ideal transformer turns ratioon the NESA formulae sheet

\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}

Symbols: $V_{p}$ voltage across the primary coilV
$V_{s}$ voltage across the secondary coilV
$N_{p}$ turns on the primary coil1
$N_{s}$ turns on the secondary coil1
Valid when: An ideal transformer on alternating current, where every field line from the primary passes through the secondary, so each turn of either coil has the same emf.
Not valid when: The supply is steady DC, which gives no changing flux and no secondary voltage, or flux linkage is poor, which makes the secondary voltage lower than the ratio predicts.
Rearranged: for V_{s}: $V_{s} = V_{p} \frac{N _{s}}{N _{p}}$ for N_{s}: $N_{s} = N_{p} \frac{V _{s}}{V _{p}}$
Where it turns up: Designing a step-down transformer for a phone charger
Finding the turns ratio needed to step up generator voltage for transmission
Where marks go missing: Inverting the ratio, so a step-up transformer is given fewer secondary turns
Applying it to a DC supply

Where this comes from

An ideal transformer loses no energy, so the power in equals the power out. Stepping the voltage up steps the current down by the same factor:

Ideal transformer power balanceon the NESA formulae sheet

V_{p} I_{p} = V_{s} I_{s}

Symbols: $V_{p}, I_{p}$ primary voltage and currentV, A
$V_{s}, I_{s}$ secondary voltage and currentV, A
Valid when: An ideal transformer, which loses no energy, so the power delivered to the primary equals the power taken from the secondary.
Not valid when: A real transformer with losses, where the output power is less than the input and the efficiency V_sI_s / V_pI_p is below 100 per cent.
Rearranged: for I_{s}: $I_{s} = \frac{V _{p} I _{p}}{V _{s}}$ for I_{p}: $I_{p} = I_{s} \frac{N _{s}}{N _{p}}$
Where it turns up: Finding the current drawn from the mains by a step-down transformer
Showing that stepping up the voltage steps down the current in a transmission line
Calculating the efficiency of a real transformer
Where marks go missing: Assuming current scales with turns the same way voltage does
Believing a step-up transformer gives out more energy than it takes in

Where this comes from

Worked example3 marks

A transformer for a halogen lamp

A $12$ V, $50$ W halogen lamp runs from $240$ V mains through an ideal transformer with $1200$ turns on its primary. Find the secondary turns, the lamp current and the current drawn from the mains.

Secondary turns

The voltage ratio equals the turns ratio.

$N_{s} = N_{p} \frac{V _{s}}{V _{p}} = 1200 \times \frac{12}{240} = 60$

Lamp current

The lamp's rating gives its power and voltage.

$I_{s} = \frac{P}{V _{s}} = \frac{50}{12} = 4.2 A$

Primary current

An ideal transformer passes on all the power, so the mains supplies 50 W at 240 V.

$I_{p} = \frac{V _{s} I _{s}}{V _{p}} = \frac{50}{240} = 0.21 A$

Answer

The secondary has $60$ turns. The lamp draws $4.2$ A, and the primary draws $0.21$ A from the mains.

Is that answer sensible?

The voltage falls by a factor of $20$ and the current rises by the same factor: $0.21 \times 20 \approx 4.2$ . If both went down, energy would be disappearing.

CheckpointAnswer before reading on.

A transformer in a phone charger steps $240$ V mains down to $12$ V. Its primary coil has $2000$ turns.

Assuming it is ideal, calculate the number of turns on the secondary coil.

Your answerturns

Give it to 3 significant figures.

Hint 1Use $\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}}$ .

Hint 2A step-down transformer needs fewer turns on the secondary.

Hint 3 $N_{s} = 2000 \times \frac{12}{240}$

CheckpointAnswer before reading on.

The ideal $240$ V to $12$ V transformer supplies a current of $3.0$ A to a device.

Calculate the current drawn from the mains by the primary coil.

Your answerA

Give it to 2 significant figures.

Hint 1An ideal transformer loses no energy.

Hint 2So $V_{p} I_{p} = V_{s} I_{s}$ .

Hint 3 $I_{p} = \frac{12 \times 3.0}{240}$

CheckpointAnswer before reading on.

An ideal transformer has ten times as many turns on its secondary as on its primary.

Which statement is correct?

AThe secondary voltage is ten times larger and the secondary current ten times smaller, so the power is unchanged.
BThe secondary voltage and power are both ten times larger.
CThe secondary voltage and current are both ten times larger.
DThe secondary voltage is ten times smaller, because the energy is shared among more turns.

Hint 1What does the turns ratio do to the voltage?

Hint 2Can a transformer create energy?

Hint 3If the voltage goes up by $10$ and power is unchanged, what happens to the current?

Real transformers

The ideal model makes two assumptions: every field line from the primary passes through the secondary, and no energy becomes heat. Neither is quite true.

Incomplete flux linkage. Some flux leaks out of the core into the air and misses the secondary, so the secondary voltage is a little below the turns ratio prediction. A closed core of soft iron keeps the flux inside, and winding the two coils on top of each other on the same limb keeps them close.
Resistive heating. The windings are long thin wires with resistance, and the currents in them dissipate $I^{2} R$ . Thicker copper, especially for the high current coil, reduces this. Large transformers are cooled with oil or fins.
Eddy currents. The iron core is a conductor in a changing flux, so currents are induced in loops inside it, heating it. Laminating the core in thin sheets insulated from each other cuts these loops short.
Hysteresis. Magnetising the core first one way and then the other on every cycle takes energy. Soft iron, which magnetises and demagnetises easily, keeps this loss small.

With all of these, large transformers are about $99$ per cent efficient, so the ideal equations are an excellent approximation for them. Small, cheap ones do noticeably worse.

Worked example3 marks

How ideal is the lamp transformer?

When the halogen lamp transformer above is measured, the mains current is $0.23$ A rather than $0.21$ A, and the lamp still receives $50$ W. Find the efficiency, and explain where the extra energy goes.

Input power

The input is what the mains actually supplies.

$P_{in} = V_{p} I_{p} = 240 \times 0.23 = 55 W$

Efficiency

Useful output over total input.

$efficiency = \frac{50}{55} \times 100 = 91 %$

Account for the loss

Evaluate questions want each loss tied to a cause and a fix.

About $5$ W becomes heat: $I^{2} R$ heating in the windings, most of it in the thick $4.2$ A secondary; eddy currents and hysteresis in the core; and a little flux leakage, which means the primary must draw more current to deliver the same output.

Answer

The transformer is about $91$ per cent efficient. The missing $5$ W heats the windings and the core.

Is that answer sensible?

A small transformer that runs warm to the touch is consistent with a few watts of loss, which is what we found. A large substation transformer handling megawatts at $99$ per cent still loses tens of kilowatts, which is why those need oil cooling.

CheckpointAnswer before reading on.

A real transformer draws $0.50$ A from the $240$ V mains. Its secondary delivers $9.0$ A at $11.5$ V.

Calculate its efficiency as a percentage.

Your answer%

Give it to 2 significant figures.

Hint 1Find the input power and the output power.

Hint 2Efficiency is output power divided by input power.

Hint 3 $\frac{11.5 \times 9.0}{240 \times 0.50} \times 100$

CheckpointAnswer before reading on.

Why is the iron core of a transformer made from thin insulated sheets rather than a solid block?

ATo reduce eddy currents induced in the core, and so the heat they produce.
BTo stop the flux passing through the core.
CTo increase the resistance of the copper coils.
DTo make the core a permanent magnet.

Hint 1The core is itself a conductor sitting in a changing flux.

Hint 2What is induced in a solid conductor in a changing flux?

Hint 3How does cutting the conductor into thin insulated layers affect those currents?

Exam question

Harder · about 9 min

5 marks

Evaluate the ideal transformer model, referring to the causes of energy loss in real transformers and the strategies used to reduce them.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1The ideal model assumes all the flux links both coils and no energy becomes heat. Which of those assumptions fail?

Hint 2For each loss, name the strategy that reduces it and explain the physics.

Hint 3Evaluate means make a judgement: how good is the model in practice?

Where students lose marks on this one

Describing the losses with no judgement.
Why it happens: Treating evaluate as describe.
Finish with a clear statement of how good the model is, backed by the efficiency figures.

Written for this site.

Transformers and the grid

Power stations are often hundreds of kilometres from the people using the electricity, and every kilometre of line has resistance. For a fixed power, the current needed is $I = P / V$ , and the power lost as heat in the line is

$P_{l oss} = I^{2} R = (\frac{P}{V})^{2} R$

Raising the voltage ten times cuts the current ten times and the loss a hundred times. Move the voltage slider and watch how steeply the loss falls.

Loss as a percentage of power sent

Both axes are logarithmic. Ten times the voltage means a tenth of the current and one hundredth of the loss. Where the line would lose more than it carries, the simple model has broken down and no value is shown.

Power sent50 MW

Line resistance8.0 Ω

Transmission voltage33 kV

Current in the line, I = P / V: 1515 A
Power lost, I²R: 1.84 × 10⁷ W
Fraction of the power lost: 36.7 %

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Worked example4 marks

Why the grid steps up

A generator produces $20$ MW at $11$ kV. It is to be sent along a line of total resistance $5.0 Ω$ . Compare the power lost if it is sent at $11$ kV with the loss if a transformer first steps it up to $220$ kV.

Current at the generator voltage

For a fixed power, the current is set by the voltage.

$I = \frac{P}{V} = \frac{20 \times 1 0 ^{6}}{11 \times 1 0 ^{3}} = 1.8 \times 1 0^{3} A$

Loss at 11 kV

The heat produced in the line depends on the current through it.

$P_{l oss} = I^{2} R = (1.8 \times 1 0^{3})^{2} \times 5.0 = 1.7 \times 1 0^{7} W$

Step up to 220 kV

A 1 : 20 transformer multiplies the voltage by 20 and divides the current by 20.

$I = \frac{20 \times 1 0 ^{6}}{220 \times 1 0 ^{3}} = 91 A, P_{l oss} = 9 1^{2} \times 5.0 = 4.1 \times 1 0^{4} W$

Answer

At $11$ kV the line would waste about $17$ MW of the $20$ MW, which is useless. At $220$ kV it wastes about $41$ kW, $0.2$ per cent of the power. Step-down transformers near the customers then bring the voltage back down to $230$ V.

Is that answer sensible?

The voltage rose by $20$ , so the loss should fall by $2 0^{2} = 400$ . And $\frac{1.7 \times 1 0 ^{7}}{4.1 \times 1 0 ^{4}} \approx 400$ . It matches.

CheckpointAnswer before reading on.

A power station sends $50$ MW along a transmission line of total resistance $4.0 Ω$ at a voltage of $250$ kV.

Calculate the power lost as heat in the line.

Your answerW

Give it to 2 significant figures.

Hint 1First find the current in the line from $P = V I$ .

Hint 2The power lost in the line is $I^{2} R$ , not $V^{2} / R$ with the transmission voltage.

Hint 3 $I = \frac{50 \times 1 0 ^{6}}{250 \times 1 0 ^{3}} = 200$ A

CheckpointAnswer before reading on.

The same power is sent along the same line, but the transmission voltage is raised from $33$ kV to $330$ kV.

By what factor does the power lost in the line change?

AIt becomes $\frac{1}{100}$ as large.
BIt becomes $\frac{1}{10}$ as large.
CIt becomes $100$ times larger.
DIt does not change, because the same power is sent.

Hint 1For fixed power, $I = P / V$ .

Hint 2Ten times the voltage means what current?

Hint 3The loss depends on $I^{2}$ .

Exam question

Harder · about 7 min

4 marks

Analyse the role of step-up and step-down transformers in distributing electrical energy from a power station to homes.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1What would happen to the losses if energy were sent at the generator's voltage?

Hint 2How does a step-up transformer change the current in the line, and why does that matter?

Hint 3Why can't homes use the transmission voltage directly?

Where students lose marks on this one

Saying high voltage is used because it carries more energy.
Why it happens: Associating high voltage with power.
The power sent is the same. High voltage matters because it allows a low current.

Written for this site.

Common mistake

Finding the line loss with V²/R, using the transmission voltage.

Why it happens: V²/R is a correct formula for power, so it looks like a shortcut.

$V^{2} / R$ needs the voltage across the resistance. The $220$ kV is the voltage the load receives between the conductors; the voltage dropped along the line itself is only $I R = 91 \times 5.0 \approx 460$ V. Using $220$ kV in $V^{2} / R$ predicts that a higher voltage makes the loss worse, which is backwards. Find the current from $I = P / V$ and use $I^{2} R$ .

Analysetypically 4 to 7 marks

Demands: Break the situation into its parts and show how they relate and affect each other.
Shape: Identify the components, then the relationships, then what follows from them.
Loses marks: Summarising the stimulus rather than taking it apart.

Through a marker’s eyes

3 marks

A magnet is dropped through a vertical copper tube and takes much longer to fall than it would in air. Explain this observation with reference to energy. (3 marks)

The attempt

1 out of 3

The magnet induces a current in the copper. Lenz's law says the current opposes the magnet, so it falls slowly.

What the marker sees

The answer names Lenz's law correctly but gives no mechanism: it does not say why a current is induced or how the current slows the magnet. The question says with reference to energy, and energy is not mentioned at all.

The same answer, fixed

3 out of 3

As the magnet falls, the magnetic flux through the rings of the tube near it changes, so by Faraday's law emfs are induced and eddy currents flow around the tube.
By Lenz's law these currents produce magnetic fields that oppose the change causing them, so they exert an upward force on the magnet, reducing its acceleration.
Gravitational potential energy is therefore transformed mostly into electrical energy in the tube, which becomes heat through its resistance, rather than into kinetic energy of the magnet.

Where the emf comes from

Two different things can change the flux, and physically they work differently. When a rod moves through a field, the charges in it move with it, and the magnetic force $q v B$ pushes them along the rod. That is the motional emf $B l v$ , and it is the motor effect force acting on free charges.

When a coil sits still in a changing field, its charges are not moving, so there is no $q v B$ force. Instead, a changing magnetic field creates an electric field that circles around it, and that electric field drives the charges. Faraday's law covers both cases with one formula, which is one of the hints that led Einstein to special relativity: whether a charge feels an electric or a magnetic force depends on the frame in which you describe it.

The minus sign is a statement about direction, not a number to carry through a calculation. Most HSC questions ask for the size of the emf, then separately for its direction or the direction of the current, which Lenz's law supplies. Writing $∣ ε ∣$ in the calculation and a sentence about direction afterwards keeps the two separate.

Where this reappears

The next topic uses induction in generators, where a coil turning at constant speed has flux $B A cos ω t$ and so an emf that varies sinusoidally, and in motors, where the spinning coil induces a back emf that limits the current. Magnetic braking in trains and roller coasters is the sliding rod simulation made large: no contact, no wear, and a braking force that fades as the speed falls.

The short version

Magnetic flux

Counting field lines through a loop

Faraday's law

A magnet pushed towards a coil

Lenz's law and energy

Why the induced current always pushes back

A rod on rails

Keeping a rod moving

Two solenoids

Ideal transformers

A transformer for a halogen lamp

Real transformers

How ideal is the lamp transformer?

Transformers and the grid

Why the grid steps up

Where the emf comes from

The sign in Faraday's law

Where this reappears