Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 14 of 21 IQ3: Fields Related

Faraday's Law of Induction

In 1831, Michael Faraday used two coils on an iron ring at the Royal Institution. Switching the current in one coil produced a brief galvanometer deflection in the other, whereas a steady current produced no sustained deflection. The experiments established that changing magnetic flux linkage induces an emf; Faraday's law expresses the quantitative rate relationship used in generators, transformers and induction systems.

Today's hook: In Faraday's ring experiment, a steady primary current produced no sustained secondary deflection, but switching the current produced a brief pulse. What changed during the switch, and why would a faster change produce a larger pulse?

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Retrieve magnetic flux, predict the observation and define the L12-L14 lesson boundary.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

A coil is connected to a sensitive galvanometer. A bar magnet begins outside the coil.

Predict the reading while the magnet remains stationary.
Predict how the deflection changes when the magnet is pushed in slowly, then quickly.
Predict what reverses when the magnet is pulled back out.

Warm-up, which of the following is required for electromagnetic induction to occur?

Learning Intentions

goals

Know, Faraday's Law

The induced emf is proportional to the rate of change of magnetic flux
$\varepsilon = -N\,\dfrac{\Delta\Phi}{\Delta t}$ for a coil of N turns
The negative sign indicates the induced emf opposes the change (Lenz's Law)

Understand, Rate of Change Matters

Fast flux change gives large emf; slow flux change gives small emf
Constant flux gives zero emf regardless of its magnitude
More turns multiply flux linkage and increase induced emf

Can Do, Calculate and Predict

Calculate induced emf given N, $\Delta\Phi$, and $\Delta t$
Use a declared reference to interpret the sign of induced emf
Recognise applications without replacing the later device lessons

Scan these before reading

vocab

Faraday's LawThe induced emf equals the negative rate of change of magnetic flux linkage: $\varepsilon = -N\,d\Phi/dt$.

Rate of change of fluxHow quickly magnetic flux changes: $\Delta\Phi/\Delta t$. Units: Wb/s = V.

Induced emfA voltage created by changing magnetic flux through a conductor. No flux change means no emf.

Number of turns (N)More turns means more total flux linkage ($N\Phi$) and a proportionally larger induced emf.

Misconceptions to fix

✗ Wrong: A large, constant magnetic flux through a coil produces a large emf.

✓ Right: Constant flux produces zero emf. Only a changing flux induces an emf. The rate of change is what matters, not the magnitude of the flux itself.

✗ Wrong: The negative sign in Faraday's Law just means the emf is small.

✓ Right: The negative sign encodes Lenz's Law, the induced emf drives a current that opposes the change in flux that caused it. It is a direction indicator, not a size indicator.

Cross-lesson boundary: L12 defined signed single-turn flux and flux linkage. L13 calculates induced emf from the rate of change of linkage. L14 determines induced-current direction and develops the conservation argument. L19 owns detailed eddy-current applications.

Core Content

Connect changing flux linkage to average and instantaneous induced emf.

Faraday's Law of Induction

+5 XP

The faster the flux changes, the bigger the induced emf

Hold a bar magnet above a coil connected to a galvanometer and the needle reads zero. Move the magnet into the coil and the needle deflects; stop it and the reading returns to zero. Faster motion gives a larger deflection because it produces a larger rate of change of flux linkage.

Faraday's Law of Induction

$$\varepsilon_{\mathrm{avg}} = -\frac{\Delta(N\Phi)}{\Delta t}$$

$\varepsilon$, induced emf (V)
$N$, number of turns in the coil
$N\Phi$, magnetic flux linkage (Wb-turn)
$\Delta t$, time interval over which flux changes (s)
The negative sign indicates the direction opposes the change (Lenz's Law)

Key insight: A small flux changing quickly can induce a large emf. A very large flux that is constant induces no emf at all. This is why alternating currents are essential in transformers, they ensure flux is always changing.

For a single loop ($N = 1$), $\varepsilon_{\mathrm{avg}}=-\Delta\Phi/\Delta t$. For $N$ identical turns that share the same flux, $\Delta(N\Phi)=N\Delta\Phi$. At an instant, $\varepsilon=-d(N\Phi)/dt$. The graph meaning is exact: induced emf is the negative gradient of a linkage-time graph.

Stop and check

A coil of 100 turns experiences a flux change from 0.020 Wb to 0.050 Wb in 0.10 s. Calculate the average induced emf. What would the emf be if the same flux change occurred in 0.010 s?

Faraday's law: $\varepsilon_{\mathrm{avg}}=-\Delta(N\Phi)/\Delta t$ and $\varepsilon=-d(N\Phi)/dt$. Constant linkage has zero gradient and zero emf. Units: Wb-turn/s = V because a turn is dimensionless.

Pause, copy the highlighted Faraday's Law formula and key proportionalities into your book before moving on.

A 50-turn coil has flux changing from 0.10 Wb to 0.30 Wb in 0.40 s. The average induced emf is:

Recognise how devices change linkage while keeping detailed applications in their owner lessons.

Applications of Faraday's Law

+5 XP

Where this law appears in everyday technology

We just saw that $\varepsilon = -N\Delta\Phi/\Delta t$ quantifies induced emf. That raises a question: which real devices actually run on this principle? This card answers it → generators, transformers, induction cooktops, wireless chargers, and magnetic brakes all rely on changing magnetic flux.

Faraday's law links these devices at one level only: each arrangement changes flux linkage and therefore produces an emf. Their circuit design, energy losses and performance limits belong to later lessons.

Generators: Rotation changes coil linkage and produces an alternating emf; current flows when a closed load is connected.
Transformers: An alternating current in the primary coil creates a changing magnetic field. This changing field passes through the secondary coil, inducing an emf proportional to the turns ratio.
Induction cooktops: A rapidly alternating current in a coil beneath the cooktop creates a rapidly changing magnetic field, inducing eddy currents in the metal pot that heat it directly.
Wireless charging: A transmitter coil creates a changing magnetic field that induces current in a receiver coil in the device.
Magnetic braking: A moving magnet induces eddy currents in a conductor, creating an opposing magnetic field that slows the magnet, used in train brakes and roller coaster systems.

Key insight

All of these applications share the same requirement: magnetic flux linkage must be changing. Increasing its rate of change increases induced emf under the model, but it does not by itself guarantee greater efficiency; resistance, magnetic losses, geometry and load also matter.

Application bridge only: generators, transformers, induction heating and magnetic braking all require changing flux. L13 owns the induced-emf rate calculation; L14 owns direction and conservation; L19 owns detailed eddy-current heating and braking behaviour.

Add the highlighted application list to your notes before the check below.

A transformer works with a steady (DC) current in the primary coil.

Magnetic braking relies on induced eddy currents opposing the motion that causes them.

Induction cooktops heat the pot by directly applying current to it via electrical contacts.

Use the optional induction model to vary turns, linkage change and time.

Interactive Tool, Electromagnetic Induction Open fullscreen ↗

Use the interactive tool. When you increase N (number of turns) while keeping $\Delta\Phi$ and $\Delta t$ constant, the induced emf:

Calculate signed average emf and interpret linkage-time gradients.

Worked Example, Applying Faraday's Law

+5 XP

Calculate induced emf in a multi-part scenario

We just saw where Faraday's Law appears in technology. That raises a question: how do we apply $\varepsilon = -N\Delta\Phi/\Delta t$ step-by-step in a calculation with multiple parts? This card answers it → compute $\Phi_i$, then $\Delta\Phi$, then divide by $\Delta t$ and multiply by $N$.

A worked example is the best way to practise the algorithm: identify knowns, calculate $\Delta\Phi$, apply $\varepsilon = -N\,\Delta\Phi/\Delta t$, and interpret the magnitude and direction.

Problem, Three-part Faraday calculation

A coil of 200 turns and area $5.0 \times 10^{-3}$ m² is perpendicular to a uniform magnetic field of 0.40 T.

(a) Calculate the initial magnetic flux through the coil.
(b) The field is reduced to zero in 0.20 s. Calculate the average induced emf.
(c) If the same change occurred in 0.020 s, what would the average emf be?

Step 1, Part (a): Initial flux

Using $\Phi = BA$ with the coil perpendicular to the field ($\theta = 0°$, so $\cos\theta = 1$):

$$\Phi=BA$$

$\Phi=(0.40)(5.0\times10^{-3})=2.0\times10^{-3}$ Wb.

Step 2, Part (b): Average emf over 0.20 s

The flux changes from $2.0\times10^{-3}$ Wb to 0.
Therefore $\Delta\Phi=-2.0\times10^{-3}$ Wb.

$$\varepsilon=-N\frac{\Delta\Phi}{\Delta t}$$

$\varepsilon=-(200)(-2.0\times10^{-3})/0.20=2.0$ V.

Taking the initial field direction as positive flux and the corresponding loop direction as positive emf, the positive result is consistent with the induced effect opposing the decrease in positive flux. L14 develops the physical current direction.

Step 3, Part (c): Average emf over 0.020 s

$\varepsilon=-(200)(-2.0\times10^{-3})/0.020=20$ V.

Ten times faster flux change produces ten times larger emf. The magnitude of $\varepsilon$ is inversely proportional to $\Delta t$.

Step order: (1) $\Phi_i=BA\cos\theta$; (2) $\Delta\Phi=\Phi_f-\Phi_i$; (3) $\varepsilon=-N\Delta\Phi/\Delta t$.
Here $\Phi_i=2.0\times10^{-3}$ Wb; $\varepsilon=2.0$ V over 0.20 s and 20 V over 0.020 s.

Pause, write the highlighted step order and worked results into your book before moving on.

Quick drill, a coil of 80 turns has flux changing from 0.040 Wb to 0.080 Wb in 0.20 s. The average induced emf (in V, as a positive number) is _____.

Essential formula, Faraday's Law

Faraday's Law: $\varepsilon = -N\,\dfrac{\Delta\Phi}{\Delta t}$

Flux: $\Phi = BA\cos\theta$ (Wb)

Units check: Wb/s = V

Key rule: Constant flux → zero emf. Changing flux → non-zero emf. Faster change → larger emf.

Three of these statements about Faraday's Law are correct. Pick the odd one out.

Average, instantaneous and circuit response

Over a finite interval, $\varepsilon_{\text{avg}}=-\Delta(N\Phi)/\Delta t$ is the negative secant gradient of a linkage–time graph. At an instant, $\varepsilon=-d(N\Phi)/dt$ is the negative tangent gradient. First choose a positive area normal and circuit direction; the minus sign then encodes Lenz's law. Coil resistance changes induced current $I=\varepsilon/R$, not the induced emf produced by a specified flux-linkage change.

Apply the law to calculations and a linkage-time graph, then consolidate model limits.

Activity 1, Faraday's Law Calculations

ApplyBand 4

Practise calculating induced emf in different scenarios

A 100-turn coil experiences a flux change from 0.020 Wb to 0.050 Wb in 0.10 s. Calculate the average induced emf. What would the emf be if the same flux change occurred in 0.010 s?
Set N = 50, $\Delta\Phi$ = 20 mWb, $\Delta t$ = 0.10 s in the interactive tool. Calculate the emf by hand and verify with the tool. Then halve the time to 0.05 s, what happens to the emf?
A student claims that if flux is constant at 0.50 Wb, the induced emf is large because the flux is large. Is this correct? Explain the error.

The negative sign in Faraday's Law $\varepsilon = -N\,\Delta\Phi/\Delta t$ is associated with:

Activity 2, Flux-Linkage Graph

UnderstandBand 5

Turn graph gradients into signed induced emf

A coil's flux linkage $N\Phi$ changes from 0 to $+0.060$ Wb-turn uniformly during the first 0.020 s, remains constant for 0.030 s, then changes uniformly to $-0.020$ Wb-turn in 0.010 s. Using the declared positive linkage direction:

Sketch $N\Phi$ against time and label each interval.
Calculate the signed average emf in each interval from the negative gradient.
Explain why a large constant linkage produces zero emf.

Wrap-up, Misconceptions and Summary

Misconceptions, final check

✗ "Increasing N increases the induced emf, so N is the most important factor."

✓ All three factors, N, $\Delta\Phi$, and $\Delta t$, multiply together in the formula. A coil with N = 1 but an extremely fast flux change can produce a larger emf than N = 1000 with very slow change.

✗ "An induced emf always means an induced current is flowing."

✓ A changing flux linkage produces an emf. Current flows only when there is a closed conducting path, and its magnitude also depends on the circuit resistance or impedance.

Copy into your books

Key Law

$\varepsilon = -N\,\Delta\Phi/\Delta t$
Negative sign = Lenz's Law direction
Units: Wb/s = V

What Increases emf?

More turns (larger N)
Larger flux change ($\Delta\Phi$)
Shorter time ($\Delta t$)

Applications

Generators (rotating coil)
Transformers (AC in primary)
Induction cooktops, wireless charging

Key Condition

Constant flux → zero emf
Changing flux → non-zero emf
Rate of change matters, not amount

Calculate average induced emf and interpret a signed linkage-time graph independently.

Quick recall, Faraday's Law and induction

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 7 marks

+5 XP

ApplyBand 4(3 marks) 1. A coil of 80 turns experiences a flux change from 0.040 Wb to 0.080 Wb in 0.20 s. (a) Calculate the average induced emf. (b) The same flux change now occurs in 0.050 s, calculate the new average emf. (c) Explain why the emf is larger in the second case.

1 mark: correct emf for (a) · 1 mark: correct emf for (b) · 1 mark: explanation links to faster rate of change

AnalyseBand 5(4 marks) 2. A flux-linkage graph rises linearly from 0 to $+0.060$ Wb-turn in 0.020 s, is constant for 0.030 s, then falls linearly to $-0.020$ Wb-turn in 0.010 s. Calculate the signed average induced emf in each interval and explain the zero-emf interval.

1 mark: $-3.0$ V in interval 1 · 1 mark: 0 V in interval 2 · 1 mark: $+8.0$ V in interval 3 · 1 mark: links emf to negative linkage–time gradient

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (3 marks): (a) $\Delta\Phi = 0.080 - 0.040 = 0.040$ Wb. $\varepsilon = N\,\Delta\Phi/\Delta t = 80 \times 0.040/0.20 = 16$ V (1 mark). (b) $\varepsilon = 80 \times 0.040/0.050 = 64$ V (1 mark). (c) The emf is larger because the same flux change occurs in a shorter time, giving a greater rate of change of flux $\Delta\Phi/\Delta t$, and since $\varepsilon \propto 1/\Delta t$, the emf is proportionally larger (1 mark).

Q2 (4 marks): Interval 1 has gradient $0.060/0.020=+3.0$ Wb-turn s$^{-1}$, so $\varepsilon_{\text{avg}}=-3.0$ V (1 mark). Interval 2 has zero gradient, so $\varepsilon=0$ V even though linkage is non-zero (1 mark). Interval 3 has $\Delta(N\Phi)=-0.080$ Wb-turn in 0.010 s, so $\varepsilon_{\text{avg}}=-(-0.080/0.010)=+8.0$ V (1 mark). The signs follow $\varepsilon=-\Delta(N\Phi)/\Delta t$ for the declared positive reference (1 mark).

Retrieve Faraday's law and revisit the observation without crossing into L19 detail.

Check what actually stuck

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quiz

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How did your thinking change?

At the start you compared a stationary magnet with slow, fast and reversed motion relative to a coil.

A stationary magnet gives constant flux linkage, so its linkage-time gradient and induced emf are zero. Moving the magnet changes linkage; faster motion gives a steeper gradient and a larger emf magnitude. Reversing the motion reverses the sign of the linkage change and therefore the sign of the emf for the same declared reference.

This lesson determines the emf from the linkage-time gradient. If the circuit is closed, its resistance and impedance also affect the induced current. L14 determines the physical current direction and L19 develops eddy-current braking and heating.

I have completed this lesson

← Previous: Magnetic Flux and Changing Flux Next: Lenz's Law and Direction →

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