Electromagnetic induction: flux, laws and transformers

Key idea: Define the linked quantity before taking its rate of change, use Lenz's law as an energy-consistent direction rule, and keep simple applications and ideal-transformer ratios within their stated assumptions.

  • H2 Physics 9478 · 2027
  • Internally reviewed by MiniEducation Team
  • Recorded selected-response study loop available

Before you start: Electromagnetic Forces objective chainCurrent Electricity objective chain

By the end, you can

  • Define and calculate magnetic flux and flux linkage with correct orientation.
  • Infer induction behaviour from experiments and apply Faraday's and Lenz's laws.
  • Explain simple motional-e.m.f. and eddy-current applications using energy conservation.
  • Explain simple iron-core transformer operation and apply ideal voltage/current ratios.

Starting-point self-check

1. Check your starting point

Attempt all four groups without notes and mark the first area, linkage, rate, polarity, application or ideal-model decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.

Simple iron-core ideal transformers 18(g)

Question 1

An ideal transformer has Np = 800, Ns = 200 and Vp = 240 V. It supplies Is = 3.0 A. Find Vs and Ip.

Check the model response

Vs/Vp = Ns/Np, so Vs = 60 V. For an ideal transformer, Ip/Is = Ns/Np, so Ip = 0.75 A; input and output powers are both 180 W.

repair

2. Repair the common breaks

Use only the correction matching an error, then retry the corresponding diagnostic.

Simple iron-core ideal transformers 18(g)

Check this idea

Misconception: A step-up transformer creates power.

Repair: An ideal transformer conserves power and trades increased voltage for decreased current.

Check this idea

Misconception: A steady d.c. primary produces a continuous secondary e.m.f.

Repair: After the switching transient, steady d.c. gives constant flux and no induced secondary e.m.f.; transformer operation requires changing flux.

worked example

3. Follow four worked models

Follow how each solution fixes area orientation, linkage sign, energy pathway or ideal-transformer assumptions before calculating.

Simple iron-core ideal transformers 18(g)

Model 1

Explain how an iron-core transformer transfers energy without conducting current between windings.

Check the model response

Alternating primary current produces changing core flux. The iron core links this flux through both windings; Faraday's law induces alternating secondary e.m.f. Turns ratio sets voltage ratio. In the ideal model, power is conserved and current ratio is inverse to voltage ratio.

guided practice

4. Guided practice

Use each hint only to select the correct perpendicular area, linkage rate, motional geometry or turns ratio.

Simple iron-core ideal transformers 18(g)

Question 1

An ideal transformer steps voltage up by factor 5. State the current factor.

Hint: Ideal power is conserved.

Check the model response

Current steps down by factor 5 because VpIp = VsIs and Ip/Is = Ns/Np.

independent practice

5. Independent practice

Solve without repair notes and state sign conventions, field geometry, circuit closure and ideal-transformer assumptions.

Simple iron-core ideal transformers 18(g)

Question 1

State the ideal-transformer assumptions behind the voltage and current ratios and explain what a step-up device does.

Check the model response

All changing core flux links both windings, winding resistance and core losses are negligible, and input power equals output power. Ns/Np = Vs/Vp = Ip/Is. A step-up transformer raises voltage and lowers current by the same ratio; it does not create power.

Practice exit check

6. Practice assessment

Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.

Simple iron-core ideal transformers 18(g)

Question 1

An ideal 240 V transformer with 1500 primary and 75 secondary turns supplies 4.0 A. Find secondary voltage, primary current and output power.

Check the model response

Vs = 240(75/1500) = 12 V. Ip = Is(Ns/Np) = 0.20 A. Output power is 12(4.0) = 48 W, equal to input power.

Re-test practice

7. Delayed re-test practice

Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.

Simple iron-core ideal transformers 18(g)

Question 1

An ideal transformer has Ns/Np = 0.10. State Vs/Vp and Ip/Is.

Check the model response

Vs/Vp = 0.10 and Ip/Is = 0.10; secondary current is ten times primary current.

Continue with established practice

Use the established six-question structured set after the delayed re-test. It samples linkage, laws, applications and transformers; the standalone flux definition remains assessed in this chain, lessons and quiz.

Open Electromagnetic Induction structured practice