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Induction Stove

Key idea: Explain how an induction stove heats a pan using changing magnetic flux and eddy currents, and why non-conducting cookware does not heat (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use magnetic flux and flux-linkage relationships.
  • Apply Faraday's and Lenz's laws to induced e.m.f. and direction.
  • Explain simple applications of electromagnetic induction, including motional e.m.f. and eddy currents.
  • Explain simple iron-core transformer operation and apply ideal transformer ratios.

1. Definitions (Must Know)

A. Induction heating (idea)

Induction stoves heat a metal pan by inducing eddy currents in the pan, which then dissipate energy as heat.

Induction stove heating a conducting panAn alternating current flows in a coil below a non-conducting cooktop. Its changing magnetic field links the metal pan and induces closed eddy-current loops in the pan base. Electrical resistance in the pan converts energy to heat there.high-frequency a.c. coilconducting paneddy currents → I²R heatingnon-conducting surfacechangingmagnetic flux
The coil's a.c. creates changing flux; induced eddy currents produce I²R heating in the conducting pan, not directly in the cooktop.

2. Key Ideas (What Earns Marks)

  • An a.c. current in a coil produces a changing magnetic field.
  • Changing flux in the metal pan induces an e.m.f. and eddy currents.
  • Eddy currents cause heating via I²R losses in the pan.
Scope note

This is a qualitative application of Faraday’s and Lenz’s laws (18f). You do not need detailed cooker electronics.

3. Detailed Explanations

A. Step-by-step chain

  1. Coil carries a.c. → changing magnetic field.
  2. Field links the metal pan → flux through the pan changes.
  3. Changing flux → induced e.m.f. → eddy currents in the pan.
  4. Eddy currents dissipate energy as heat in the pan.

4. Common Mistakes

  • Saying the coil “heats the pan directly” (the pan heats mainly from eddy current losses in the pan itself).
  • Thinking any material heats equally well (a good conductor supports larger eddy currents).

5. Exam Tips

  • Use the phrase “changing flux induces eddy currents”.
  • Mention Lenz’s law only if direction/opposition is asked.

6. Worked Examples

Modelled example 1

Why does the surface stay relatively cool?

Core

Problem

Explain why the cooking surface can remain relatively cool compared with the pan.
Study the worked solution

  1. Locate the induced current

    Method

    Eddy currents are induced mainly in the conducting metal pan.

    Reason

    The changing magnetic flux links the pan, whose conducting paths allow circulating currents.

    Working

    changing flux ⇒ eddy currents in pan

  2. Locate the energy dissipation

    Method

    The main electrical heating occurs in the pan.

    Reason

    The eddy currents dissipate energy through the pan’s resistance.

    Working

    P = I_eddy^(,2)R

  3. Compare the surface

    Method

    The non-conducting cooking surface does not support comparable eddy currents, so it remains relatively cooler.

    Reason

    It is not the main site of induced-current heating; it can still warm later by thermal contact with the pan.

    Working

    I_(eddy,surface) ≈ 0

Guided practice 2

Why must the current be a.c.?

About 4 min

Problem

Explain why an induction cooker needs an alternating current in the coil rather than a steady d.c. current.

Try this before viewing the solution

Required field behavior

Hints

Hint 1: start from Faraday's law
Ask what must happen to magnetic flux for a sustained induced e.m.f.
Hint 2: compare long-term fields
After switching is complete, steady d.c. gives a steady field; a.c. continues changing direction and magnitude.
View solution step by step

  1. Follow steady d.c.

    Method

    After a brief switching transient, steady d.c. produces a steady magnetic field.

    Reason

    The coil current is then constant.

    Working

    I = constant ⇒ B = constant

  2. Apply Faraday's law

    Method

    A steady field gives no sustained induced e.m.f. in the pan.

    Reason

    The linked flux is not changing.

    Working

    dΦ/dt = 0 ⇒ E = 0

  3. Use a.c.

    Method

    Alternating current supplies the continuing flux change needed for eddy-current heating.

    Reason

    Its changing magnetic field produces a continuing induced e.m.f.

    Working

    I_(a.c.) ⇒ B(t) ⇒ Φ(t) ⇒ E

Common misconception 3

Pan removed

Find and correct the mistake

Learner claim

Many induction hobs stop heating when the pan is removed. A learner says the empty hob should keep heating because the coil still produces a changing magnetic field. Diagnose the claim.

Try this before viewing the solution

Missing condition

View solution step by step

  1. Retain the correct part

    Method

    The coil can still produce a changing magnetic field.

    Reason

    Removing the pan does not by itself make the alternating coil current steady.

    Working

    I_(a.c.) ⇒ B(t)

  2. Identify the missing path

    Method

    There is no metal pan providing effective closed conducting paths for large eddy currents.

    Reason

    The non-conducting surface does not replace the cookware as the intended induced-current load.

    Working

    no pan ⇒ I_(eddy,pan) = 0

  3. Correct the conclusion

    Method

    The pan’s I²R heating disappears, so the hob stops delivering its intended heating.

    Reason

    Energy dissipation depends on the induced current in the cookware, not merely on the existence of a field.

    Working

    Pₚₐₙ = I_eddy^(,2)R = 0

Examiner practice 4

Effect of frequency (qualitative)

3 marks

Examination question

The coil’s a.c. frequency is increased while its current amplitude and geometry stay the same. Explain the effect on the rate of flux change, induced e.m.f. and tendency to heat the pan. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Compare flux-change rates

    1 mark

    Method

    The magnitude of the rate of flux change increases.

    Reason

    The same flux variation is completed in a shorter cycle time.

    Working

    f↑ ⇒ |dΦ/dt|↑

  2. Apply Faraday's law

    1 mark

    Method

    The induced e.m.f. magnitude tends to increase.

    Reason

    It is proportional to the rate of change of linked flux.

    Working

    |E| ∝ |dΦ/dt|

  3. Infer heating

    1 mark

    Method

    Eddy-current heating tends to increase, all else equal.

    Reason

    The larger induced e.m.f. can drive larger eddy currents in the pan.

    Working

    |E|↑ ⇒ I_eddy↑ ⇒ P↑

Challenge 5

Which pans work best? (qualitative)

Minimal support

Independent transfer

For the same induced e.m.f. and comparable geometry, explain why a good electrical conductor pan generally heats more effectively than a poor conductor in an induction cooker.

Try this before viewing the solution

Effect of lower path resistance

Hints

Hint 1: hold induced e.m.f. fixed
Combine I = E/R with electrical power rather than treating both I and R as independently fixed.
View solution step by step

  1. Compare resistance

    Method

    The better conductor has lower effective eddy-current path resistance.

    Reason

    Higher conductivity means less opposition to current for comparable geometry.

    Working

    σ↑ ⇒ R↓

  2. Compare induced current

    Method

    The same induced e.m.f. drives a larger eddy current.

    Reason

    Current is inversely proportional to resistance for the stated comparison.

    Working

    I_eddy = E/R

  3. Compare dissipation

    Method

    The induced-current power is larger for the lower resistance.

    Reason

    With the induced e.m.f. held fixed, P = EI = E²/R.

    Working

    P = E²/R

7. Mind Stretchers

Mind stretcher 1: Why do slits reduce heating?Extension

Suggest why cutting radial slits in a metal disc would reduce induction heating.

Show Answer

Slits break up large circular eddy current loops into smaller loops and increase the effective resistance of the paths.

That reduces eddy current magnitude and therefore reduces I²R heating.

Mind stretcher 2: Current scaling (approximate)Extension

If the coil current amplitude is doubled (same frequency and geometry), what happens to the heating power in the pan approximately? State your assumptions.

Show Answer

Assume induced e.m.f. is proportional to the rate of change of flux, and flux is proportional to the coil current. Then doubling coil current roughly doubles induced e.m.f.

If the effective resistance of eddy-current paths stays roughly constant, eddy current magnitude roughly doubles, so heating power P ∝ I²R increases by a factor of 2² = 4 (about quadruples).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027