There is an ongoing discussion about induction cooktops on Seasoned Advice.

The situation in nutshell is:

  • About 1 cm of aluminum on the cooktop.
  • About 1 cm of iron on top of the aluminum.

My question for Physics: is the total heat generated significantly less than if the aluminum was replaced with same thickness of plastic? I'd consider anything more than 20% reduction as significantly less.

I think most will agree on these basic assumptions, but feel free to question them:

  • Aluminum will heat up a bit. The energy going there is away from the energy going to iron, but nonetheless ends up in the pot.
  • Iron will heat up less at 1 cm distance than it would at 0 cm distance.
  • Energy doesn't disappear, so if the heat generated is less, either the efficiency of the cooktop must decrease or it must simply be unable to "push" more energy into the magnetic field.

1 Answer 1


As I commented there, the key issue (as described by https://en.wikipedia.org/wiki/Induction_cooking#Cookware) is skin depth at the ~24kHz most stoves use: much lower in ferromagnetic metals.

The same current in the induction coil induces the same current in the aluminum as it would in a steel pot, but (unless the frequency is much higher) the skin depth is deep enough that that fixed amount of induced current doesn't make nearly as much heat. I2R heating, where I2 is ~constant but R is much smaller.

And yes, the induced current in the aluminum does cancel out the magnetic field, shielding the iron disk above the aluminum. This is how a Faraday cage works; it's well known that Faraday cages don't have to be ferromagnetic to shield stuff inside them from induced currents.

Energy doesn't disappear, so if the heat generated is less, either the efficiency of the cooktop must decrease or it must simply be unable to "push" more energy into the magnetic field.

Yes, this is the key part of the question. But it's not exactly energy in the form of a magnetic field that's in question, it's energy / power coupled into a piece of metal. The magnetic potential energy held in the field through the coil (an inductor) at peak strength is highest in the open-circuit case (no pan), I think.

For a constant-current power supply to the induction coil, the total power dissipated in the aluminum and the induction coil (resistive losses) is less than the total power with an unshielded iron disc directly over it. (I think).

In the open-circuit case for a transformer, or for an induction pad with nothing conductive above it, the load on its supply is purely inductive. There will be current but it's out of phase with voltage so the true power is 0 (plus resistive losses), even though VA apparent power may be high.

For a constant-voltage supply to a transformer, the current vastly increases when you short out the secondary winding vs. leaving it open. But I assume there must be some current-limiting in the stove to prevent burning itself out if you put a copper pan on your induction stove. The power coupled into the load is low; most of the total power dissipation is in the induction coil.

Shorting the secondary brings the V and A phases in line with each other, drawing true power. A constant-current supply would greatly reduce voltage to keep current the same when the load suddenly lost most of its inductance, keeping only its resistance and a small bit its inductance, when you short the transformer secondary (put a Cu or Al pan on the stove).

In the good case (iron pan), most of the electrical power is dissipated in the pan, turning into heat where you want it. Voltage in the primary side of the transformer can stay high, inducing a higher voltage in the secondary.

If it takes much more voltage to make the current actually flow (because of skin effect and higher resistivity), we get lower induced current but that current is doing more work because of V*I = I2R = power.

  • $\begingroup$ Thanks for the in depth explanation! $\endgroup$
    – jpa
    Mar 12, 2020 at 9:00

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