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Imagine we have an induction coil which is strong enough to heat a sheet of metal. We can put a sheet of ferromagnetic metal close to the coil at distance $h_1$, and it gets heated to temperature $t_1$, or at distance $h_2 > h_1$ so that the sheet gets heated to temperature $t_2 < t_1$.

I want to know what happens if we have two identical sheets at once, one at distance $h_1$ and one at distance $h_2$, on the same side of the coil, with some insulator between the sheets (the insulator does not conduct electricity, is not ferromagnetic, and does not conduct heat well). Will the sheet farther away from the coil heat up at all, or will the closer sheet shield it from the electromagnetic field in some way? What will the temperatures $t_1'$ and $t_2'$ of the sheets in this experiment be, higher, lower or the same as $t_1$ and $t_2$?

Does the answer to the above change if we have a small conducting connection between the two sheets of metal, e.g. a wire which touches both the close and the far one, but most of their surface is still separated?

The application for this question: I am thinking of getting a cast iron waffle iron to use on my induction stove, and I am trying to imagine how this will function. By the way, I know that I will get some heat conducted through the waffle itself, and I will probably turn it anyway so both plates get hot, but please ignore these effects when answering the question and tell me the effects of induction only.

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A wire will not work, (and the waffle iron is too thick, you will have to turn it) but theoretically if the first surface were thin enough, and you had a substance in between which was like a dielectric for a magnetic field, and that was also thin enough, you could also heat the second surface. This effect is mathematically exactly like quantum tunneling or frustrated TIR. – daaxix Jan 12 '13 at 3:56
up vote 4 down vote accepted

The answer is no. The penetration depth of the magnetic field in the first sheet is too small. Read this for example. The penetration depth $\delta$ is typically given by a formula looking something like this:

$$ \delta=\sqrt{\frac{\rho}{\pi\mu f}}\approx\sqrt{\frac{1\cdot10^{-7}}{\pi\cdot8.8\cdot10^{-4}\cdot20\cdot10^3}}\approx4.3\cdot10^{-5}\mathrm{m}=0.043\mathrm{mm} $$

Where $\rho$ is the resistivity of the material (I've assumed steel), $\mu$ is the magnetical permeability (I've assumed steel) and f is the freequency of the magnetic field (20 kHz is on the lower end of the range used in induction stoves. Higher frequencies would give even shorter penetration depths.) Assuming each side of the waffle iron is considerably thicker than 0.043 mm, your plan won't work.

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Interesting. I had always assumed that an induction stove heats both the upper and lower surface of a pan on the stove (but not the inside, due to skin effect). Now your answer makes me think that it must be the lower surface only after all, and that any heat on the upper surface conducts through the pan itself. Which is correct? – rumtscho Jan 11 '13 at 21:46
I was going to say it would work in theory if you could get close enough, but as stated here there is exponential decay of the fields, so heating is really all within the skin depth, then the heat diffuses from there. This is basically how Faraday cages work for electric fields... – daaxix Jan 12 '13 at 3:52

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