# What is the main mechanism for induction heaters (cookers)?

I thought that induction heaters (as used in cooking) relied on Joule heating losses due to Eddy currents induced by an oscillating magnetic field, produced by an inductor through which AC runs through. That is basically what I can understand from the Wikipedia's article.

However, Pieter, a materials scientist, claims that the main mechanism is not the above one, but instead is due to magnetic hysteresis losses. Microscopically, the heat is released when the domain walls are set in motion by the applied magnetic field. This viewpoint conflicts with the one described above, in that non ferromagnetic materials do not posses magnetic walls and therefore are not susceptible to suffer from any magnetic hysteresis loss. But it is possible to melt aluminum via induction heating, though I suppose by using a very high frequency AC compared to a regular induction cooker. But still, a model based solely on ferromagnetism seems lacking.

Now, a quick analysis. The skin depth goes like $$\sim 1/ \sqrt \mu$$ where $$\mu$$ is the magnetic permeability. If we take nickel vs copper, Ni has a skin depth roughly 10 times greater than that of Cu. Furthermore, it also has a resistivity of about 5 times than that of Cu. This means that, for a given AC frequency, Ni has an impedance roughly 50 times that of Cu. Now, if we use Ohm's law $$V=RI$$ where $$V$$ is the induced emf by the induction cooker and does not depend on which material one heats, then this means that $$I$$ the Eddy currents generated in Ni will be about 50 times lesser than those generated in Cu. And since the Joule heat is $$I^2R$$, it means that there is about 50 times more Joule heat losses in Cu than in Ni.

This is quite counter intuitive at first to me, I would have expected Ni to dissipate more heat through Joule losses than Cu, but it is the other way around. Now, I think I understand that it is because Cu conducts electricity so much better than the current is much higher and due to how Joule heat scales with current, no wonder that Cu dissipates much more heat than Ni via the Joule effect. However this clashes with the common Wikipedia claim that induction heaters work mainly via Joule heating due to Eddy currents. Because if that was the case, then copper should work much better than nickel, and I suppose, than stainless steel since it should be similar to nickel.

Hence in the end it looks like Pieter's viewpoint might be the most accurate if we consider a stainless steel pan in the kitchen. But I seek an answer that will show the math very concisely and compare the two views, mention in which case(s) each view is more accurate, etc. Feel free to compute the skin depth for $$\sim 1 cm$$ thick pans with the AC frequency used in houses, to use in your answer.

• I am no expert on cookware, so I am googling and I found: "Attempts to determine which process plays the more important role have been known to cause screaming arguments between induction-cooktop engineers." popularmechanics.com/home/how-to/a5966/… – Pieter Jul 11 at 22:24