Electronic – How induction heating works

On the internet there is unbelievable amount of contradicting information. The textbooks don't explain it in depth enough, so I have to ask here.

I have an induction heating system which has a coil trough which there is applied an AC voltage which frequency I can regulate and I use cooking pots as load.

I'm interested in power consumption of different cooking pots. I'm most interested in why aluminium pots can't be heated at the same frequencies as pots with higher permeability.

One reason could be the hysteresis losses, but many papers say that hysteresis losses only contribute up to 10 % of total power losses inside the pot. So if this is true, this couldn't be the reason.

They then say that the losses increase if the resistance of the pot increases. But that doesn't make sense because then plastic pots would have the highest losses because of high resistance. My thinking is that if the pot has a low resistance, then the currents will be higher which will increase the losses. If a pot has high resistance, less current will flow and the losses will be lower. But this is not what we observe.

So what is the reason for this? Low resistance materials induce poor losses and high resistance materials induce poor loses. If resistance is not what determines the losses, then what is?

EDIT for mkeith: I am more interested in how the eddy currents are induced inside the material and how it leads to losses, than an actual product.

But to answer your question, induction cookers can only heat aluminium pots at high frequencies (around 100 kHz), while steel can be heated equally well at 18 kHz and lower (they use 18 kHz+ to avoid noise in the audible range).

Now Inductance with steel pots is indeed a lot higher, and so is resistance felt by the coil. Now what happens if we put on an aluminium pot? The resistance and inductance decrease. This effects resonant frequency of the resonant LC circuit. The resonant frequency increases. Also because of decrease in resistance, the current increases (if we increase the frequency to match the new resonant frequency). This would mean that because we have a low resistance, the current drawn would be very big and because of so low resistance the heating would still be smaller. Big current increases the negative losses because of internal resistance of the coil and the switching elements can only take a limited amount of current.

This explains why we can't heat aluminium at low frequencies. We increase the frequency to increase the resistance of aluminium pots because of increased skin effect which allows us to have lower current running through the switching elements for the same amount of heating power.

Now I understand this, but the real question I'm interested in is the following. If we have a current source that maintains 10 A at a fixed frequency. And in one case we have an aluminium pot causing 0.3 ohms increase in resistance over primary and in the other a steel pot which adds 5 ohms of resistance over primary. In both cases the internal resistance of primary is neglectable.

We see that aluminium pot is taking 30 W of power while steel pot is taking 500 W. But how does this make sense from the following perspective?

The current is the same in both cases, so the electromotive force created in both pots is the same. Because the pots have different electrical resistances, the eddy currents inside of them should be different as well.
The losses are calculated by current flowing inside the pots squared times the pots resistance.

Because resistance of aluminium is lower, the currents inside them is larger, which means more losses should be created in the aluminium pot. But the actual result is just the opposite. This is the main part of my confusion.