# energy of eddy current with increasing frequency

As far as I know, eddy current losses increase with the increase of change of magnetic field.

That means that they increase with a square of a peak of magnetic field and with a square of frequency.

Now what I don't understand is from where does the energy come from if I just increase the frequency of the power delivery system. The energy input is the same at both frequencies. But the eddy current losses actualy increase with a square of frequency.

That means that at some point you could excede the 100 % efficiency. Why this doesn't happen?

• What do you mean by "the energy input is the same"? You mean the max energy you can provide? – mikuszefski Mar 8 '17 at 10:31
• The power is calculated by U * I. The frequency doesn't matter. But in case of eddy currents it does. So if I increase the input frequency, the power input will stay the same, yet output energy caused by eddy currents will not. What is wrong with my thinking? – MaDrung Mar 8 '17 at 10:47

In detail eddy currents are very complicated and a 100% answer might not be possible. First of all, it should be clear that the efficiency will not be greater than 100%. The Eddy current is taking energy from your system. This is due to the fact that your osculating field is actually radiating. (have a look at Pointing Vectors) This energy loss is small compared to, e.g., ohmic losses, so you do not care very much. If you could increase frequency to any arbitrary value, your radiated energy will increase, eventually limiting your energy input, i.e. there will probably be a max frequency. A second point is to consider that the current distribution changes with frequency.The article on Eddy Currents mentions for the formula, you are probably talking about, neglecting the Skin Effect. Finally, you have to consider that large eddy currents will produce very large fields themselves, which then will couple in your field generating coil and power system, making things even more complicated. Long story short: For ever increasing frequencies your $E \propto f^2$ formula becomes invalid.

Edit

Due to the ongoing discussion in the comments, let me try to give another hint.You can look at your system:

Power source $\rightarrow$ Coil $\rightarrow$ Field $\rightarrow$ Metal object $\rightarrow$ Eddy current $\rightarrow$ Loss

as a transformer with load. The metal object is the second coil, the Eddy current the current flowing if a load is provided and the resistance acting on the Eddy current the load.

• Thank you for answering. I am aware of these effects, but let's say we have a coil and an aluminium pot on top of it. If I put it on an electricity grid that has the same voltage amplitude, the average current Will be the same as if I put it on a same network but with a twice higher frequency. If I then calculate the power draw from the grid, it should be the same according to U * I. But the actual energy transfer because of the eddy currents Will be 4 times bigger because of twice higher frequency. There is something wrong here. What? :P – MaDrung Mar 8 '17 at 11:05
• Is it a back EMF that is decreasing the current that Will flow, so we get less current at bigger frequencies because of that, but the power it takes from the grid is more or less equivelant to that used by the eddy currents? – MaDrung Mar 8 '17 at 11:10
• Dear @MaDrung I have some difficulties to understand your example. Maybe you update your question with an according sketch. My understanding of your comment is: In the limits you are discussing the $f^2$ behaviour is approximately correct. It is just not valid for $\lim f \rightarrow \infty$. Moreover, it should be clear that the power taken by the Eddy current is provided by the power source driving your field coil. It is just that the simple approach does not account for that. A nice example here are brakes that "eat up" kinetic energy. – mikuszefski Mar 9 '17 at 13:53
• Sorry, I was in a bit of hurry and thought it would be clear what I'm asking. It's just that with frequency the eddy currents Will increase. That means that the power input by the grid must also increase. Yet I don't see how if the current and voltage stay practically the same, only the frequency changes which adds nothing to the power input. I want to learn where does this increase in power draw come from (pls don't say the power grid, I'm how in terms of current, voltage etc.) :P – MaDrung Mar 10 '17 at 6:54