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?

  • $\begingroup$ What do you mean by "the energy input is the same"? You mean the max energy you can provide? $\endgroup$ Commented Mar 8, 2017 at 10:31
  • $\begingroup$ 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? $\endgroup$
    – MaDrung
    Commented Mar 8, 2017 at 10:47

2 Answers 2


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.


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.

  • $\begingroup$ 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 $\endgroup$
    – MaDrung
    Commented Mar 8, 2017 at 11:05
  • $\begingroup$ 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? $\endgroup$
    – MaDrung
    Commented Mar 8, 2017 at 11:10
  • $\begingroup$ 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. $\endgroup$ Commented Mar 9, 2017 at 13:53
  • $\begingroup$ 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 $\endgroup$
    – MaDrung
    Commented Mar 10, 2017 at 6:54

While I cannot provide any mathematical prove, I can give you a diagram from an FEM simulation of a coil and a conductive plane placed in parallel. The coil is excited by a constant current density and therefore it is not affected by the counter field generated from the eddy-currents. The current density in the plane is analyzed and shown in the following diagram. Eddy current density vs. frequency

You can see that after 1MHz the current density saturates.

Maybe I can provide some intuition at least from my perspective. First of all, if you analyze it from an energy perspective it's logical to say that you should not be able to create more energy than you provide. However I believe that this is not the case here. In most situations it will probably be the width of the conductive material that limits the current density.

Basically the current depth reduces drastically with higher frequencies according to the following expression: \begin{equation} \delta = \frac{1}{\pi f\mu\sigma} \end{equation} where $f$ is the frequency of the magnetic field. The more you increase the frequency the less space the current has to flow in. Although the excitation is greater at higher frequency, the resistance that the current goes through is much higher due to limited space. Take this image as an example: Eddy-currents distribution vs. frequency which I procured from here. If the penetration depth is more than the width of the material, then current distribution spreads only horizontally. If penetration depth is less than the width of material, then current distribution shrinks horizontally and vertically. This drastic attenuation of space probably increases the resistance of the area and induced currents reach limit.

Now you might say, but what if the material width is infinite? Than I'd probably say that the eddy current limit would be defined by the energy of the magnetic field. The current distribution depth would go towards infinite width of the material with ever increasing attenuation such that the created energy increases towards the input energy, but never really reaches it.


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