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.

  • 2
    $\begingroup$ 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/… $\endgroup$
    – user137289
    Jul 11, 2019 at 22:24
  • $\begingroup$ According to cookware sites magnetism is essential. However, that fact does not decide the question, as magnetic permeability enhances induction current but also implies Weiss regions. $\endgroup$
    – my2cts
    Dec 21, 2021 at 11:11
  • $\begingroup$ Induction cookers don't use mains frequency! $\endgroup$
    – ProfRob
    Dec 21, 2021 at 11:43

4 Answers 4


The resistance per square metre of surface is given by (or at least is proportional to) the resistivity (in $\Omega$ m) divided by the skin depth (in m). This will be true so long as the material is much thicker than the skin depth. $$ R \propto \frac{\rho}{\delta} = \sqrt{\frac{\omega \mu \rho}{2}}$$

In order to have a high heating effect you want a combination of high frequency, high permeability and high resistivity.

For a given frequency (induction hobs work at tens of kHz so that the skin depth is a fraction of a mm in steel and the power is dissipated within a thickness of 1-2 mm - my pans are not 1 cm thick!) the resistance of stainless steel will be about 250 times greater than a copper pan.

The skin depth of copper at these frequencies (which depends on $\sqrt{\rho/\mu}$ is only a few times bigger but still smaller than the pan thickness. It is the combination of higher resistivity and smaller skin depth for the same frequency and resistivity (due to the high permeability) that leads to very high heating losses in steel.

Taking the example of Nickel - the resistivity is a few times higher than copper and the permeability is a few hundred times higher, so the skin depth will be about ten times smaller than for copper (and smaller than for steel). The resistance will be about 50 times higher than copper at the same frequency, but still 5 times lower than stainless steel.

It seems the root of your problems with these arguments is a miscalculation of the skin depth for nickel compared with copper?

Hysteresis losses can be estimated from a version of the Steinmetz equation. $$ P = k f B^{1.6}\ {\rm W m}^{-3}\ ,$$ where $f$ is the frequency and $B$ to magnetic flux density (e.g. see here). The Steinmetz constant $k$ is I think of order 10 for steel in SI units (e.g. Takacs et al. 2010).

At 25 kHz and magnetic flux densities of $10^{-5}$ T (e.g. here) the hysteresis losses are only of order $2 \times 10^{-3}$ W m$^{-3}$. For a pan of thickness 5mm and diameter 200 mm, the hysterisis losses will be a few $10^{-9}$ W. I don't think this calculation is very accurate, but accurate enough to say that hysteresis losses are negligible.


However, Pieter, a materials scientist, claims that the main mechanism is not the above one, but instead is due to magnetic hysteresis losses

It is a strange claim, especially for a materials scientist. Unfortunately, your link is broken now. Looks like the Peter's profile was removed for some reason (hopefully, that's not due to what he said regarding hysteresis). I can only assume that his words were taken out of context or he didn't explain his point of view more clearly because of time constraints.

non ferromagnetic materials do not posses magnetic walls and therefore are not susceptible to suffer from any magnetic hysteresis loss

True. And it's not just a viewpoint. It's a scientific fact.

The very definition of hysteresis implies we are dealing with a system whose current state depends on the past events. So-called memory effect. No memory - no hysteresis. Easy peasy.

But what exactly can ferromagnetics (iron, nickel) memorise that paramagnetics (aluminum) or diamagnetics (copper) can't? To answer the question we need to take a closer look at hysteresis loops of different materials.

Hysteresis Loop Analysis

Here is a detailed figure depicting a hysteresis loop of a ferromagnetic material (image from Magnetic Particle Testing lecture of Technische Universität München):

Hysteresis loop of a ferromagnetic material


  • An external magnetic field is applied to the ferromagnetic material
  • Initially the ferromagnetic material is magnetized up to a saturation point 1 (orange curve)
  • When the external magnetic field is being decreased, the ferromagnetic demagnetization follows different curve. This means the ferromagnetic doesn't give up its magnetization state so easily: we need to apply the external field up to a point 4 to fully demagnetize the material
  • This phenomenon is called hysteresis
  • Paramagnetics and diamagnetics do not posses the same ability to remember the magnetization state - their respective magnetization graph follows the same curve regardless of whether we increase or decrease the external field

In-Depth Review

1. System Description

First of all, let's define the system we are analysing to prevent any ambiguity (image from Chapter 12, Fundamentals of Power Electronics, 2nd edition, R.W. Erickson and D. Maksimovic):

Magnetic circuit model

As you can see in the figure, the main components of our system are:

  • Ferromagnetic core with permeability $\mu_c$
  • Coil with n turns

An alternating current $i(t)$ flows through the coil turns producing fluctuating magnetic field strength H.

H interacts with the ferromagnetic core causing its internal magnetic dipole moments to be realigned in a specific way.

The following figure depicts the effects of the applied external magnetic field to different materials very clearly (from Chapter 8, L.E. Smart and Moore, "Solid State Chemistry An Introduction", 4th edition):

Applied magnetic field effects

Note that the terms spins and magnetic moments are used interchangeably here because we are interested only in macroscopic magnetic effects (fortunately, we are not going down the quantum mechanics route). Wikipedia provides more details about the linear relationship of an electron's magnetic moment and spin.

From the above figure it is apparent that H applied to ferromagnetics produces the following changes:

  • Magnetic domain walls of the material are moved
  • Magnetization M is increased

Paramegnetic materials don't have magnetic domains. Conceptually that means their atoms rotate upon applying external field without moving around and rubbing crystal structure.

On the other hand, movement of domain walls in ferromagnetic materials causes some energy to be dissipated as heat.

As for M, it changes because of applied H in both types of materials. However, in paramagnetics the relationship between M and H is linear, while in ferromagnetics - nonlinear.

An easy way to think about this:

  • When H increases or decreases, the paramagnetic atoms say: "Okay, we will change our M as you please, our master!"; while ferromagnetic atoms say: "Okay, we will change our M, but only when we want to."

If you combine H and M using a little bit of math you will obtain magnetic field density B.

Some textbooks present hysteresis loops as relationships between B and H, some - between M and H. But, as you will see below, it doesn't really matter what you choose - it's the final shape of a hysteresis loop what is important for our analysis.

2. B vs H vs M

Before describing what's happening in the given hysteresis loop we should get a deeper understanding of the things which are depicted on x and y axes.

The relationship between B, H, and M can be very confusing. Especially, taking into account the fact that scientific community still struggles to name them consistently (e.g, H aka Magnetic field intensity aka Magnetic field strength aka Magnetic field etc).

The next formula from Chapter 11.10, Electricity and Magnetism, Purcell, Edward M., third edition provides, in my opinion, a good intuition about the above-mentioned phenomena: $$ curl\,\mathbf{B} = \mu_0(\mathbf{J}_{bound} + \mathbf{J}_{free}) = \mu_0\mathbf{J}_{total} $$

where B is the magnetic field we are interested in, $ \mu_0 $ is the magnetic permeability of space where we are measuring B, $ \mathbf{J}_{free} $ is a density of the free current (the one that flows through the coil in the figure above), $ \mathbf{J}_{bound} $ is a density (macroscopic average) of the bound currents (induced inside the ferromagnetic core). Both free and bound currents will be discussed in more details below.

It should be easier to see now that B actually consists of two components whose values are related to different currents. What are these components? Yes, that's H and M.

In our system H can be expressed as

$$ curl\,\mathbf{H} = \mathbf{J}_{free} $$

and M as

$$ curl\,\mathbf{M} = \mathbf{J}_{bound} $$

All this company (B, H, and M) can also be defined in terms of relations involving volume magnetic susceptibility and magnetic permeability (as described in Wikipedia). However, it's easier to analyse our system and the corresponding hysteresis loop thinking about currents. You can find a comprehensive discussion of this topic in the book I mentioned.

Let's return to the magnetic core material. These are the directions of B, H, and M (image from H-field and magnetic materials, Wikipedia):

B, H, and M lines

The free current $ i(t) $ flowing through the coil with $n$ turns generates the magnetic field strength H in accordance with Ampère's law. H in turn changes the magnetization M of the ferromagnetic core. As a result, the bound volume and surface currents are starting to flow with their own magnetic fields opposing H. You can see in the figure above how M lines oppose H lines. It's worth noting there are no M lines outside the core. Since M is represented as the average magnetic dipole moment of the core region, there is no M outside the core boundaries by definition.

B can be described as a combination of H and M magnetic effects.

The problem with B, H, and M relationship is that it's nonlinear for ferromagnetic materials. That means you need to do different measurements and calculations for different core composites.

Fortunately, magnetic material manufactures usually provide carefully prepared hysteresis loop graphs (e.g., MnZn Ferrite Material Datasheet) that allow us to quickly analyse required magnetic properties.

Look again at the first figure of the hysteresis loop of a ferromagnetic material. By now it should be obvious what all those curves mean.

Initially B increases with H up to the saturation point 1. The saturation point can be changed by precise air gap manipulation (topic for another discussion).

However, when H is being decreased (because of alternating current) B is following a different path. This is where the nonlinearity arises and its cause is hysteresis. Hysteresis means that our core material stays magnetized even when the external field is decreasing. That's why we need to apply H with opposite direction up to a point 4 where B finally becomes zero.

The shape of the resulting hysteresis loop allows to estimate a proportion of magnetic losses of the material. Ferromagnets with large and wide loop area will have more eddy current losses than hysteresis losses. That's because the wider the area the slower is the change of B and, consequentially, magnetic domain walls move less rapidly. Also less steep B curve indicates more stable eddy currents.

But how will a hysteresis loop look like for a paramagnet or a diamagnet?

Turns out it will be just a single curve without any loop area altogether.

Here is a nice figure (from DOI 10.1007/s10853-012-6724-4 article) illustrating how mixing pure aluminum with a cobalt ferrite material changes B-H curve:

Fig. 8 Magnetization-hysteresis loops of pure aluminum and cobaltferrite-reinforced aluminum composite

3. Free vs Bound Currents

One more thing that needs clarification is the distinction between free and bound currents.

To prevent any confusion during our analysis a scale of the electromagnetic phenomena must be defined unambiguously. What are the effects we are interested in: macroscopic or microscopic?

Since we are not investigating atomic interactions on a quantum mechanics level it should be apparent that it's the outcome of the microscopic activities that defines the parameters we are trying to measure and calculate.

Can you identify where are the free and bound currents in our system?

The names free and bound give good enough hints already. There are different definitions in the textbooks, but for all our purposes the most important points are:

  • Free means an external current that can be freely measured with an ammeter and started and stopped with a switch (more details in Chapter 11.10, Electricity and Magnetism, Purcell)
  • Bound current is a resultant current caused by movements of electric and magnetic dipole moments per unit volume of our ferromagnetic core (check out equations in Wikipedia).

It would be accurate to think about a density of the bound current as an average effect of a repeated application of Ampère's law to every single volumetric current that arises in the magnetic material.

Why should we pay attention to the bound current density in our analysis? Because it is this density that is responsible for the core losses which in turn define how much heat is generated in the system.

So, as we've just found out, the free current is the current $ i(t) $ running through the coil with $ n $ turns. There is nothing special about it.

The bound currents are more interesting.

On a microscopic scale there are a lot of volumetric currents generated because of magnetization. The following figure shows the circulating volumetric currents (black) and a surface current (red) induced in the magnetic material (image from Magnetization article in Wikipedia):

Volumetric and surface currents

There is a non-uniform probability of the currents distribution throughout the material volume. Some internal currents will be cancelled out because of opposite directions of adjacent orbitals. On the other hand, there are no currents on the material surface which can balance out the induced currents. That's why after taking into account all the microscopic interactions the material surface will have larger current density than the inner volume.

Bound volumetric and surface currents are also called eddy currents.

Microscopically the electrons which comprise the bound volume and surface currents are bound to their respective atoms. Macroscopically the magnetic material volume defines their boundaries. On the contrary, we don't care about the origin and destination of the free current charges in the analysis.

Math behind the bound currents is explained in the lectures of Loomis Laboratory Of Physics, University Of Illinois.

Just like in ferromagnets, bound currents are also induced in paramagnets and diamagnets. For example, check out a model where paramagnets and diamagnets mentioned specifically in doi.org/10.1119/1.4773441.

4. Hysteresis vs Eddy Current Losses

As was stated earlier in the discussion, the hysteresis losses are defined only for ferromagnetic materials.

Eddy current losses are applicable to all magnetic materials.

Now, if we use Ohm's law 𝑉=𝑅𝐼

It's easy to get confused by Ohm's Law.

What is usually overlooked by students (and sometimes even by professional scientists) is a cause and effect relationship.

In a nonlinear system (such as the one involving an inductor) Ohm's Law won't tell us the exact response of $ I $ to $ R $ changes.

It may be unintuitive at first, so let's go step by step.

Ni has a skin depth roughly 10 times greater than that of Cu

Check out the magnetic permeability table. Permeability of Ni is represented as a range $ 1.26\times10^{-4} – 7.54\times10^{-4} $ (value varies with field strength). Cu: $ 1.256629\times10^{-6} $. So, actually it's Cu who has a greater skin depth.

What does it mean for a material to have a greater skin depth?

Current density formula gives us the answer: the greater skin depth - the less current density near the surface.

it also has a resistivity of about 5 times than that of Cu


this means that 𝐼 the Eddy currents generated in Ni will be about 50 times lesser


You may wonder how on earth can bigger resistance and bigger currents coexist at the same time and place? Try to think about the cause and effects.

Does increasing resistance mean a current is decreasing? Yes. But what current?

As was discussed in the previous section, there is a lot of currents in the magnetic material.

On a microscopic level larger resistance can decrease bound currents. Conversely, on a macroscopic level the larger current density per unit volume actually increases the effective (or average) current.

Imagine a bound current with a value of 0.01 A. Increased resistance decreases it to 0.001 A. But if the current density at the same time is increased (because of different magnetic effects) the sum of currents per unit volume will be bigger: fifty more currents yield 0.05 A (0.001 * 50).

As you can see, it is quite possible to increase resistance and current at the same time. And more resistance and current per the same area means more heat.

Useful exercise. Try to answer yourself: What will happen if you reduce the area where the eddy currents are flowing like this (image from Magnetic Core Laminations, Wikipedia):

Magnetic core laminations

Using the resistance formula we can conclude that the smaller the area - the greater the resistance.

How our system will behave this time? Will there be more eddy currents as in the previous case? No.

Previously, comparing Ni with Cu, we had two systems with different parameters (resistance, current density, permeability etc.). Different systems - different causes and effects.

Now we are changing the area of the same material, so all other parameters are staying constant. That's why in this case it would be correct to say that bigger resistance reduces eddy currents.

If you are still confused on how to compute Joule heating without Ohm's Law: try to use the right tool for the job.

Look at the differential form of the Joule heating equation:

$$ \frac{𝑑P}{𝑑V} = \frac{J^2}{\sigma} $$

where J - current density, $ \sigma $ - the material conductivity.

Now it is easy to compare heating output of Ni and Cu.

From the conductivity table: $ \sigma_{Ni} = 1.43\times10^7 $, $ \sigma_{Cu} = 5.9\times10^7 $

Okay, Ni is ~4 times less conductive than Cu, Ni has ~10 times less skin depth than Cu $ \Rightarrow $ ~100 times more current density $ \Rightarrow $ $ P_{Ni} \gt\gt P_{Cu} $

I seek an answer that will show the math very concisely and compare the two views

In accordance with Fundamentals of Power Electronics, 2nd edition, R.W. Erickson and D. Maksimovic the hysteresis losses for a ferromagnetic material can be obtained using the next equation:

$$ P_H = K_HfB^\alpha_{max}V$$

And eddy current losses for magnetic materials can be found with the help of

$$ P_E = K_Ef^2B^2_{max}V$$

Here $ B_{max} $ is a peak magnetic flux density, $ f $ is an alternating current frequency, $ V $ is a core volume.

Problem is the coefficients $ K_H, K_E $, and $ \alpha $ are determined experimentally. So it's hard to show you the math without an appropriate equipment.

If you are really into this topic, I'm sure, you'll find with the help of your favourite search engine a lot of scientific papers with eddy current and hysteresis losses measurements and computations for various systems.

Here is a nice example of core losses decomposition at 0.5T into different components (from DOI:10.1109/ECMSM.2017.7945864):

Decomposition of iron losses at 0.5T

Hope at least some of this info will help you. Good luck.


Wow, as I'm an analog engineer you have all over analyzed and haven't considered it from the inducing coil and high frequency switching driving circuit design perspective. I haven't designed the induction circuit, but I am familiar with a lot of the switcher design concept tradeoffs.

Although the previous discussions are a valid analysis of the physics of why certain losses occur and how to calculate the them. It is a waste of time for experienced induction designers do to do this type of analysis. Rather, intelligent trial and error processes are used, incremental optimizing on highly accurate simulators. It's all about cost and energy efficient switcher design, hob design vs existing pan types, and useful consumer features.

Typically, the hob coil has an inductance which is driven by a high frequency switcher at a fixed frequency in an FCC approved ISM band.

The switcher driving the hob coil creates an oscillating inducing magnetic field. When operating, on one half of the electrical cycle, power is transferred to the hob coil as an increasing magnetic field in one direction, while on the other half the energy in the magnetic field is recovered as it collapses and/or shifted in opposite direction to a rising magnetic field in the other direction, depending on switcher design.

Normally, energy is transferred to the pan due to what looks like resistive losses looking into hob coil which are non orthagonal phase shifts of current from voltage. Normally, hob coil and switcher are designed to have low losses relative to pan losses.

In circuit simulations the hob coil and pan load can be modeled as a transformer driving a resistive load where the primary characteristics are fixed and coupling and secondary inducatance and series resistance are dependent on pan types. Normally, the hob transformer would be simplified to just a load inductance with a series and parallel resistance. The values of inductance and parallel resistance are dependent on pan types.

Typically, the designer would have a set of inductive resistive loads representing extreme values of pan types. Converting pan types into these models is tricky, but doable. More difficult with a non sinusoidal driving waveform. One issue, is that a simple LRC model might be inadequate if it changes significantly with frequency, temperature, and power levels.

Magnetic hysteresis losses are non-linear, switch type losses, increasing with frquency. Commonly, the designer just has worst case pans and tries them on an induction breadboard test circuit that accurately measures energy into pan vs energy into driving circuit. Even metal pan resistance and iron hysteresis changes with temperature.

To accomodate a wide range of heatable pans, both the switching circuit and hob coil might require switching modes, like changing frequency or driving a different taps on a hob coil.

The type of pan effects the apparent inductance and efficiency of switching circuit. If the pan is magnetic (like iron, but typical not stainless), the inductance will be higher due to the effect of iron increasing magnetic field, the current will ramp slower. If the pan is copper, the inductance will be low, the current will ramp up quickly due to large induced eddy currents from low resistance of copper. The iron pan also has eddy currents, but they tend to be lower amplitude due to higher iron eddy resistance, but also magnetic hysteresis losses.

So, generally iron bottom pans will have higher losses (heating) relative to hob coil and circuit losses. Because heat loss on iron pans is also from magnetic hysteresis losses, the switchers can operate at lower, more efficient frequencies which minimize loss in switch on-off transistion. While copper bottom pans will require higher switching frequency and higher peak hob coil current which tends to be less efficient.

Probably early hob coil and switcher designs required the use of iron magnetic bottom pains for greatest heating, since they are most efficient at lower switching frequencies, necessary due to the inadequacy of old switcher components. (Old specs never die, because most writers don't understand that why specs become obsolete.)

Newer designs will probably accomodate a wider variety of pans to deliver maximum heat. Use of incorrect pan on old inductive design usually just limits heating or might even turn off to avoid damage.

Probably the switcher output power into load goes up and down with 60/50Hz power mains so that stove looks like old resistive electric stove, to keep power factor looking resistive.

Not to mention FCC and international conducted radiative emission limits...


This paper answer the question.

Improving the energy efficiency of induction cooking

Eddy–current, hysteresis, and total loss in the new and the traditional pots, and their rates versus inductor current

In lower power(total loss 200w), the eddy-current loss $P_j$ and the hysteresis loss $P_h$ are balance. In high power, eddy-current loss $P_j$ is main.


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