# Why can induction cooktops not use aluminium or copper utensils?

Why can induction cooktops not use aluminium or copper utensils, when both materials are conductors and hence conducive to eddy currents, which are used by these devices to cook?

Edit 1 (10 Nov 2016): Following argument is common to all the answers including deleted one. Since it does not seem to be correct, it is important that this matter is resolved. It is being argued that ferromagnetic material have thinner skin depth due to high permeability, leading to eddy currents being confined to a thinner surface, which obviously will have higher resistance and hence higher power dissipation following P=I^2R. However, to me it seems that the argument is incorrect. It is the voltage which is being induced. Because of skin depth the induced electric field will be confined to a thinner surface. this should lead to lower power dissipation in ferromagnetic materials due to P=V^2/R. If my line of argument is correct then the answer to the original question lies somewhere else.

• If you change the resistance of the pan this also changes the voltage. It has to be seen as if it were the secondary circuit in a transformer, not as a simple resistance with a fixed voltage. Nov 10, 2016 at 18:04

In principle all metals can be used in induction cook tops.

However, iron-based pans (or at least bases) work much better, because of the magnetic properties of the metal.

When an AC current is flowing in a conductor, the current distribution is not uniform throughout the volume. Instead, current tends to be confined to a layer near the surface through something called the skin effect. This happens because the currents induce magnetic fields in the conductor, which in turn induces currents that try to counter the flux change (a variation of Lenz's law is at play, basically). Because of this, the effect will be stronger when the size of the magnetic field induced by a given current is greater. The thickness of the conduction layer can be calculated as

$$\delta = \sqrt{\frac{2\rho}{\omega\mu}}$$

where
$\delta$ = skin depth
$\rho$ = resistivity
$\omega$ = frequency
$\mu$ = $\mu_0 \mu_r$, the product of the permeability of free space and the relative permeability of the material

For a ferromagnetic material, $\mu_r$ is very high (if can be 200,000 for annealed iron). This means the skin depth will be much smaller, and the resistance (which depends on the cross sectional area of the conductor) will be higher.

The induction heater imposes a varying magnetic field on the pan, which will result in an induced current. The magnitude of this current is largely independent of the material of the pan - it just depends on the currents in the induction coil and the mutual inductance between the coil and the pan. For the same induced current, higher resistance will give more heating. And that is why you need pans (with a base) made of iron.

• "The magnitude of this current is largely independent of the material of the pan - it just depends on the currents in the induction coil and the mutual inductance between the coil and the pan." Is there a formula capturing how much current is induced? I thought voltage is induced as per Faraday's law, resulting in current as per Ohm's law.
– akm
Nov 7, 2016 at 17:21
• If equal voltages are induced in different material, then your argument does not hold, since power dissipated would be V^2/R.
– akm
Nov 7, 2016 at 17:39

Induction cookers work by inducing eddy currents in the pan that is placed upon them. Typically the cooker consists of a coil through which an alternating current in the range 20-100 kHz is passed. The alternating magnetic field of the same frequency induces an alternating electric field in the pan. The electric field drives currents, which then dissipate as heat, hence heating the contents of the pan.

Heat is generated most efficiently if the power inherent in the electromagnetic fields is dissipated in a thin outer layer of the pan. The work done per unit volume by the fields is $\vec{E} \cdot \vec{J}$, which for a linear conductor is equal to $J^2/\sigma$, where $\sigma$ is the conductivity of the metal. But for a given eddy current $I$, the current density $J$ is inversely proportional to a linear dimension of the pan $L$ multiplied by the effective thickness in which the alternating electric field is confined - this is known as the skin depth. $$\delta = (\pi f \mu_0 \mu_r \sigma)^{-1/2}$$

Thus $J \propto L^{-1}\delta^{-1}$ and thus the total heating effect per unit volume is $J^2/\sigma \propto L^{-2} \delta^{-2} \sigma^{-1}$. The total heating effect is then obtained by multiplying by the pan area $L^2$ and the thickness of material in which the eddy currents ($\delta$ again). So the total heating effect $$H \propto J^2 \sigma^{-1} L^2 \delta \propto L^{-2} \delta^{-2} \sigma^{-1} L^2 \delta \propto \sigma^{-1} \delta^{-1}$$

For two materials of similar conductivity then the one with the smaller skin depth will result in the greatest Ohmic heat losses. Looking at the formula for skin depth, we can see that $\delta^{-1} \propto \mu_r^{1/2}$, so ferrous materials with a high relative permeability have a much smaller skin depth and thus the eddy currents dissipate much more power.

EDIT: As your edited question points out, this argument only works if the currents in the pan are equivalent. However, the coil plus pan can be viewed as a step down transformer where the secondary load resistance is $R \propto \sigma^{-1} \delta^{-1}$. The $\mu_r$ of a ferromagnetic pan could be a few thousand, so for similar conductivity, this increases $R$ by factors of $\sim 50$. If the resistance in the coil is $R_C$ then the fraction of power "usefully" dissipated in the pan is $$f_U \simeq \frac{a^2R}{a^2R + R_C},$$ where $a>1$ is the ratio of the voltage in the coil to the EMF induced in the pan (see ideal transformer). If $R \gg R_C$ then the transfer of power is very efficient. If you reduce $R$ by a factor of 50 (by using an aluminium pan with a factor 50 larger skin depth) then that is probably not going to be the case.

An additional point which rarely gets a mention is that there are hysteresis losses in ferromagnetic materials. But I am unsure about the relative magnitudes of these effects.

• your calculations are for a given eddy current. but for two different material, equal currents are induced or equal voltages?
– akm
Nov 7, 2016 at 17:23
• @Amit The changing magnetic field produces an EMF/voltage drop around a loop, $\mathcal E = -d\phi/dt$, so it's the voltage that's the same. However, while copper is about six times more conductive that iron, iron has more than a thousand times larger permeability.
– rob
Nov 10, 2016 at 10:37
• @rob so ???????
– akm
Nov 10, 2016 at 11:26
• @Amit So the conductivity is about the same, and the same EMF produces more or less the same amount of eddy current. But, this answer argues, the large magnetic permeability of iron concentrates that current in a much smaller volume of the metal than in aluminum or copper. You can't directly compare $I^2R$ to $V^2/R$ because the resistance $R$ seen by the eddy currents also depends on the volume of the metal involved. That's why iron, with its shallow skin depth, is more efficient at converting eddy currents to heat than non-ferromagnetic conductors.
– rob
Nov 10, 2016 at 14:19
• @rob Actually, I understand Amit's concern and am wrestling with it myself. That is why I tried to reduce it to the microscopic level. The same EMF will not produce the same current; the same electric field will produce the same current density. But that isn't the same thing and isn't what I've used in my answer. Nov 10, 2016 at 14:37