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I'm interested in figuring out the approximate peak electrical field strength inside a microwave oven. The question is not so much about the exact number, but rather what reasoning to use to work it out approximately. I'm interested in getting the rough order of magnitude without doing a full physics simulation.

Scenario 1: Microwave oven with output of 1kW, running with nothing inside, 1 cu ft chamber. I assume in this case most of the 1kW (except for very small amount absorbed in walls or lost as leakage) is reflected back into the magnetron where it must be dissipated as heat. There will be an oscillating electrical and magnetic field inside the chamber, possibly with some standing waves (peaks and nulls). What determines the height of the peaks?

Scenario 2: Same, but with an object inside (let's say a cup of water, 1 pint). What fraction of power is absorbed in the object? What is the electrical field inside the object now, and what is it in the rest of the chamber? I assume now most of the power would be absorbed in the object, because (essentially) the waves bounce around until they are absorbed; the entry aperture is relatively small so the fraction hitting that and going back to the magnetron is much less than hitting the object (thinking of the waves as rays as an approximation). The electrical field in the object would seem to be related to dielectric constant, dielectric loss factor, etc.

What simplifying assumptions can be made here? For example, is it possible to model the chamber as a waveguide? Etc.

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    \$\begingroup\$ It is modeled very similar to a waveguide, if you look up microwave resonator you should get a variety of helpful resources. Essentially it's like a rectangular waveguide but with both ends closed, so there are standing waves in each direction. Additionally, the object inside and the coupling into the resonator act as small perturbations to the cavity, if you look up pertubation theory for microwave resonators this is also well documented. \$\endgroup\$ – jramsay42 Nov 12 '17 at 10:36
  • \$\begingroup\$ "...is reflected back into the magnetron where it must be dissipated as heat.." Why it should be? The potential difference is now zero, so it can't emit more energy into the space. \$\endgroup\$ – Marko Buršič Nov 12 '17 at 14:34
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    \$\begingroup\$ @MarkoBuršič: The potential difference is only zero (or reduced) if the reflection is exactly 180 degree out of phase. I think in a microwave it may come back with random phase because of mode stirring. Then it could be either re-reflected or absorbed... I don't know enough about magnetrons to know how they handle reflections which are in-phase or slightly shifted instead of perfectly out of phase. \$\endgroup\$ – Alex I Nov 12 '17 at 21:24
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    \$\begingroup\$ at 180 deg, you got nodes of a standing wave, but these are two waves back and forth, so at the transmitter are random phase, as you call it. These phases are equal coming from standing wave and from the transmitter, they change at same frequency/phase angle. However, there is a phase shift of 90 deg between E and H field, so the power is zero. The voltage is doubled - risk of ceramic isolation break, the current is doubled - higher heat dissipation due to the internal resistance. If there is a load inside the cavity, then the phase shift goes to zero - 100% TEM wave. \$\endgroup\$ – Marko Buršič Nov 13 '17 at 12:13
  • \$\begingroup\$ @MarkoBuršič Interesting, thanks! I still don't know how to calculate peak E field in the cavity with no load. \$\endgroup\$ – Alex I Nov 13 '17 at 21:11
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I will try. The power level of the microwave in Watts will tell you what it might transfer to your cup of coffee. The energy density will be \$\frac{e_0\cdot E^2}{2}\$ , where \$e_0\$ is the electric constant, 8.854187817E-12 Farad/meter and the electric field, E, is in volts/meter.

$$\text{EnergyDensity }= \frac{e_0 \cdot E^2}{2}$$

If the microwave is empty, I think that most of the power used is heating the microwave generator, and there will be some absorption in the walls, glass plate, moisture in the air, door, etc. If there is water, then you might take a look at the dielectric loss chart (blue) in this description of electromagnetic absorption spectrum of water. https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water

I thought that the power should be low when the microwave is empty, since it is not doing any work if there is nothing to heat, but the watt meter says 1267 watts whether it is empty or full. I would have it transmit energy and only heat the water. Seems kind of wasteful to me.

But to answer your question, it is easy enough to google [electric field inside microwave oven] to find http://physicsed.buffalostate.edu/pubs/TPT/TPTMay02Microwave/TPTMay02MicrowaveOrig.pdf where he says 2000 volts/meter and 2.8 megawatts/meter2.

Wikipedia's article on Intensity can give you the Intensity in Watts/m2 and its relation to the electric field. https://en.wikipedia.org/wiki/Intensity_(physics)

$$I = \frac{c \cdot n \cdot e_0 \cdot E^2}{2}$$

in Watts/m2.

$$Intensity = SpeedOfLight \cdot IndexOfRefraction \cdot EnergyDensity$$

The link to the refractive index gives 1.333 for the refractive index for water. \$e_0\$ is also called the "vacuum permittivity" as above. Plugging in 2000 Volts/meter gives 7.07668E3 watts/m2. To get his 2.8E6 Watts/m2 would require 39.783 kilovolts/meter2. So something is not quite right.

Googling [power absorbed in a microwave] leads to https://www.emu.dk/sites/default/files/physics_of_microwave_oven.pdf where that blue curve in the Wiki article is explained. It is the frequency dependent dielectric constant of water (that is not the proper name, but the easiest way to remember it). To get his 2.8E6 watts/m2 would require the refractive index to be about 53. Since that does not seem to be the case, keep going.

I will leave you to read his equation (8) which gives the power absorbed as follows:

$$P = 2\pi f e_0 e_2 E^2 VolumeOfObject$$ $$Power = 4 \pi \cdot Frequency \cdot e_2 \cdot EnergyDensity \cdot VolumeOfObject$$

\$e_2\$ is the blue curve. It is the absorption part of the dielectric constant. The blue curve that is somewhere between 20 and 40 depending on the temperature.

Omega (ω) is 2×π×frequency. Radians per second mean nothing to me here, so I use frequency. \$e_0\$ is still the electric constant, or electric permittivity. The volume has to be in meter3. After 50 years I learned to always do the calcuations in SI units, regardless of the convenience when working in some small domain. E is the electric field in volts per meter and the power is in Watts.

Not one to give up easily, I googled [power absorbed in a microwave "electric field"] and found http://www.pueschner.com/en/microwave-technology/basic-calculations. He uses the same expression for the power, but gives his slant on the problem. He says the \$e_2\$ is about 12 at 20 Celsius. And he did not multiply by the volume, but calculated the power density (watts per meter3, watts per cubic meter).

One could simply take the effective wattage of the microwave (my 1267 watt microwave only produces 900 watts effective), and divide by the volume of the oven (30×30×20 cm for mine is 0.018 m3) to get 50,000 watts/m3. I think that will lead somewhere useful, but it is time for lunch. I hope I have left some useful clues.

These kinds of problems are much better handled in a spreadsheet or calculator, rather than text formulas. I normally take the energy density (Joules/m3), multiply by twice the magnetic permeability (1.25664E-6 Newtons/Ampere2) and take the square root to get the the magnetic field in Tesla. Then multiply that by the speed of light or gravity to get the electric field in volts per meter. Or take the energy density (Joules/m3), multiply times the speed of light and divide by 4 to get the intensity in Watts/m2.

$$EnergyDensity = \frac{e_0 E^2}{2}$$ $$Intensity = EnergyDensity\frac{SpeedOfLight}{4}$$

in Watts/m2.

I use energy density to convert gravitational energy density to electromagnetic units. Then the 2000 volts/meter would be equivalent to an acceleration field of 172.34 nanometers/second2. And its effect would depend on the absorption.

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    \$\begingroup\$ From what I've been told, the glass plate in a typical microwave oven is engineered to absorb microwave radiation significantly less well than typical food or water, but better than other parts of the oven, which is a big part of the reason oven manuals warn not to use the oven without the plate. The thermal mass and heat dissipation of the plate are generally not sufficient to handle all of the power from an oven for a sustained period without damage, but would provide significant protection if one were to e.g. heat something enough to drive off all the moisture. \$\endgroup\$ – supercat Feb 21 '18 at 23:44

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