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A friend of mine is concerned about electromagnetic field negatively affecting their health and got a "DMF meter" to measure the field strength in their house in units of milliGauss. They are trying to compare the reading to a background level measured in mW/cm^2.

I am skeptical if such a measurement makes sense, but this is NOT meant to be a debate on whether cell phone towers, microwaves impact health: I just want to know if milliGauss and mW/cm^2 could possibly be referring to the same thing, and if so, what it might mean in terms of an electromagnetic field inside one's house.


P.S. The background rate they are using is 0.000000126mW/cm^2.

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I can't see how this "background level" is measured with which instrument. @ wsc: Webers were my first thought too. – Georg Jan 20 '11 at 19:32
I am not sure how they got the background level either... – hpy Jan 21 '11 at 2:07
up vote 4 down vote accepted

No, they do not. The short answer is that the Gauss (or milligauss) is a unit of magnetic field strength, but milliwatts per centimeter squared is a measure of intensity (technically known as irradiance), the amount of energy delivered per unit time per unit surface area.

For electromagnetic radiation, it is possible to convert between the two measurements, though. The time-averaged magnitude of the Poynting vector, denoted $\langle S\rangle$, gives you the irradiance of an electromagnetic wave when the wave impacts a surface perpendicular to its direction of travel. Assuming the wave has only a single frequency, you can calculate $\langle S\rangle$ from the electric or magnetic field as

$$\langle S\rangle = \frac{c}{2\mu_0}B_0^2$$

where $B_0$ is the amplitude of the magnetic field oscillations and $\mu_0$ is the magnetic constant.

Now, I'm not sure if this device your friend has actually measures the magnetic field amplitude. But if it does, you can plug that measurement in for $B_0$ and use the numerical value

$$\frac{c}{2\mu_0} = 0.119\frac{\text{mW}}{\text{cm}^2\,\text{mGs}^2}$$

(obtained from Google's calculator function) to calculate $\langle S\rangle$ in milliwatts per square centimeter.

EDIT: Something I just thought of thanks to Stan's comments: the value you plug in for $B_0$ in the formula is the magnetic field amplitude of just the EM wave, but if the device does measure magnetic field amplitude, it'll probably be picking up the Earth's magnetic field and any other contributions from, say, electronic devices in the area. So there might be a rather involved process to get from the reading on the device to the number that should be plugged into the formula. Depending on how the device works, it may not be as simple as subtracting off the value of a background reading.

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Ah I see, I will refer my friend to this information. Thank you!! – hpy Jan 20 '11 at 0:15
Isn't that the $\frac{c}{2\mu_0}B^2$ the energy flux due to the magnetic field? For electromagnetic waves, the electric field contributes just as much, so what you have for $<S>$ should be doubled. – Stan Liou Jan 20 '11 at 4:45
@Stan: since it's a sinusoidal oscillation, the time-averaged RMS value of the magnetic field is $B = B_0/\sqrt{2}$. – David Z Jan 20 '11 at 6:10
Ah, sorry--I was inattentive and assumed the device was average. – Stan Liou Jan 20 '11 at 6:47
@Stan: actually I think you are paying attention ;) the question doesn't say (and I presume penyuan doesn't know) whether the device measures the amplitude or the average or some other metric of the field strength. That's why I was careful to say in the answer that $B_0$ in the formula I quoted is not necessarily just the number that comes out of the device. – David Z Jan 20 '11 at 7:34

Are you sure they aren't webers?

1 mWb/cm$^2$ is 10 Tesla. Which is ridiculously huge. 1 milligauss, by comparison, is $10^{-7}$ Tesla.

Using @David's conversion, we would have a flux density

$B_0$ = $sqrt(\frac{\mu_0}{2c} * 1.26 * 10^{-7} mW/cm^2) \approx 5*10^{-11} T$

which is $5*10^{-4}$ millgauss... this doesn't seem right at all -- though the websites I've found all seem like they're trying to sell something and thus not worth linking to, this doesn't seem like the usual advertised background, which seems to be on the order of milligauss. (Though I'd be glad to hear what these levels should be from someone who knows!)

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Apparently the background level is: 0.000000126mW/cm^2, would that shed light on whether it is watts or webers??? I edited the original question to include this, too. – hpy Jan 20 '11 at 0:07
This seems like it might be more appropriate as a comment... – David Z Jan 20 '11 at 0:11
(although it is a good point) – David Z Jan 20 '11 at 0:19

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