# $\mu$T in to $\frac{W}{m^2}$ (for interpreting EMF readings)

I am considering purchasing an EMF reader, to collect data about what is being thrown off of power lines and various other sources in the house to reach some conclusions.

An issue is, the meter can only read in Teslas or gauss. This is fine, however a Tesla is equal to $\frac{Wb}{m^2}$ and one Weber being one volt-second or joule/ampere and I get confused.

My intention is just to derive the total power at that specific point that the source is radiating.

Is there a direct conversion from µT to $\frac{W}{m^2}$ I am missing?

Since Teslas are measuring the magnetic flux density of the low frequency radiation coming off of power lines, does that skew my result if it is not also measuring the electric field? Would I be better off creating an antenna/inductor and measure µW (or similar) induced directly in different areas?

Update: Seems I cannot get reasonable numbers with the following formulas (from "in plane waves, poynting vector, see ):

$B = 0.000004$ (4μT)
$B_0 = \frac{1}{c}E_0$, so, $E_0 = \frac{B_0}{\frac{1}{c}} = 1199.169832\frac{V}{m}$

so.. $S = \frac{1}{\mu_0} E\times B = 3817.07613$ which = $\frac{3.8kW}{m^2}$)

or.. $S = \frac{cB_0^2}{2\mu_{0}} = 1908.538065$ which = $\frac{1.9kW}{m^2}$

I am probably doing something grossly wrong, I need to realise my mistakes so I can calm my mind and learn how to apply these formulas. I assume the $B_0$ means at time 0, which I can consider the peak field amplitude in my purposes, no?

Tesla unit resource: http://en.wikipedia.org/wiki/Tesla_%28unit%29
Poynting vector: http://en.wikipedia.org/wiki/Poynting_vector

• Yes, $B_0$ means peak value. Perhaps the problem is that you measure combined constant Earth's magnetic field and harmonic electromagnetic wave's magnetic field $B = B_\text{Earth} + B_0 \sin(k x - \omega t)$? Edit: probably not, $B_\text{Earth} \approx 50 \mu$T. – Pygmalion May 5 '12 at 6:22
• Calculation seems fine to me. 2 kW/m$^2$ doesn't look like an impossible result to me. Depends on where (distance from the source and power of the source) and how you measure magnetic field. – Pygmalion May 5 '12 at 6:32
• See perhaps this: lipower.org/pdfs/company/pubs/brochures/emf.pdf – Pygmalion May 5 '12 at 6:39
• @Pygmalion, great for comparing, but I may as well just be comparing 2kW/m^2 of radiation over 300 metres to a fridge magnet (which has a more powerful B field for size), Teslas by themselves seems not very useful for energy or safety, wonder why all sorts of "EMF detectors" are advertised for safety or detection or leakage, when magnetic radiation could mean anything. 10000T might mean nothing, 10000W does. Grr, maybe I'll have to ditch EMF reading, not listen to people complaining about uT's bombarding their house, and come up with my own Watts. :) – Alexander. May 5 '12 at 9:13
• Still widely unsure of the results I've written. I can't imagine ever presenting this to someone, or keeping those general numbers in mind for later reference. Unless someone can guide me in what exactly is wrong (or right), why those numbers differ, and tell me if it is at all possible to get Watts from Teslas, they're not useful to me. – Alexander. May 5 '12 at 9:19

You might want to calculate Poynting vector (which corresponds to intensity [W/m$^2$] of electromagnetic waves):
$$\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B},$$
$$\langle S \rangle = \frac{c B_0^2}{2 \mu_0}.$$
• I can derive E from B (apparently, as $B_0 = \frac{1}{c}E_0$) to use in the ExB formula with same results, S = ~2.9 billion. Wiki says B can be "magnetic flux density" which is Teslas, so I am unsure why this result is so high. This could help me with future formulas if I figure what is wrong here. – Alexander. May 4 '12 at 21:25