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In summary: Yes. The electric field and magnetic field are both in volts per meter. They are both measured in Tesla.

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So 100uV/m corresponds to a magnetic field strength of E/377 = 38mA/m approx.

It does not depend on the type of transmitting antenna provided you are more than about lambda/6 away from it. The ratio may vary if the receiver is surrounded by buildings, trees etc as these can alter the impedance of the medium in that area.

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$$\vec{E}=\vec{E}_0(\vec{x}) \exp(-\mathrm{i} \omega t), \quad \vec{B}=\vec{B}_0(\vec{x}) \exp(-\mathrm{i} \omega t),$$

where the physical fields are understood to be the real parts of this complex field. From Faraday's Law

$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$$

you get

$$\frac{1}{\mathrm{i} \omega} \vec{\nabla} \times \vec{E}_0(\vec{x}) = \vec{B}_0(\vec{x}),$$

i.e., a more complicated relation.

For a plane wave with wave-vector ##\vec{k}## and ##\omega=c |\vec{k}|## you have

$$\vec{E}_0(\vec{x})=\hat{\vec{E}} \exp(\mathrm{i} \vec{k} \cdot \vec{x})$$

with ##\hat{\vec{E}}=\text{const}##. Then

$$\vec{\nabla} \times \vec{E}_0(\vec{x})=-\hat{\vec{E}} \times \vec{\nabla} \exp(\mathrm{i} \vec{k} \cdot \vec{x}) = +\mathrm{i} \vec{k} \times \hat{\vec{E}} \exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$

From this you get

$$\vec{B}_0(\vec{x}) = \frac{\vec{k}}{\omega} \times \hat{\vec{E}} \exp(\mathrm{i} \vec{k} \cdot \vec{x})=\frac{1}{c} \frac{\vec{k}}{|\vec{k}|} \hat{\vec{E}} \exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$

The most general field you get by Fourier integrals wrt. ##\vec{k}## from these plane-wave modes.

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OK thanks. I need my units to be in Tesla. So, I went and used a conversion tool. 38mA/m is 0.047652 micro Tesla. That's the magnetic field strength of an electromagnetic wave whose electric field strength is 100 microvolts/m. If I made no errors. The page I found for the conversion tool deals with electric and magnetic field, not electromagnetic fields. Hope that does not represent a problem.tech99 said:

So 100uV/m corresponds to a magnetic field strength of E/377 = 38mA/m approx.

It does not depend on the type of transmitting antenna provided you are more than about lambda/6 away from it. The ratio may vary if the receiver is surrounded by buildings, trees etc as these can alter the impedance of the medium in that area.

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It's a good thing you don't use Gaussian units. There, B=E and this thread would not exist.

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My understanding was that the electric field strength and magnetic field strength in the Gaussian cgs system were related by a number which happens to be the speed of light. That is what gave Maxwell the original idea about c.Meir Achuz said:It's a good thing you don't use Gaussian units. There, B=E and this thread would not exist.

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In an electromagnetic wave, the energy is split evenly between the magnetic and the electric component of the wave. I suppose this is the reason that it's possible to take an electric field strength value, in volts per meter, apply a conversion factor (number) and state the corresponding magnetic field strength in Tesla. Is it this even split that allows for the conversion? Yes or no will do. I am conscious that we are talking about electromagnetic waves, not simply an electric or a magnetic field. Thanks.

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Yes, the energy is divided equally.richard9678 said:

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Heaviside-Lorentz units is the best comprimise for theoretical physics. If you set in SI units ##\mu_0=\epsilon_0=1## and then also automatically ##c=1##, both systems are indeed the same ;-)).Meir Achuz said:It's a good thing you don't use Gaussian units. There, B=E and this thread would not exist.

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As to #13 above. The anwser I was seeking is, is the fact that you can convert an electric field strength into a corresponding magnetic field strength down to the fact that energy in the electromagnetic wave is equally split? If the answer were to be "yes" that might mean conversion was kosher for electromagnetic waves, but not really in the case of converting an electric field strength into a magnetic field strength. I'm just not sure whether it does make sense seeking to convert an electric field strength into a magnetic field strength (not electromagnetic wave), but find no problem at all when an electromagnetic wave is involved.

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richard9678 said:converting an electric field strength into a magnetic field strength

BvU said:Maxwell's equations ##\qquad## !

##\ ##

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I keep talking about converting, when what I'm actually after is not so much a conversion but establishing or calculating. If I take the value of the electric field, of an electromagnetic wave, I can manage to take that value and establish the magnetic field strength. As it stands, I've established that a wave with an electric field strength of 100 microvolt / m has a magnetic field strength of 0.047652 micro Tesla. Unless I've made a mistake somehow. Can anyone tell me if my figure is indeed correct? Thanks. EDIT: Non-the-less, I took "So 100uV/m corresponds to a magnetic field strength of E/377 = 38mA/m approx." and I think I then converted it (38mA/m) to Tesla. And hoping I've not made an error.

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Can folks please appreciate that I have not studied much math and physics, and that I've been somewhat confused in some ways because of my initial failure to recognise that it's a calculation to determine the magnetic field strength of a certain value of electric field strength (waves), but a matter of a conversion thereafter to express the magnetic field in terms of Tesla. Now, it has been stated: "So 100uV/m corresponds to a magnetic field strength of E/377 = 38mA/m approx.". I'm not responsible for this statement, so it's no way right that I should be held responsible, if it's incorrect in any way. I'm not an expert. Can I be given some space rather than being criticized? Although maybe any criticism that I have seen was not directed at me. If so, OK and I don't have a problem.

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I thought at first, you were getting on my case, but now I don't think you are. Apologies. So, it's not possible to measure the value of the electric field, in any circumstances, of a real electromagnetic wave (like from a radio transmitter) and calculate from that the value of the magnetic field? That's what you are saying.vanhees71 said:

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tech99 said:My understanding was that the electric field strength and magnetic field strength in the Gaussian cgs system were related by a number which happens to be the speed of light. That is what gave Maxwell the original idea about c.

The number c was found to be the ratio of the emu charge to the esu charge.

The Gaussian system uses this to use the esu charge (statcoulomb) and write I/c for magnetic effects.

There is a c in Curl E=-(1/c)d_t B, which makes B=E in plane waves.

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I think you should use what @BvU said way back in post 4. For a plane wave, you haverichard9678 said:

$$E=cB \quad \Rightarrow B = \frac{E}{c} = \frac{100~{\rm \mu V/m}}{3.00\times 10^8~{\rm m/s}} = 3.33\times 10^{-13}~\rm T.$$

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I think what is being said is that if your radio transmitting antenna (radio wave, not light wave) consisted of a dipole way out in space, using no transmission line feed (Tx in between the dipole elements) and the receiving antenna was also a dipole, again with no coax, the Rx between the dipole elements, the distance between dipoles being significantly greater than the wavelength- then in such a situation there would be a very very good aproximation to a plane wave at the receiving dipole. But, in the what we might call the real world, in a case where the electric field strength (of a radio signal from a terrestrial Tx) is accurately measured, it's highly unlikely that the magnetic field strength will actually be as calculated. This is what I'm getting from comments. There would be a need to actually measure the magnetic field strength, "in the real world".

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It only takes a few wavelenghs of propagation away from the TX antenna to get to a plane wave, so for most real-world RX EM signals, that's what interacts with the RX antenna. You can measure the E field strength via the RX voltage (and a frequency conversion factor), and from that calculate the B-field strength.richard9678 said:But, in the what we might call the real world, in a case where the electric field strength (of a radio signal from a terrestrial Tx) is accurately measured, it's highly unlikely that the magnetic field strength will actually be as calculated. This is what I'm getting from comments. There would be a need to actually measure the magnetic field strength, "in the real world".

Sorry if you already said it and I missed it, but is there a reason that you want to be able to calculate the B-field of an EM waveform?

https://interferencetechnology.com/antenna-fundamentals/

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I think that ##E=cB## holds for all kinds of waves (not only plane) because we can easily prove it for plane waves and any wave is a superposition of plane waves.vela said:I think you should use what @BvU said way back in post 4. For a plane wave, you have

$$E=cB \quad \Rightarrow B = \frac{E}{c} = \frac{100~{\rm \mu V/m}}{3.00\times 10^8~{\rm m/s}} = 3.33\times 10^{-13}~\rm T.$$

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This isn't correct; ##|\mathbf{E}| = c|\mathbf{B}|## is only true forDelta2 said:I think that ##E=cB## holds for all kinds of waves (not only plane) because we can easily prove it for plane waves and any wave is a superposition of plane waves.

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tech99 said:

Excerpt from the citation:berkeman said:It only takes a few wavelenghs of propagation away from the TX antenna to get to a plane wave, so for most real-world RX EM signals, that's what interacts with the RX antenna. You can measure the E field strength via the RX voltage (and a frequency conversion factor), and from that calculate the B-field strength.

Sorry if you already said it and I missed it, but is there a reason that you want to be able to calculate the B-field of an EM waveform?

https://interferencetechnology.com/antenna-fundamentals/

The categorization of antennas in this way is somewhat artificial, however, since the actual mechanism of radiation involves both electric and magnetic fields no matter what the construction.

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Hi eto, glad to see you.ergospherical said:This isn't correct; ##|\mathbf{E}| = c|\mathbf{B}|## is only true forplane, monochromaticwaves. (Recall the non-linearity of the ##|\bullet|## function).

Yes I think in the most general case you are correct, however what made me wrote this is that I had in mind the so called linearly polarized waves, where the direction of the fields is everywhere constant and at all times. I think then the modulus function becomes linear and the relation holds for any kind of wave shape, not only sinusoidal(monochromatic) .

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I don’t understand your point. Your claim is untrue.

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Yes my original claim is untrue. Hold on why I make my point more clear...

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