Relation between electric & magnetic fields in terms of field strength

[richard9678]

Hi. A electromagnetic wave consists of an electric and a magnetic component. I believe that the electric field strength is measured in volts per meter. The magnetic field I think is measured in Tesla. Let's imagine that I measure the electic field strength of two different radio stations and the electric field strength measured is 100 microvolts per meter. However, let's say one station is employing a dipole and the other a magnetic loop antenna. My question is, would the strength of the magnetic component of the wave be the same for each station? Thanks.

 

[BvU]

Maxwell's equations say yes !

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[richard9678]

If the magnetic field strength, associated with a wave having an electric field strength of 100 microvolts, is always of the same value, irrespective the type of transmitting antenna used, then it would be possible to draw up a table with one column consisting of electric field strength values and the other magnetic field strength. Is there such a table? Thanks.

 

[BvU]

You mean a table for ${E\over B} = c$ ?

Or in Hyperphysics: here

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[richard9678]

I may be looking for a formulae that takes the value of the measured electric field strength, of an electromagnetic wave, which gives me the magnetic field strength of that wave.

 

[Ibix]

Yes. That's what @BvU gave you. It really is as simple as dividing by $c$ for a wave in free space.

 

[tech99]

The ratio of the electric field strength to the magnetic field strength, E/H, is equal to the impedance of free space, which is 377 Ohms approx. https://en.wikipedia.org/wiki/Impedance_of_free_space
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.

 

[vanhees71]

But this simple relation holds for plane-wave modes only. Take the more general case of fields of harmonic time dependence. Then you can represent the fields with complex valued fields of the form
$$\vec{E}=\vec{E}_0(\vec{x}) \exp(-\mathrm{i} \omega t), \quad \vec{B}=\vec{B}_0(\vec{x}) \exp(-\mathrm{i} \omega t),$$
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.

 

[richard9678]

The ratio of the electric field strength to the magnetic field strength, E/H, is equal to the impedance of free space, which is 377 Ohms approx. https://en.wikipedia.org/wiki/Impedance_of_free_space
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.

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.

 

[Meir Achuz]

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

 

[tech99]

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

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.

 

[richard9678]

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.

 

[tech99]

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.

Yes, the energy is divided equally.

 

[vanhees71]

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

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 ;-)).

 

[richard9678]

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.

 

[BvU]

converting an electric field strength into a magnetic field strength

Maxwell's equations $\qquad$ !

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[richard9678]

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.

 

[vanhees71]

Once more! The simple equal split of electric and magnetic energy (density) holds only for plane-wave modes, it's not a generally valid property of real-world electromagnetic waves. A plane-wave mode can never be realized in nature, because it's containing an infinite total energy.

 

[richard9678]

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.

 

[richard9678]

Once more! The simple equal split of electric and magnetic energy (density) holds only for plane-wave modes, it's not a generally valid property of real-world electromagnetic waves. A plane-wave mode can never be realized in nature, because it's containing an infinite total energy.

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.

 

[Meir Achuz]

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.

 

[vela]

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.

I think you should use what @BvU said way back in post 4. For a plane wave, you have
$$E=cB \quad \Rightarrow B = \frac{E}{c} = \frac{100~{\rm \mu V/m}}{3.00\times 10^8~{\rm m/s}} = 3.33\times 10^{-13}~\rm T.$$

 

[richard9678]

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".

 

[tech99]

The only report I have seen of the magnetic and electric fields differing from 377 Ohms is for long wave reception in an urban area. Otherwise, ignore the issue. Similarly for the wave not being truly plane. Other propagation issues are much greater.

 

[berkeman]

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".

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/

 

[richard9678]

The reason I was wanting to calculate the magnetic field is that I'm examining ferrite rod antennas, and have assumed that published field strengths for radio stations have traditionally or by convention only been expressed in terms of the electric field. Whereas I needed the magnetic field strength value. Also, I was not aware at the time that field strength meters can give you the magnetic field strength as well as the electric.

 

[tech99]

I have used broadcasting stations to obtain a known field strength for calibrating a short dipole for E-field and a small multi-turn loop for H-field. If you go to an open position 1km from a 1kW transmitter you will have 300mV/m and 300/377 mA/m.

 

[Delta2]

I think you should use what @BvU said way back in post 4. For a plane wave, you have
$$E=cB \quad \Rightarrow B = \frac{E}{c} = \frac{100~{\rm \mu V/m}}{3.00\times 10^8~{\rm m/s}} = 3.33\times 10^{-13}~\rm T.$$

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.

 

[ergospherical]

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.

This isn't correct; $|\mathbf{E}| = c|\mathbf{B}|$ is only true for plane, monochromatic waves. (Recall the non-linearity of the $|\bullet|$ function).

 

[Klystron]

I have used broadcasting stations to obtain a known field strength for calibrating a short dipole for E-field and a small multi-turn loop for H-field. If you go to an open position 1km from a 1kW transmitter you will have 300mV/m and 300/377 mA/m.

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/

Excerpt from the citation:

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.

 

[Delta2]

This isn't correct; $|\mathbf{E}| = c|\mathbf{B}|$ is only true for plane, monochromatic waves. (Recall the non-linearity of the $|\bullet|$ function).

Hi eto, glad to see you.

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) .

 

[ergospherical]

@Delta2, for example, consider two oppositely traveling plane waves which interfere to produce a standing wave. Clearly $|\mathbf{E}| \neq c|\mathbf{B}|$.

 

[Delta2]

@Delta2, for example, consider two oppositely traveling plane waves which interfere to produce a standing wave. Clearly $|\mathbf{E}| \neq c|\mathbf{B}|$.

ehm yes right but only if their E-fields are of different directions.

 

[ergospherical]

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

 

[Delta2]

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