# Relationship between electric and magnetic fields

• Voncarsteine
I'm still trying to wrap my head around this concept.In summary, the electric field strength at a certain point will be determined by voltage and current, but the electric field strength at any other point will be different because it will be influenced by the magnetic field.f

#### Voncarsteine

TL;DR Summary
I'm a bit confused; some say the relationship between E and B is E/B=c but how can that be?
So, basically I can follow the math deriving E/B = c from Maxwell.

And I can calculate B and H from I: H = I/2*pi*r and B=uH. Easy.

So, for example I take a 2000 A, 50 Hz, current and a distance of 2 meter from that current in a round conductor.
H and B are set: H = 160 A/m and B = 0,2 mT.
Note there is no voltage or E field under consideration here. Ok. So, next step is E=B*c = 60 kV/m.

But, since there is no E or V in the equations for B and H, does this mean that whatever voltage I apply the H and B are not affected by this? So, whether I apply 100 V or 400 kV, the E field is the same and not dependent on the voltage...

How is that possible? What am I missing

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Unit of E is volt/meter. Unit of B is weber/m^2= volt sec /meter^2 = volt/ meter * sec/meter. They have different units. In order both to have same unit to form electromagnetic tensor F, c is used, i.e.
$${\displaystyle F^{\mu \nu }={\begin{bmatrix}0 & -E_{x}/c & -E_{y}/c & -E_{z}/c\\E_{x}/c & 0 & -B_{z} & B_{y}\\E_{y}/c & B_{z} & 0 & -B_{x}\\E_{z}/c & -B_{y} & B_{x} & 0\end{bmatrix}}.}$$

It does NOT imply
$$E_x/c=B_x$$
They can have independent values of unit weber/meter^2.　 But as for amplitudes of electromagnetic waves
$$E/c=B$$

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Unit of E is volt/meter. Unit of B is weber/m^2= volt sec /meter^2 = volt/ meter * sec/meter. They have different units. In order both to have same unit to form electromagnetic tensor F, c is used, i.e.
$${\displaystyle F^{\mu \nu }={\begin{bmatrix}0 & -E_{x}/c & -E_{y}/c & -E_{z}/c\\E_{x}/c & 0 & -B_{z} & B_{y}\\E_{y}/c & B_{z} & 0 & -B_{x}\\E_{z}/c & -B_{y} & B_{x} & 0\end{bmatrix}}.}$$

It does NOT imply
$$E_x/c=B_x$$
They can have independent values of unit weber/meter^2.　 But as for amplitudes of electromagnetic waves
$$E/c=B$$

Thanks for your reply!
https://physics.info/em-waves/#:~:text=The ratio of the electric,to the speed of light.&text=This knowledge can then be,expression containing the electric field.
E/B = c here. With a derivation I can understand.

What is derived there is NOT a general relationship between E and B. The only relationship that always holds is Faraday's law (which is included in the text in the link) which is obviously quite a bit more complicated.
Hence, the calculation in your first post make no sense.

What is derived there is NOT a general relationship between E and B. The only relationship that always holds is Faraday's law (which is included in the text in the link) which is obviously quite a bit more complicated.
Hence, the calculation in your first post make no sense.
Hmm, ok I can live with that. But Faraday's law does state that I can calculate E from B?

So, let's take a 20 A current in a round copper wire on a 15 Vac voltage. On a distance of 2 meter from that wire, in vacuum, this current will create a B field of (4*pi)^-7 * 20 / (2*pi*2) = 20^-7 T or 0,2 uT.

This current is an ac current, thus B field changes with the same (50 Hz) frequency. It goes from 0,2 uT to -0,2 uT in 10 ms. dB/dt is then 0,4 uT/10 ms = 40 x 10^-6 T/s.

This is where it gets hazy for me. Faraday's law says this dB/dt is triangle * E. Ergo, E must be dB/dt / triangle
E = 40E^-6 / triangle. I don't get that triangle. Just don't understand what it does.

I tried just going '15 Volts at 2 meters = 15/2 = 7.5 V/m and electric field goes brrrrr' but that seems to be too easy.

Now I am not sure what you mean by E. Electro motive force for wire which Faraday's law says, or electric field at the same position you take for B ?

Now I am not sure what you mean by E. Electro motive force for wire which Faraday's law says, or electric field at the same position you take for B ?
E to me is the electric field caused by applying a voltage on the wire. That field radiates out from the wire with 1/r. EMF is the force on a charged particle CAUSED by that field.

My original question was about calculating the field strength at a certain point. I can calculate the magnetic field B and heard the rumour that E/B = c. It turned out from replies here it's not that simple.

I don't know how to calculate the electric field strength at a certain point now, with knowing only distance, voltage and current.

How can you calculate the electric field (V/m or N/C) as a function of distance knowing only voltage and current? Assume a circular (cylinder) conductor. I'm also not too good with the sphere :)

Don't shoot me though, I'm a lawyer primarily focussed on construction law in Europe. We had an issue awhile back where our client was supposed to prove the electric field was within limits, but couldn't. Their customer refused payment, and so it ended up on my desk. And I had no clue what it was so I read up a bit on the subject and found it very interesting. So here I am.

In this figure the components are current I, distant radius r and magnetic field B. That's all. E has nothing to do with it. E=0 everywhere.

You may claim that current I needs E or V in ohm's law I=V/R where R is electric resistance of wire. But V or E is not essential, e.g., we can apply superconductor as wire which carry current with zero electric field and zero voltage.

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Indeed, You wouldn't normally expect to have any E field generated by a long wire; i.e. E should be zero everywhere outside of the wire.
Note this is of course not entirely true in real life simply because no wire is infinitely long; if the frequency of the alternating current is such that the wavelength i say a quarter of the length of the wire you effectively have a radio antenna which will radiate meaning there is a free space E field.
However, in this case the wavelength at 50 Hz is 3e8/50 = 6000 km meaning this never happens.

berkeman
Another problem has another solution.
However, the wavelength at 50 Hz is 3e8/50 = 6000 km meaning this never happens.
You do not have to compare wire length with the wavelength of radiation unless you need 100 meter antenna for your AM radio.

We had an issue awhile back where our client was supposed to prove the electric field was within limits, but couldn't.
To identify the electric field due to a current carrying conductor, you must first identify the return current path for the circuit. The electric field appears between the current carrying conductor and the current return conductor. That is the circuit voltage.

To minimise electric fields and eddy currents, the wiring rules require that the circuit currents follow the same parallel path and pass through the same conduits. For that reason the distance between the conductors will be kept small and the insulated wires in the cable will probably be twisted together, which at 50 Hz, significantly reduces the stray electric and magnetic fields outside the cable. The fields due to the circuit can be simulated, mapped or measured.

Note that it is the AC voltage in a circuit that is defined, which generates the electric field, while it is the equal and opposite load dependent currents that vary, to produce the stray magnetic field.

For a radiated EM wave in free space, well away from any conductors, the relationship between E and H is through the characteristic impedance of free space;
E / H = Zo = 377 ohms.
https://en.wikipedia.org/wiki/Impedance_of_free_space

Klystron
You do not have to compare wire length with the wavelength of radiation unless you need 100 meter antenna for your AM radio.

True, what I was getting at was that you don't normally have a to worry about free space far-field radiation from 50 Hz lines.
You can obviously still have capacitive or inductive pick-up from a 50 Hz line, but that is quite different

Indeed, You wouldn't normally expect to have any E field generated by a long wire; i.e. E should be zero everywhere outside of the wire.
Note this is of course not entirely true in real life simply because no wire is infinitely long; if the frequency of the alternating current is such that the wavelength i say a quarter of the length of the wire you effectively have a radio antenna which will radiate meaning there is a free space E field.
However, in this case the wavelength at 50 Hz is 3e8/50 = 6000 km meaning this never happens.
If the wire is well away from other conductors, we expect an E-field jumping between points on the same wire, even if radiation is small. This arises because of the potential caused by current flowing through the inductance per unit length.

If the wire is well away from other conductors, we expect an E-field jumping between points on the same wire, even if radiation is small. This arises because of the potential caused by current flowing through the inductance per unit length.
That electric field gradient is parallel to the wire conductor and is typically very small.
V = L ⋅ di / dt.
A 50 Hz, 10 amp RMS current, has di/dt ≈ 4250 A/sec.
The inductance of 1 metre of 10 mm² wire will be ≈ 1.5 uH.
Therefore; V = 1.5 uH * 4250 A/s ≈ 6.4 mV per metre.

But at 50MHz it should be 10^6 times greater, or 6.4 kV/m, which is noticeable.

But at 50MHz it should be 10^6 times greater, or 6.4 kV/m, which is noticeable.
It would also be very inefficient at containing the energy.

At 50 MHz, λ = 6 metre, and you are considering the near field of a traveling wave antenna element, radiating tens of kilowatt. At 50 MHz, you cannot ignore the capacitance per unit length, and so must consider the transmission line characteristics of the conductor in it's environment, yet unspecified.
The E field is then parallel to the wire, with peak wave voltage differences every λ/2 = 3 metres apart. H will be perpendicular to E and the Poynting vector will be radial, outwards from the conductor. It will radiate efficiently.

E field is the same and not dependent on the voltage...

If you use the formula ## E=cB ##, you are dealing with radiated electromagnetic fields. In short, the power emitted by the antenna should be proportional to the square of the input voltage, and the power of the radiated electromagnetic field should also be proportional to the square of its own electric field strength, so the conclusion is that the electric field strength of the radiated electromagnetic field should be proportional to the input voltage of antenna.

Reference link https://www.antenna-theory.com/basics/main.php

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Voncarsteine
If you use the formula ## E=cB ##, you are dealing with radiated electromagnetic fields. In short, the power emitted by the antenna should be proportional to the square of the input voltage, and the power of the radiated electromagnetic field should also be proportional to the square of its own electric field strength, so the conclusion is that the electric field strength of the radiated electromagnetic field should be proportional to the input voltage of antenna.

Reference link https://www.antenna-theory.com/basics/main.php
I see, thanks. Makes sense as well.

To identify the electric field due to a current carrying conductor, you must first identify the return current path for the circuit. The electric field appears between the current carrying conductor and the current return conductor. That is the circuit voltage.

To minimise electric fields and eddy currents, the wiring rules require that the circuit currents follow the same parallel path and pass through the same conduits. For that reason the distance between the conductors will be kept small and the insulated wires in the cable will probably be twisted together, which at 50 Hz, significantly reduces the stray electric and magnetic fields outside the cable. The fields due to the circuit can be simulated, mapped or measured.

Note that it is the AC voltage in a circuit that is defined, which generates the electric field, while it is the equal and opposite load dependent currents that vary, to produce the stray magnetic field.

For a radiated EM wave in free space, well away from any conductors, the relationship between E and H is through the characteristic impedance of free space;
E / H = Zo = 377 ohms.
https://en.wikipedia.org/wiki/Impedance_of_free_space
The client depended on this document:
https://www.ijareeie.com/upload/2015/august/87_Analytical.pdf
Requirements were a maximum electric field strength of 5 kV/m and my client was responsible for design and proving requirements were met.

However;
From what I can tell, there is an error in equation 11, where top and bottom part should be switched. I checked that with taking sine instead of tangent.

Equation 17 has variable alpha in the bottom part, but no deduction on how to calculate alpha. alpha kn is clear, it's the bottom alpha that's not clear.

Equation 18 brings in variable 'r' in the right-most part. I can't find that variable anywhere in the document.

Could anybody explain these to me? Because I can't see the calculation given in the document as sufficient proof, even if it's from a 'renowned website' .

Could anybody explain these to me? Because I can't see the calculation given in the document as sufficient proof, even if it's from a 'renowned website' .

That journal appears on several lists of "predatory" journals which typically means pay-to-publish. It does not necessarily mean that the content of the paper you linked to is wrong but I would certainly not use it as a source.

That journal appears on several lists of "predatory" journals which typically means pay-to-publish. It does not necessarily mean that the content of the paper you linked to is wrong but I would certainly not use it as a source.
Hmm that's too bad. My problem is the client is now blaming me for losing their legal fight. Despite the contract being crystal clear. It's for my own satisfaction. I don't understand the mentioned equations. If somebody could please work out this calculation for me I'd sleep much better. If the calculation is correct and the requirement in the current design is actually met, it's just a case of bruised ego (mine) and I can live with that. If the calculation is inherently wrong, my client just messed up and it's no wonder I lost their case.

If somebody could please work out this calculation for me I'd sleep much better.
I have lost sleep looking at it.

The model in that paper may not be applicable to the existing situation. The question I have is about the conductivity of the construction. The electric field inside a building, or under a roof, may well be very small because the conductive shell of the building "short circuits" the many circulating 3PH electric fields. That may also enhance the electric field above the building by reducing the "ground" clearance.

1. Can you give some idea of line voltage, wire height above the ground, and the distance separating the wires.

2. Can you describe the construction type, material and size.

3. Was the field specified to be measured inside, next to, or above the structure while standing on the roof ?

4. Why could the AC electric field not be measured to verify compliance ?

I have lost sleep looking at it.

The model in that paper may not be applicable to the existing situation. The question I have is about the conductivity of the construction. The electric field inside a building, or under a roof, may well be very small because the conductive shell of the building "short circuits" the many circulating 3PH electric fields. That may also enhance the electric field above the building by reducing the "ground" clearance.

1. Can you give some idea of line voltage, wire height above the ground, and the distance separating the wires.

2. Can you describe the construction type, material and size.

3. Was the field specified to be measured inside, next to, or above the structure while standing on the roof ?

4. Why could the AC electric field not be measured to verify compliance ?
Thank you! I can:

1) It's 150 kV stations, where the requirement is that the electric field outside the fence is max 5 kV/m. The wires are 5.4 m above the ground. The wires are a bit weird: there's two wires per phase hanging 10 cm from each other. The distance between sets of these is wires is 2.1 meter.
2) it's a basic aluminium wire, 2x910 mm^2 only 3 meters long.
3) The requirement is maximum 5 kV/m field outside the fence. The wires are 5.8 meter from the fence (horizontal distance). I assume the requirement is about ground level (not top of fence, since it's not specified how to read that requirement). The field is not measured, it's not even built yet. The project is still in design phase, partly because the contractor has not demonstrated the design meets requirements. One of these requirements has become my problem.
4) I'm sure it has to be measured to confirm the design was in fact correct, but there is literal requirement that the design has to verify that the electric field is 5 kV/m max, and B field is 500 uT max.

If you could help me out here, i'll be sure to paste a picture of a box of chocolates for you here

If this is for a real-life project it needs to be designed by good engineers using the right tools.
You can't (or at least shouldn't) try to design this using formulas from random sources. The proper way to do this would -in my view- be to load the CAD drawing for the station into COMSOL, Maxwell or similar software and actually run the calculations.
There is nothing wrong with using analytical calculations as a guide as well; but real life is always complicated. Example: You mention a fence, it this metallic and grounded it could easily dramatically change the field distribution.

If this is for a real-life project it needs to be designed by good engineers using the right tools.
You can't (or at least shouldn't) try to design this using formulas from random sources. The proper way to do this would -in my view- be to load the CAD drawing for the station into COMSOL, Maxwell or similar software and actually run the calculations.
There is nothing wrong with using analytical calculations as a guide as well; but real life is always complicated. Example: You mention a fence, it this metallic and grounded it could easily dramatically change the field distribution.
It should be designed by engineers with proper tools, I totally agree with that. I don't think the contractor used software for this; they relied on math as I displayed in the original post. Their client, a HV network operator, claimed the contractor failed to comply with the contract. Specifically because they failed to demonstrate requirements in the design were met. Client refused payment. Then it landed on my desk, and I basically advised the contractor to just fix the design because they have no legal standing. They decided to go forward, and we lost the case. Now they blame me for losing their day in court. I'm literally losing sleep because of this. And I'm hoping to get help here showing me how the calculation can be done. Even better would be a calculation showing the design does in fact meet requirements. In that case my ego is bruised but the project can move forward.

I see no problem with exploring the math or model. An engineer can do the real design.
Example: You mention a fence, it this metallic and grounded it could easily dramatically change the field distribution.
If the fence is conductive, we need to know the height of the fence.
Does the fence run parallel to the wires ?

The wires are short compared to their height above ground, so the ground image will not be ideal. The support structure for the wires will affect the field, and we don't know how much is conductive and how much is insulator.

The two wires bundled together, reduce corona discharge from the line by increasing the radius of the line and reducing the electric field at the wire.

I see no problem with exploring the math or model. An engineer can do the real design.

If the fence is conductive, we need to know the height of the fence.
Does the fence run parallel to the wires ?

The wires are short compared to their height above ground, so the ground image will not be ideal. The support structure for the wires will affect the field, and we don't know how much is conductive and how much is insulator.

The two wires bundled together, reduce corona discharge from the line by increasing the radius of the line and reducing the electric field at the wire.
The fence is `1.8 meters high. Support structure is steel and earthed. It does run parallel, at a distance of 5.8 meters from the wires of the outer most phase.
I thought bundling was done to meet current requirements, 2000A i think. But thanks!