# Poynting Vector of a wire

• I
• niko_.97
In summary: I mean, couldn't the fields be generating the energy inside the wire too?The E field is generated by surface charges and the B field is generated by the current. And in turn, both of those are generated by the battery.The fields are carrying the energy in because Poynting's theorem states that the energy per unit time per unit area is flowing into the surface of the wire. The fields are not generating the energy inside the wire.

#### niko_.97

If we have a some wire (length L) with a PD of V from one end to the next and a current I moving along it we can work out the Poynting vector. It's pointing radially inwards and so tells you the energy per unit time per unit area flowing into the the surface of the wire.
What I don't understand is how the energy can be passing into the wire from the outside? At first I was thinking the fields outside carry the energy in, but that just makes it even worse because it would imply that the fields are generate outside the wire, away from the charge and current which makes even less sense (not to mention the E field is zero outside). My professor told me that it can't be the fields because it's a statics problem. Though I didn't really understand his explanation for what the actual meaning of the poynting vector here is. I know it's defined as the energy passing per unit time per unit area, but what is carrying it in a problem where the fields are static?

Dale
Excellent question. Trust the math.
niko_.97 said:
we can work out the Poynting vector. It's pointing radially inwards and so tells you the energy per unit time per unit area flowing into the the surface of the wire.
Yes that is correct.

niko_.97 said:
At first I was thinking the fields outside carry the energy in,
Yes that is also correct.

niko_.97 said:
it would imply that the fields are generate outside the wire, away from the charge and current which makes even less sense
The E field is generated by surface charges and the B field is generated by the current. And in turn, both of those are generated by the battery.

niko_.97 said:
My professor told me that it can't be the fields because it's a statics problem.
It is the fields, the Poynting’s vector is very clear about it. There is no requirement in Poynting’s theorem that restricts it in any way to non-static fields.

niko_.97 said:
what is carrying it in a problem where the fields are static?
Energy

Here is a relevant paper that I particularly like http://depa.fquim.unam.mx/amyd/arch...ia_a_otros_elementos_de_un_circuito_20867.pdf

cnh1995
Yes that is correct.
So how are the fields carrying the energy in from the outside? Especially for the electric field since it is zero outside.

The E field is generated by surface charges and the B field is generated by the current. And in turn, both of those are generated by the battery.
I'm pretty sure there's meant to be no surface charge on the wire. The electric field is created by the potential difference. A surface charge on a cylinder would create a circumferential field, not one pointing along the axis of the wire. Also, if we were to work out the pointing vector at a radius b<a, then it's magnitude would be:
S = (VI/2bLpi)(b^2/a^2) = VIb/2La^2pi. So it's non- zero at a surface with a smaller radius, so if it's a surface charge at each point contributing to that then it would become a volume charge.

Energy
What do you mean? How can you say the energy is carrying the energy?

Also, should I be thinking about these fields in the same way as electromagnetic radiation (even though its not an electrodynamics question)?

niko_.97 said:
Especially for the electric field since it is zero outside.
The E field is most definitely not zero outside the wire. See the paper that I linked to. There is a surface charge in the wire that produces a nonzero E field.

niko_.97 said:
A surface charge on a cylinder would create a circumferential field, not one pointing along the axis of the wire.
The E field is mainly radial outside the wire and the B field is mainly circumferential so the energy flow is mainly along the wire. As the paper shows, however, there is also some deviation from pure radial and pure circumferential which directs a portion of the energy flow into the wire.

niko_.97 said:
I'm pretty sure there's meant to be no surface charge on the wire.
There is surface charge there and it is essential for the proper operation of the wire. I don’t know what “meant to be” means in this context, but it is there anyway.

niko_.97 said:
What do you mean? How can you say the energy is carrying the energy?
No, the fields are carrying energy (I misread the question). That is what Poynting’s Theorem shows. It is not in any way limited to electrodynamics.

cnh1995
Ahh yeh, sorry, I don't know what I was thinking. I forgot you could derive the surface charge from the continuity equation, and of course the surface charge produces a radially outward field. Thanks for the link to the paper, it explains it pretty well!
One thing I didn't quite understand is this: if the energy comes from the battery, and the fields carry this energy along the wire and into the wire, then why do the fields not diminish as z increases (in fact in one of their models it increases), since the energy going into the wire should be lost by the fields.

It is not in any way limited to electrodynamics.
So, the fields are static. But are you saying they do in fact propagate in the direction of the poynting vector? (this leads me to ask, they can be said to be radiation...?)

niko_.97 said:
if the energy comes from the battery, and the fields carry this energy along the wire and into the wire, then why do the fields not diminish as z increases (in fact in one of their models it increases), since the energy going into the wire should be lost by the fields.
In the paper the battery is connected at the top, so the strongest fields are near the top. The fields increase as you go towards the battery, which is therefore increasing as z increases.

niko_.97 said:
So, the fields are static. But are you saying they do in fact propagate in the direction of the poynting vector?
The fields are static so the fields do not propagate at all. But energy continually flows in this static field configuration as described by Poynting. Furthermore, since they are static there is no change in the field energy, so any place where there is no current the Poynting vector must be divergence free.

cnh1995
OK, thank you for clearing this up for me! I think a lot of my confusion came from the fact that I couldn't imagine how the fields can carry the energy into the wire without them themselves moving.

Dale
I don't know, whether it helps, but here's a complete analysis of the DC current trough a (very long) coaxial cable. The only approximation used is that the (self-consistent) Hall effect is neglected, i.e., Ohm's law is used in the non-relativistic approximation $\vec{j}=\sigma \vec{E}$, which however is an excellent approximation since the drift velocity of the electrons within the wire is of the order of mm/s.

It's very clear that indeed the only energy transport along the cable is in the conductor-free space between the inner cable and the outer "shell". Within the cable the Poynting vector is purely radially inward. You find the analysis in my new manuscript on electrodynamics. Since it's written for high-school-teacher students, it's entirely using the SI units (sorry for that):

https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

For the coaxial cable with DC currents, see Sect. 3.5. The non-zero components of the Poynting vector are plotted in Fig. (Abbildung) 3.5.

vanhees71 said:
I don't know, whether it helps, but here's a complete analysis of the DC current trough a (very long) coaxial cable. The only approximation used is that the (self-consistent) Hall effect is neglected, i.e., Ohm's law is used in the non-relativistic approximation $\vec{j}=\sigma \vec{E}$, which however is an excellent approximation since the drift velocity of the electrons within the wire is of the order of mm/s.

It's very clear that indeed the only energy transport along the cable is in the conductor-free space between the inner cable and the outer "shell". Within the cable the Poynting vector is purely radially inward. You find the analysis in my new manuscript on electrodynamics. Since it's written for high-school-teacher students, it's entirely using the SI units (sorry for that):

https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf

For the coaxial cable with DC currents, see Sect. 3.5. The non-zero components of the Poynting vector are plotted in Fig. (Abbildung) 3.5.
Thanks, but the link you posted is all in German.

niko_.97 said:
Thanks, but the link you posted is all in German.
I will ask my Austrian friend for translation assistance. I am amused you suggested SI units were a language problem. I believe a certain Albert we all know wrote many papers in German

Kurt Littlewood said:
I will ask my Austrian friend for translation assistance. I am amused you suggested SI units were a language problem. I believe a certain Albert we all know wrote many papers in German
Welcome to PF, but @niko_.97 hasn't posted here since 2019...

## What is the Poynting Vector of a wire?

The Poynting Vector of a wire is a mathematical quantity that describes the flow of electromagnetic energy in a certain direction. It is represented by the symbol S and is measured in watts per square meter.

## How is the Poynting Vector of a wire calculated?

The Poynting Vector of a wire is calculated by taking the cross product of the electric field vector and the magnetic field vector at a specific point along the wire. This calculation gives the direction and magnitude of the energy flow at that point.

## What does the Poynting Vector of a wire represent?

The Poynting Vector of a wire represents the rate at which electromagnetic energy is being transferred through a specific area of space. It gives information about the direction and amount of energy flow in a certain direction.

## How does the Poynting Vector of a wire relate to the direction of energy flow?

The direction of the Poynting Vector of a wire is perpendicular to both the electric and magnetic fields, and is also in the direction of energy flow. This means that the vector points in the direction that the energy is being transferred.

## How is the Poynting Vector of a wire used in practical applications?

The Poynting Vector of a wire is used in a variety of practical applications, such as in the design of antennas and in the study of electromagnetic radiation. It is also used in the analysis of energy transfer in electrical circuits and in the development of new technologies.