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I Poynting Vector of a wire

  1. May 16, 2018 #1
    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?
     
  2. jcsd
  3. May 16, 2018 #2

    Dale

    Staff: Mentor

    Excellent question. Trust the math.
    Yes that is correct.

    Yes that is also correct.

    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.

    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.

    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
     
  4. May 16, 2018 #3
    So how are the fields carrying the energy in from the outside? Especially for the electric field since it is zero outside.

    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.

    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)?
     
  5. May 16, 2018 #4

    Dale

    Staff: Mentor

    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.

    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.

    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.

    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.
     
  6. May 17, 2018 #5
    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 in to the wire should be lost by the fields.

    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...?)
     
  7. May 17, 2018 #6

    Dale

    Staff: Mentor

    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.

    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.
     
  8. May 17, 2018 #7
    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.
     
  9. May 18, 2018 #8

    vanhees71

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    Gold Member
    2017 Award

    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.
     
  10. Jun 5, 2018 #9
    Thanks, but the link you posted is all in German.
     
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