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I How does the Poynting vector factor into a normal circuit?

  1. Jan 15, 2017 #1
    So, the Wiki page on the Poynting vector has this image:

    1920px-Poynting_vectors_of_DC_circuit.svg.png

    I remember hearing/reading somewhere that the energy transmission in a circuit like this is actually not travelling through the wire, but that it actually happens through the electromagnetic field, I.e. essentially the Poynting vectors.

    At the same time, I feel this is too wild a claim given how something like Ohm resistance is due to "pushed" electrons bouncing against atoms.

    So, is this a mixed influence, or is the usual analysis of circuits just a convenient, but eventually inaccurate way of looking at things?
     
  2. jcsd
  3. Jan 15, 2017 #2

    Dale

    Staff: Mentor

    I wouldn't say that circuit analysis is inaccurate. It is based on certain simplifying assumptions, but as long as those assumptions are met it is accurate.

    The problem is that people read too much into it. Circuit theory describes how much energy is transferred, but it never makes any claim about where energy is located within a circuit element nor where energy crosses a lumped element's boundary. There is no conflict here because circuit theory makes no claim on the question.
     
  4. Jan 15, 2017 #3
    So, what is the role of the wires, if not the means of energy transmission? Is it simply creating the field which then transmits the energy straight from the battery?
     
  5. Jan 15, 2017 #4

    Dale

    Staff: Mentor

    Yes. In particular, it contains the currents which are the source of the magnetic field.
     
  6. Jan 15, 2017 #5
    How does this work transfer happen in a DC circuit like this? Since the magnetic field is constant, nothing gets induced in the resistor, so what part of the field actually transmits the energy to the resistor? That is, how does the "electrons bouncing off atoms" play into this which eventually transmits the energy to the atoms in the resistor?
    If the electrons got accelerated by the field, thus providing the energy that makes them bounce against the atoms, I could see that. But a static magnetic field won't accelerate electrons.
     
  7. Jan 15, 2017 #6

    Dale

    Staff: Mentor

  8. Jan 15, 2017 #7
    Ooh, that is an excellent paper, thanks a lot!
     
  9. Jan 16, 2017 #8

    Baluncore

    User Avatar
    Science Advisor

    In a nutshell.

    The potential difference between the conductive wires guides the electric field.
    The currents that flows along the conductive wires guide the magnetic field.

    The cross product of the electric and magnetic fields gives the Poynting vector = energy flow.
    The product of the voltage and the current gives the rate of energy flow = power towards the load.

    The resistance of the wires, multiplied by the square of the current that is needed to guide the magnetic field, results in a loss of energy along the wires before that energy can reach the load.
     
  10. Jan 16, 2017 #9

    anorlunda

    Staff: Mentor

    As @Dale said, simplified is a better word than inaccurate. I say thank god that we can do the simplification or else we could never design circuits.

    Consider the following four pictures. They depict four different ways to lay out the wires implementing the simple circuit you showed in the OP. According the simplifications of circuit analysis, the could all be equivalent, and when checked with a voltmeter and ammeter, give identical results. Yet when analyzed with Poynting vectors, they are wildly different. It might take a lifetime of analysis to prove that the four are equivalent using Maxwell's Equations.

    slask.jpg
    slask2.jpg
    slask3.jpg
    slask4.gif
     
  11. Jan 16, 2017 #10
    So, if I understand that paper @Dale linked to correctly, while it is true that the Poynting vector permeates all space around the setup, due to the surface charges on the wires, the electric and magnetic fields are guided in a way that a large fraction of the energy travels parallel to the wire, just outside of it. At the resistor these vectors then point inward, which means the energy flows inside the resistor.

    @anorlunda , I think that guiding action is probably also the reason why all those circuits can be analyzed in the simplified manner, since virtually all energy flows right outside the wire.

    On a side note, reference 8 of that paper quoted Feynman, who was stymied by this too. Amazingly, it seems only in recent years have people figured out how this works even for the most basic circuit.
    EDIT: This lecture:. http://www.feynmanlectures.caltech.edu/II_27.html. The relevant section starts at 27-5.
     
    Last edited: Jan 16, 2017
  12. Jan 16, 2017 #11

    Dale

    Staff: Mentor

    Yes, you are correct.
     
  13. Jan 16, 2017 #12
    So, to *some* degree this makes sense, but what is still unclear to me is how this eventually relates to the atomic level. Is the idea here that instead of the usual view of electrons being pushed along the wire, it is actually the external electric field enters the resistor and pushes those electrons? (since it is a resistor and not a wire, the electric field actually enterering the resistor)

    Is the "electrons pushed along the wire" idea entirely wrong, or just negligible to the energy transmission?

    Sorry for the million questions, but I am mesmerized by that despite my Master's in EE, I apparently have no clue how even the most basic circuit operates.
     
    Last edited: Jan 16, 2017
  14. Jan 16, 2017 #13

    Dale

    Staff: Mentor

    I don't know enough quantum electrodynamics to answer anything at the atomic level. I can only help at the classical level. However, if you are willing to simply accept Ohm's law, ##J=\sigma E##, as a valid classical law in a resistor then I can help.

    Personally, I find Ohm's law to be conceptually sufficient and don't see any need or value to talk about atoms or electrons. So I deliberately avoid mixing classical and quantum concepts.

    The key is to recognize that the surface of the conductors and resistors have surface charges. A surface charge leads to a discontinuity in the E field. Outside the conductor the E field is roughly perpendicular to the surface of the conductor, but inside the conductor the E field is in the direction of the current flow.

    Here is a reference showing this in a rather qualitative manner.
    https://www.tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf
     
    Last edited: Jan 16, 2017
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