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What are the fields produced around a current carrying conductor?

  1. Oct 12, 2014 #1
    If we consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e, it would not have any net positive or negative charge). Remember dipole has zero net charge, but it does have electric field around it. So, if net charge is zero, it doesn't mean there is no electric field.

    It is important to notice that, if we assume only electrons to be moving, and kernels (positive nuclei) to be static, magnetic field will be produced only due to electrons.

    Does it mean that electric field and magnetic field exists around the current carrying conductor?
    Or
    Does it mean that only magnetic field exists around the current carrying conductor?

    The question is simple, but I have found varied answers until now.
     
  2. jcsd
  3. Oct 12, 2014 #2

    Dale

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    No, Gauss' Law ( http://en.wikipedia.org/wiki/Gauss's_law ) says that if there is no net charge inside a closed surface then the total electric flux through that surface is 0. There may still be an E field, but it will point inwards at some points and outwards at other points such that the total flux is 0.

    A conductor with some finite resistivity will have a slight E-field as well as a B field. A superconductor carrying a steady current will have only a B field.
     
  4. Oct 13, 2014 #3
    Yes, that is what I said before.

    Is there any source for this statement?
     
  5. Oct 13, 2014 #4

    sophiecentaur

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    Which part of the statement? (There are two.)
    If there is Resistance then there must be a Potential Drop so there will be an E field (Volts per Metre). That's normal text book stuff.

    If it's a superconductor then the charges will arrange themselves on the conductor to fit the minimal potential condition (same as if it were not carrying a current). Any net E field round the conductor will be due to its absolute potential (i.e. its place in the external circuit.
     
  6. Oct 13, 2014 #5
    This is OK.

    Not clear for me.

    Particularly, Dalespam stated that "A superconductor carrying a steady current will have only a B field." I wanted the source for this statement.
     
  7. Oct 13, 2014 #6

    Dale

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    The E field for a current carrying material is ##\sigma E=J##. The source of this is Ohm's law: http://en.wikipedia.org/wiki/Ohm's_law. Ohm's law is also covered in every textbook that covers EM physics or circuit analysis. This E field is small for ordinary conductors (large ##\sigma##) and 0 for superconductors (##\sigma = \infty##).

    EDIT: corrected for mistake indicated by cabraham. See below.
     
    Last edited: Oct 13, 2014
  8. Oct 13, 2014 #7

    sophiecentaur

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    I don't think a "source" is necessary. Is it not just very basic? There will be no PD across a superconductor . There can be no fields parallel with the surface. What field can their be, apart from a field due to any net charge?
     
  9. Oct 13, 2014 #8

    Dale

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    It is OK, asking for sources is always appropriate here on PF. I provided a Wikipedia link, and Ohm's law is also in every textbook on basic EM physics and circuit analysis.
     
  10. Oct 13, 2014 #9

    ehild

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    It depends whether the current is stationary or it changes with time.
    The stationary current produces stationary magnetic field around the wire: the field lines are concentric circles in planes perpendicular to the wire.
    The time dependent current produces time dependent magnetic field. The time dependent magnetic field induces electric field.
    Time dependent electric and magnetic field are inter-related, described by Maxwell's equations.

    ehild
     
  11. Oct 13, 2014 #10
    I believe you have the right general idea, but you typed ##\sigma## where you should have typed ##\rho##. I recall Ohm's law in 3-D form as:

    ##J=\sigma E##, or ##E=J/\sigma##, or ##E=\rho J##. For a superconductor ##\sigma## is not zero, but infinite, whereas ##\rho## is zero. I assume you meant that. No big deal. Best regards. :)

    Claude
     
  12. Oct 13, 2014 #11

    Dale

    Staff: Mentor

    Oops, you are 100% correct. I will go back and correct my post.

    I make that mistake often because I think in terms of resistivity but I dislike the usual symbol for resistivity since it is the same as the usual symbol for charge density. I really should think in terms of conductivity.
     
    Last edited: Oct 13, 2014
  13. Oct 14, 2014 #12
    J=E/ρ, if ρ=0, J=∞, i.e current per unit area is infinite, so infinite charge particles are moving per unit time per area. Do superconductors have infinite charge particles?

    Let's restrict our discussion to stationary currents for this time.
     
  14. Oct 14, 2014 #13

    davenn

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    "stationary current" is a contradiction

    the word current by definition is something that flows

    do you really mean a static charge ?
     
  15. Oct 14, 2014 #14
    I meant what echild meant above. To be explicit I didn't mean static charge.
     
    Last edited: Oct 14, 2014
  16. Oct 14, 2014 #15

    ehild

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    Stationary current means constant current.
    It produces magnetic field.
    Stationary current is maintained by potential difference between the end of the wire, that means electric field existing along the wire. The tangential component of the electric field is the same at both sides of the interface: both inside and outside of the wire. But I would not say that the electric field is produced by the current.

    ehild
     
  17. Oct 14, 2014 #16

    davenn

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    so you mean DC rather than AC ?
    if so, why not just say DC ?
     
  18. Oct 14, 2014 #17

    Dale

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    J is finite. E is 0.

    There is a critical current density above which a superconductor becomes resistive.
     
  19. Oct 15, 2014 #18

    ehild

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    I mean stationary current which is not the same as DC. DC - direct current - means current flowing in the same direction all time. But it can change with time.
    AC - alternating current - alternates its direction so the time average of the current is zero.

    ehild
     
  20. Oct 15, 2014 #19
    Not the correct answer mathematically.

    The equation suggests J to be ∞ . Your explanation should match with the equation results.
     
    Last edited: Oct 15, 2014
  21. Oct 15, 2014 #20

    Dale

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    The math is correct. Your claim here is not right. Your statement would only be correct if E were nonzero.

    The observed facts are 0 E and finite J. What value of ##\sigma## in ##J=\sigma E## do you think matches those observed facts? If not ##\sigma=\infty## then there must be some finite ##\sigma## which satisfies it. What do you think is that value?
     
    Last edited: Oct 15, 2014
  22. Oct 16, 2014 #21
    @DaleSpam:

    Considering your claims to be true, (i.e J=finite value, ρ=0, E=0), we have

    J ≠ E/ρ

    Finite value ≠ 0 / 0, your values are not agreeing with the equation.

    Your claim here is not right.
     
  23. Oct 16, 2014 #22
    Dale's claim is quite right. ##E## = ##\rho####J##. Do you agree? If ##J## is finite and ##\rho## is zero, would you expect ##E## to be anything other than zero? As far as the "0/0" thing goes, that ratio can be finite. In a superconductor, ##E## & ##\rho## are both zero while ##J## is finite. No conflict, because a ratio of 2 quantities can be finite when the numerator & denominator both vanish. Again, "0/0" could be finite, or zero, or infinite. In calculus, there is a rule known as "l'Hopital's rule". If you've studies calculus, you already know that a ratio where top & bottom both vanish can be zero, infinite, or finite. I suggest you review the rule of l'Hopital. Best regards.

    Claude
     
  24. Oct 16, 2014 #23

    Dale

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    0/0 is undefined as is ##\infty ~ 0##. It certainly may be finite. It only tells you that you need a different method to find the value. In this case J and E are measured experimentally. J is finite E is 0. Those are known values. The unknown is ##\sigma## and there is no finite value of ##\sigma## for which ##J=\sigma E## holds. Therefore, we say ##\sigma=\infty##.
     
  25. Oct 16, 2014 #24
    To illustrate this issue using a mechanical example, consider the equation ##F = ma##. If an object of mass ##m## is at rest, then the net force acting on it must be zero. The acceleration is zero as well. If we tried to ascertain the mass from this information we get ##m = F/a = 0/0##, which is indeterminate. The mass is non-zero, non-infinite, but force (summation) and acceleration are both zero. With any "##y=mx##" type of equation this is applicable. Does this help? :-)

    Claude
     
  26. Oct 16, 2014 #25
    Read the article from Arvix.org: http://arxiv.org/pdf/1202.1851v1.pdf

    Here is a passage from that article:

    "For a spherical geometry, infinite cylinder and infinite
    plane, no electric fields exist outside the superconductor.
    For other geometries however electric field lines do exist
    outside the superconductor, implying that the surface is
    not equipotential"

    So, I am under impression that electric field may exist around a superconductor.
     
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