Something wrong with Ampere's Law?

Discussion in 'Physics & Math' started by eram, Oct 1, 2015.

  1. eram Sciengineer Valued Senior Member

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    We use ∇ x B = μ0 J

    Imagine a thin metal wire. We measure the curl at some distance from the wire and from the Biot-Savart law we know that it is not zero.

    However, as this point is at some distance from the wire, the current density at that point is definitely zero.

    I'm confused as to why this is the case.
     
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  3. Q-reeus Banned Valued Senior Member

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    You 'know' the curl exterior to wire is not zero? How so? You can cite a credible published paper or recent experiment backing that up? We are assuming a DC current in wire, I take it? That sort of thing needs to be carefully specified, eram. Just based on opening statement "We use ∇ x B = μ0 J", that is in fact implied. My guess is you are confusing non-uniformity with 'curl'. The former does not imply the latter.
     
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  5. James R Just this guy, you know? Staff Member

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    Yeah, I'm puzzled about the same thing.

    How do you know that?

    Are you sure you understand what curl B means?
     
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  7. Q-reeus Banned Valued Senior Member

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    Well eram, are you like many others here smitten with the appalling combo of egotism and cowardice, or are you prepared to show some gumption and meaningfully respond?
     
  8. danshawen Valued Senior Member

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    To have "current density", you must have electrons capable of electric current flow in a conducting, semiconducting or even an insulating medium. Normally, air is not a conductive medium, but ionized air is. It's called "lightning", or "electrical discharge". But you would need a very large magnetic field in order to produce this. Even an MRI resonance magnetic field of a few Tesla by itself would ordinarily not be sufficient to produce a breakdown of the dielectric insulating properties of dry air at STP.

    But the current density in the case of lighting is very high.

    What exactly are you attempting to do with this relation? The vector cross product is simply a vector means of providing a convention of direction where none actually exists. Understanding what current density means is one way to mitigate this conceptual inadequacy, but it is by no means the only way.
     
  9. James R Just this guy, you know? Staff Member

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    danshawen,

    eram wants to know why curl B = 0 near a wire carrying a constant current, and how this relates to the non-zero result for B derived from the Biot-Savart law (or, perhaps, the integral form of Ampere's law).
     
  10. Q-reeus Banned Valued Senior Member

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    That is simply adding confusion here - evidently you also do not understand the meaning of vector curl. eram made an opening mathematical statement reprenting the low frequency limit of the generalized Maxwell-Ampere law, and proceded to make a false inference via Biot-Savart expression. There is no point in introducing extraneous situations outside of the scenario given in #1.
     
  11. James R Just this guy, you know? Staff Member

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    A related example is one involving the differential form of Gauss's law for electric fields:
    \(\nabla \cdot E = \rho / \epsilon_0\)
    Consider a point in space somewhere near a point charge q. Coulomb's law says there is an electric field there. But the charge density there is zero because we're talking about a point in free space. Gauss's law in differential form says that the electric field has zero divergence at this point in space. Yet there is a finite electric field at the same point.
     
  12. danshawen Valued Senior Member

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  13. The God Valued Senior Member

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    So what ? Can't the cause for magnetic field and associated swirl be away fom that point ?

    Please try to understand the significance of Curl operator, please refer to following two links in that order, The secodn link is provided to emphasize the need of area for curl calculation, point alone will not make any sense. if confusion still persists do, come back on this point.


    http://citadel.sjfc.edu/faculty/kgreen/vector/block2/del_op/node9.html

    http://mathworld.wolfram.com/Curl.html
     
  14. danshawen Valued Senior Member

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    a point charge at that point would "feel" the "current density" of the wire, would it not?
     
  15. Q-reeus Banned Valued Senior Member

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    Only if said point charge is in relative motion. And again, nothing to do with eram's trivially correct observation there is no current or current density exterior to the wire.
    But in a way congrats - your involvement has 'ignited' this thread back to life.
     
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  16. The God Valued Senior Member

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    In case of point charge, the electric flux through a small surface is not zero if that point charge lies on that surface, otherwise it is zero. So its a matter of identifying the gaussian surface for del calculations, both del and curl are area operators.
     
  17. Q-reeus Banned Valued Senior Member

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    Careful. As per James R's expression in #8, div E (your 'del') is notionally defined at a point - as the limit of shrinking an enclosed area containing a continuous charge density. Owing to the 'granularity' of charge, that is an idealization.
    In the case of curl, a similar thing applies and it is also defined at a point. When an associated time varying E field is involved, an in-vacuo point definition of curl B is exactly correct.
     
  18. danshawen Valued Senior Member

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    And yet, there is the Hall effect.
     
  19. Q-reeus Banned Valued Senior Member

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    Last time I checked, Hall effect involved *motion* of charge carriers subject to an applied B field. Which has what to do with eram's issue in #1?
     
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  20. danshawen Valued Senior Member

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    The vector mathematical partition between electric, magnetic fields is evidently not rendered entirely by Maxwell's equations, and the Hall effect is a case in point.

    A magnetic field which penetrates a conductor or a semiconductor will cause unlike charges to separate and build a potential difference across the conductor. If the charges were initially not moving, they still separate.

    In the real world, everything moves and geometry is only an exact solution when it doesn't. Get used to it.

    Maxwell didn't mention this because Faraday did not describe it. Your math is only as good as your model. If Maxwell had really been on the ball, relativity would have been discovered much sooner.

    My humble apologies if this demolishes any philosophical need for everything to work as well in reality as it does in your head, or as described by symbols on paper.
     
    Last edited: Oct 2, 2015
  21. James R Just this guy, you know? Staff Member

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    Do you have an answer for eram, danshawen? Or are you going to keep posting irrelevancies to this thread?
     
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  22. Q-reeus Banned Valued Senior Member

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    I suggest (shudder shudder) starting a new thread challenging the validity of ME's danshawen. But be prepared for a lot of flak if doing so!
     
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  23. The God Valued Senior Member

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    He has not included dE/dt (do E by Do T) component, that can be left out for simplification.
     

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