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Why isn't the magnetic flux of a loop infinite?

  1. May 4, 2012 #1
    If B~1/r2, then if we have a simple loop, B near the inner edge of the loop will be infinite (or close to it). Why then, would our flux not be infinite?

    I also get infinity if I take
    ∫ ∇ X A *da =∫B*da =∫(closed)A*dl
    Since A ~1/r and r~0 at the limit of our surface integral.

    I know I am missing something simple here, help! If possible, show where I'm wrong in both of my attempts.
     
  2. jcsd
  3. May 4, 2012 #2

    Philip Wood

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    You're taking the wire carrying the current as having no thickness. If you consider the wire as nested cylindrical shells, you'll find (e.g. using Ampère's law) that each of these contributes a B outside itself but not inside. Therefore as you approach the wire from outside it, and then enter the wire, the field does not go to infinity.
     
  4. May 4, 2012 #3
    That would seem to indicate that flux through a wire loop is very dependent on the thickness of the wire. That doesn't seem right to me. Furthermore, I would be able to make a similar argument for a solenoid for which (to any reasonable approximation) wire size makes no difference. Are you sure that's the only factor I'm missing?
     
  5. May 4, 2012 #4
    Your paradox is more general than just the flux case. It is about integration. As you put it, in distances very close to the wire, the field density is very height ( not really infinite), but when you integrate for the flux, you also multiply the high value by an infinitesimal surface. The integral is not necessarily large then.
     
    Last edited: May 4, 2012
  6. May 4, 2012 #5
    The line integral for x-2 diverges near zero, was I wrong in assuming the surface integral (I guess for [r0 - r]-2) similarly diverged?
     
  7. May 4, 2012 #6

    Philip Wood

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    I haven't calculated the flux for the loop, but I have considered a simpler case; the flux produced by a long straight wire (length L) inside the wire itself (radius a) and in the vacuum between it and a cylindrical metal sheath (radius b) acting as return wire. I find, using elementary maths, that
    [tex]\Phi = \frac{\mu_0 I L}{2\pi} ln\frac{b}{0.61a}[/tex]

    So, in this case, the flux is indeed crucially dependent on the radius of the wire. [The factor of 0.61 is due to the flux inside the wire, and assumes a uniform current density over the wire's cross-section, so no skin effect, for example.]

    Some quick thoughts about the solenoid... The local B on the inside wall of the solenoid due just to the adjacent wire is of the order of [itex] B_{loc} =\frac{\mu_0 I}{2\pi r_{wire}}[/itex], whereas the general field in the solenoid is [itex] B_{gen} =\frac{\mu_0 N I}{L_{sol}}[/itex].

    But for a single layer coil [itex] N2r_{wire}=L[/itex].

    Thus [itex]\frac{B_{loc}}{B_{gen}}= \frac{1}{\pi}[/itex]

    suggesting that perturbations to B near the solenoid wall due to the discreteness of the turns
    of wire won't be very significant.

    [This just an order of magnitude estimate. I'd be the first to acknowledge the crudeness of this treatment.]
     
    Last edited: May 4, 2012
  8. May 4, 2012 #7
    To my knowledge, close to a thin wire, B is proportional to 1/r and A to ln(r) ( because [itex]B=\nabla \times A =\frac{\partial A}{\partial r}[/itex]) . As we move toward the center of the loop, the dependence become less . But now I think their integrals are till infinite as you said.

    I found in Wikipedia that the inductance of a circular coil depends on ln(r/a) where r is the loop radius and " a" is wire radius. So as "a" goes to zero, the inductance becomes infinite which, for a finite current, makes the flux infinite too.
     
  9. May 4, 2012 #8

    Philip Wood

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    Yes, the integrals are infinite, but only because you're letting a go to zero (which I think is an odd thing to do)!
     
  10. May 4, 2012 #9
    This is actually very interesting ( and odd!). I thought thicker power transmission lines have a higher inductance than thinner ones.
     
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