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Contradiction in Maxwell?

  1. May 30, 2010 #1
    Okay, the title is promising something big. I'm sorry, it's probably not big, but it does seem important, although for some it will be unsignificant, I suppose it depends on your interests in physics:

    So if you have an infinite plane with current density J (current per meter width) (say pointing up in the y direction, in the xy-plane), you have a magnetic field independent of distance, pointing in the +/-x direction (dependent on which side) with formula:

    [tex]B = \frac{\mu J}{2}[/tex]

    This is easily derived from the Ampère-Maxwell law (Bds = mu*I will suffice).

    But what if we use the general result that if we have one (infinite) conducting wire with current I, the magnetic field at a distance a is:

    [tex]B = \frac{\mu I}{2 \pi a}[/tex]

    Then if we go back to our infinite J-plane, and we fix ourselves on a certain point on the z-axis looking at the infinite plane, say (0,0,d), we can say a dx-segment of the plane contributes a dB field with

    [tex]dB = \frac{\mu J dx}{2 \pi \sqrt(d^2+x^2)}[/tex]

    since dI = Jdx (from the definition of J) and we use the fact that the distance 'a' from a certain dI to the fixed point on the z-axis is a = sqrt(d² + x²) with d the distance to the closest dI.

    However, taking the integral from x = - infinity .. infinity, it diverges!

    Why can't I calculate it this way?
     
  2. jcsd
  3. May 30, 2010 #2

    Born2bwire

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    You are ignoring the fact that the magnetic field is a vector. If we have an infinite line of current along the z-axis, the magnetic field is directed along the \hat{\phi} direction. You have assumed that all the magnetic fields contributing from your line currents are directed along the same direction, thus adding up without any cancellation. You need to take this into account by converting the magnetic field to a vector in cartesian coordinates and then finding the appropriate direction based on the position of the wire.
     
  4. May 31, 2010 #3
    you need an extra factor of

    [tex]
    \cos \theta = \frac{d}{\sqrt{d^{2} + x^{2}}}
    [/tex]

    for the x component. The other components will cancel and the integral converges now.
     
  5. May 31, 2010 #4
    Oh of course :) how silly, thank you!
     
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