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Faraday's Law - paradox

  1. Sep 27, 2015 #1
    There's something very curious about Faraday's Law that results from considering a closed curve in space (and any surface whose boundary is that curve). Forget about conducting wires and EMFs: Faraday's Law gives the result of the integral of E along the curve in terms of the rate of change of flux through the surface. This is true for ANY curve, including an imaginary one without any actual wires, loops or conductors.

    Consider any kind of magnetic field in space (such as the one around the Earth, or the Milky Way Galaxy, or just a hypothetical constant magnetic field), and assume it is constant in time. If there is no electric charge in the vicinity, there does not seem to be any source of any electric field (in the laboratory frame of reference) since the magnetic field is constant. But imagine a closed curve (say a circle, for simplicity) rotating about its diameter, or with a radius that is shrinking and expanding. By Faraday's Law, the integral of E along this curve is nonzero, so there ARE electric fields. In fact, since there are infinitely many possible curves, with different motions and distortions that can be imagined, there must be infinitely many electric field directions and magnitudes at every point in space, to produce all these various nonzero line integrals.

    This doesn't make much sense. Can anybody explain it? Thanks.
     
    Last edited: Sep 27, 2015
  2. jcsd
  3. Sep 27, 2015 #2

    andrewkirk

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    To say the magnetic field is constant requires the assumption of some rest frame O. If the circle is stationary relative to frame O, Faraday's law gives a zero integral of E around the circle. If the circle is moving relative to O, as with what you have described, the magnetic field is no longer constant, ie ##\int_\Sigma \frac{\partial\mathbf{B}}{\partial t}d\mathbf{A}\neq 0## where ##\Sigma## is the interior of the circle, because ##\mathbf{B}## is measured relative to the rest frame of the moving circle, which is not O.
     
  4. Sep 27, 2015 #3
    Oops I see my mistake. Faraday's law for a _moving_ loop requires that the "E" in the line integral be the "E" field in the instantaneous frame of reference of the line element dl, which involves a (v x B) term, where v is the velocity of the line element dl. So in fact there is an EMF but the electric field in the lab frame is zero. The EMF arises from the motion of dl in the field B.
     
  5. Sep 27, 2015 #4
     
  6. Sep 27, 2015 #5
    Andrew: thanks. The rest frame of the moving loop is not an inertial one, of course. But in the instantaneous rest frame of a line element dl, there IS an electric field. It's clear now.
     
  7. Sep 27, 2015 #6

    Dale

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    @vanhees71 I thought you had a really good post addressing this question recently, but I cannot find it right now. You showed the differential and the integral form of Maxwells equations and how the integral form is modified for moving boundaries.
     
  8. Sep 28, 2015 #7

    vanhees71

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  9. Sep 28, 2015 #8

    Dale

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    Yes, that is the one I was thinking of. Thank you!
     
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