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Stock's theorem and calculation of current density

  1. Jun 3, 2013 #1

    I'm trying to use stock's theorem with the following magnetic field -
    [itex]B=1/r\hat{\theta}[/itex] on Cylindrical coordinate.

    From one side I get -
    [itex]\nabla X B=0 = \mu \int\int J \cdot dA[/itex], means that the current density is zero.

    From the other side I get -
    [itex]\oint B \cdot dl = 2 \pi r \cdot 1/r = \mu I[/itex]
    and hence
    [itex]I = 2 \pi / \mu[/itex]

    So... How the current density could be zero?

  2. jcsd
  3. Jun 3, 2013 #2


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    Hi adamp. Let me first show you a "simpler" example from electrostatics and hopefully from this example you can see what goes wrong in your calculation. Consider the Coulomb field of a point charge ##E = \frac{q}{4\pi \epsilon_0} \frac{\hat{\mathbf r}}{r^{2}}##. If we naively calculate its divergence, we find that ##\nabla\cdot E = 0## hence by the divergence theorem ##\int _{\Sigma}(\nabla\cdot E )dV = \int _{\partial \Sigma}E\cdot dA = 0##. But if we calculate the surface integral from the start we find that ## \int _{\partial \Sigma}E\cdot dA = \frac{q}{\epsilon_0}## so what's the deal?

    The problem is that our system consists of a localized point source. Recall that ##\nabla\cdot E = \frac{\rho}{\epsilon_0}## so the divergence will vanish everywhere except at the localized point source because this is the only point in space where there is a charge. As such, when we calculated the divergence of the Coulomb field, we naively ignored what happens at ##r = 0## which is where the charge density "spikes up".

    What we really have to do is represent the charge density in terms of a Dirac delta distribution i.e. ##\rho = q \delta^{3} (\mathbf r)##. Consequently, ##\nabla\cdot \frac{\hat{\mathbf r}}{r^{2}} = 4\pi \delta^{3}(\mathbf r)## i.e. this will respect the statement of the divergence theorem: ##\int _{\Sigma}(\nabla\cdot E)dV = \frac{q}{ \epsilon_0}\int _{\Sigma}\delta^{3}(\mathbf r) = \frac{q}{\epsilon_0} = \int _{\partial \Sigma}E\cdot dA##.

    Can you now extrapolate from this and see what goes wrong in your calculation?
  4. Jun 3, 2013 #3

    Thank you very much for your response.

    I still don't understand why when we are calculate [itex]\nabla\cdot E[/itex] we "naively" ignored what happens at [itex]r=0[/itex].
    Is it comes up from the divergence definition?
  5. Jun 3, 2013 #4


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    ##\nabla\cdot E = \frac{q}{4\pi \epsilon_0}\frac{1}{r^{2}}\partial_{r}(r^{2}\frac{1}{r^{2}})## we have a problem at ##r = 0## as you can see :)
  6. Jun 3, 2013 #5
    So, can we say that it impossible to use the divergence for fields which are not defined at [itex]r=0[/itex] and we need to use stokes' theorem for getting the right answer using Path integral?
  7. Jun 3, 2013 #6


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    No we can still use the divergence theorem as long as we include the dirac delta distribution, as I have done in post #2. The problem of things "spiking up" at ##r = 0## comes from us having a vanishing charge density everywhere in space except at a single point. The dirac delta distribution takes care of that.

    In your case we have a vanishing current density at all points in space except for the infinitely thin current carrying wire along ##r = 0## (now in cylindrical coordinates); we again have a sudden "spiking up" problem. The situation is analogous to what I talked about in post #2.
  8. Jun 3, 2013 #7
    Now it's all clear :-)

    Thank you again
  9. Jun 3, 2013 #8


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    Anytime. Good luck with your studies!
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