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I'm sure that this question was addressed here before, but I failed to find any other instances, so with your permission I ask the question myself.

I'm experiencing a very disturbing glitch, there is a fallacy somewhere in my logic, but I can't put a finger on it.

I was considering a cross section of a circular wire (radius $R$) carrying DC current with magnitude I. Let's say for now that this wire is a perfect conductor.

If we calculate the magnetic intensity adjacent to the wire's surface, we get $H = I/2\pi R$.

Since the fields inside the wire are zero, one could easily calculate the surface current density $K$:

$\oint K\,R\,\mathrm{d}\phi = \oint H\,R\,\mathrm{d}\phi \Rightarrow K = H = I/2\,\pi\,R$. ($\mathrm{d}\phi$ is the differential of the polar angle). Since $K = I/2\,\pi\,R$, that means that the surface currents hog all the current that flows through the wire. Thus, inside the wire the current density is zero.

Please someone explain to me how one arrives at the conclusion that DC current is uniformly distributed across a wire's cross section, taking Maxwell's equations as a starting point.

Thank you in advance.

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  • $\begingroup$ Electric fields must be zero inside a perfect conductor, and thus magnetic fields must be constant (or else their change would induce an electric field), but stationary magnetic fields are okay. $\endgroup$
    – tparker
    Commented Jun 15, 2016 at 23:26
  • $\begingroup$ @tparker that is true for a PEC but a PMC allows E fields to form inside. $\endgroup$
    – M Barbosa
    Commented Jun 15, 2016 at 23:32
  • $\begingroup$ For DC currents, the fields inside the wire are not zero, and the current density as well as the E-field within the wire is uniform across the cross section. See this poster , and study the simulation mentioned therein. It's a simulation starting with Maxwell's equations, but not an analytical solution. $\endgroup$
    – garyp
    Commented Jun 16, 2016 at 2:27
  • $\begingroup$ @MBarbosa What is PEC and PMC? $\endgroup$
    – garyp
    Commented Jun 16, 2016 at 2:28
  • $\begingroup$ @garyp a PEC is a perfect electrical conductor, a theoretical material that has infinity conductivity, zero permittivity and zero skin depth (see below as well). And since magnetics has an analog to everything electric, there is also a theorized perfect magnetic conductor which has infinite magnetic conductivity (there is also a magnetic current $M$) zero permeability and zero skin depth for magnetic source points (if such things exist). $\endgroup$
    – M Barbosa
    Commented Jun 16, 2016 at 3:07

2 Answers 2

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The flaw in your reasoning is in assuming that there are no magnetic fields inside the conductor. Equilibrium electric fields cannot exist in a perfect conductor, but magnetic fields can as long as they are not changing in time. Empirically, the charge density is quite uniform through the wire's cross-section, so the current does indeed flow through the bulk, and there are concentric loops of magnetic field: see here.

Explaining this theoretically is quite complicated, because there are two competing effects: the fact that parallel currents attract tends to concentrate the currents at the center of the wire, but is counterbalanced by the fact that the moving charges' electrical repulsion resists such charge clustering. Things only get more complicated when you consider the relativistic Lorentz contraction of the moving charges. See here for a simple model that tries to incorporate these effects.

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There's something called the skin effect which is roughly what you described in your question. The skin effect forces charges to aggregate on the surface of a conductor. How far from the surface the charges exist is called the "skin depth", which has a low-frequency approximation of $\delta = \sqrt{\frac{2\rho}{\omega \mu}} $. For a perfect electrical conductor (PEC) the resistivity ($\rho$) is 0 so its skin depth is 0, meaning the charges really are clinging to the surface. For DC, one can say that the wavelength is infinite, which means the frequency is 0. Plugging directly into that formula will cause a division by 0 error, but without analyzing we can see that skin depth goes up as the frequency goes down. So the skin depth for DC is sort of infinity, but this really just means that the skin will cover the whole cross-sectional area.

Edit: for the DC the equation is actually in indeterminate form (as pointed out by @tparker). Regardless, the skin effect is what determines where the charges flow.

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    $\begingroup$ Thank you for your responses. Unfortunately, that doesn't make things clearer for me. Qualitatively I understand why the current is uniform inside the wire. If we consider a non-ideal conductor, like copper wire, how would one go about calculating current density? I need math, please... $\endgroup$ Commented Jun 15, 2016 at 23:36
  • $\begingroup$ @SergeyTheCobaltChoob no problem, but I showed you qualitatively and quantitatively. The skin depth approaches $R$ as the frequency approaches 0. So the closer we get to DC, the closer the skin depth becomes to $R$. Think of $\delta$ as a limit on how far the charges can go into the conductor. If that limit is equal to the radius, then the charges will spread out through the entire cross-section. $\endgroup$
    – M Barbosa
    Commented Jun 15, 2016 at 23:40
  • $\begingroup$ For a perfect conductor carrying DC current, both $\rho$ and $\omega$ are $0$ so the expression is indeterminate, and this equation is not at all useful. $\endgroup$
    – tparker
    Commented Jun 16, 2016 at 1:05
  • $\begingroup$ @tparker ah! Good catch, that is true in the second case, don't think it was worth the down vote though. The equation is still valid regardless of what inputs it gets, it IS the low-frequency approximation for $\delta$ after all. And also I acknowledged that I wouldn't plug in anyway because of the division by 0. $\endgroup$
    – M Barbosa
    Commented Jun 16, 2016 at 1:12
  • $\begingroup$ The problem is that you can use that exact same formula to argue that the charge is clustered around the surface, or spread uniformly throughout the bulk. In fact, you did argue both positions in your post. Since the entire question was to determine which of these cases is true, then this expression gives you absolutely zero information relevant to the OP's question. $\endgroup$
    – tparker
    Commented Jun 16, 2016 at 1:53

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