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Using relevant equations for E and J, show that the current in a steady current I in a cylindrical conductor with uniform conductivity $\sigma$ is uniformly distributed across its cross-section.

I think the relevant equations are the divergence of the E field from the Maxwell equations and $\sigma$E=J but calculating the divergence of J using the symmetry of the problem doesn't seem to work at all. I think I might be confused about the definition of the variables or using something I should be.

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    $\begingroup$ Terribly phrased assignment. What are "relevant" equations for E and J? And the claim of constant J is true only if differential Ohm's law is assumed valid. However, Ohm's law ignores magnetic force of current on itself. How are you supposed to know magnetic force is not relevant? There are no numbers in the assignment. I would complain to the person who assigned this task to you. $\endgroup$ Mar 7, 2015 at 11:46

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Since the cylinder is an ideal conductor, the electric field inside the cylinder must be parallel to the axis of the conductor, and thus no charge should be moving radially inward or outward.

Now consider the electric field at a certain radius r from the central axis of the conductor. By your equation, $ \sigma E = J$. Sigma and E are constants no matter what the radius is, because of the problem's specifications and the fact that the cylinder is a conductor, so J must be constant. That is the general idea behind the proof, but you might need more rigor especially regarding the first paragraph.

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  • $\begingroup$ I did not want to be too specific regarding the proof because I was unsure if this might be a homework problem. $\endgroup$ Mar 7, 2015 at 6:38
  • $\begingroup$ This just assumes the answer to the question, $\endgroup$
    – ProfRob
    Nov 3, 2019 at 11:53
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By symmetry there can be no gradient of the electric field with respect to $z$ or $\phi$ (using the $R$, $\phi$, $z$ cylindrical coordinates). If the conductor is net neutral then Gauss's law tells us that the divergence of the electric field is zero. This means there can also be no electric field in the $R$ direction.

If the current is in a steady state then there will be no changing electric or magnetic fields. From Faraday's law we also know then that the curl of the electric field is zero.

We know that in an ideal conductor that ${\bf J} = \sigma {\bf E}$ and so if the current flows along the wire, there must be an $E_z$ component. But if the curl is zero then there cannot be a $\partial E_z/\partial R$ and so this component is uniform across the wire and hence the current density is also uniform across the wire.

Maxwell's equations do permit an $E_{\phi}$, as long as it depends on $R^{-1}$, which would produce a circulating current density in the same direction. But that cannot be a steady-state solution because the current would ohmically dissipate without a source of EMF to keep it going.

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If there were a uniform current density, then there would be a tangential magnetic field within the wire by Ampere's Law. Electrons traveling through that field (opposite the direction of the current) would experience an inward radial magnetic force due to current at a smaller radius. With electron density nearer the center increased, an outward electric force develops to balance the inward magnetic force on the electrons. With this balance maintained, electric flux and magnetic flux remain constant in all directions. The fault lies in assuming the conductor is ideal. If this were so, then resistivity would be zero and conductivity would be infinite. If a constant current density is proportional to electric field, then resistance must exist. This in turn allows electric energy to dissipate as heat energy. Electromagnetic energy flows inward according to ExB, thus causing the wire temperature to increase. If the wire does not heat and has no resistance, then current can flow without a driving electric field.

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