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I came across this question. The question asks for a solution to this problem:

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.

An answer is given below it, but it is not very formal (in the mathematical sense). The problem is that I do not understand why $\mathbf E$ is uniform in the conductor and the answer assumes this without proof.

Any hints on how to prove $\mathbf E$ is uniform and parallel to the axis (of the cylinder)?

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If it's a steady current and there is no changing charge density, then the continuity equation says the divergence of the current density is zero. That means current density cannot have a radial component in the wire, but the axial (z) component (which it obviously must have) could still depend on position in the wire.

Gauss's law tells us that must also be true for the electric field if there is no free charge density.

However, if it is a steady current then there is no changing magnetic field and the electric field is curl-free. This means that the z component of the electric field does not depend on R. If we then assume $\vec{J} =\sigma \vec{E}$, then the same is true for the current density and the z components of both must be uniform.

What about a phi component to the electric field circulating in the wire? That could be set up and could be curl-free (if it depends on R$^{-1}$). But the corresponding current would not exist in a steady state due to Ohmic dissipation ,as I assume any external EMF is applied to the ends of the wire.

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Assume that E is parallel to the axis, say along $\hat z$, but depends on $x$. This means that $0\neq {\bf \nabla} \times {\bf E} = - \partial_t {\bf B}$. Therefore the current is uniformly distributed only if $\partial_t {\bf B}$ is zero. Note that if $\partial_t {\bf B}$ is parallel to $\hat z$ then $\bf E$ cannot be parallel to $\hat z$, so this is excluded by assumption.

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  • $\begingroup$ Why is this true: "Therefore the current is uniformly distributed only if ∂tB is zero."? $\endgroup$
    – Ma Joad
    Commented Nov 3, 2019 at 12:03
  • $\begingroup$ If it is not zero $\bf E$ depends of $x$ or $y$, or both. A non-uniform $\bf E$ gives a non-uniform current via ${\bf J} = \sigma {\bf E}$ $\endgroup$
    – my2cts
    Commented Nov 3, 2019 at 12:11

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