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$\Delta V_{ABCDA} = - \int_A^A \vec{E} \dot{}d\vec{l}$

The requirement that the round-trip potential difference be zero means that $E_1$ and $E_2$ have to be equal. Therefore the electric field must be uniform both along the length of the wire and also across the cross-sectional area of the wire. Since the drift speed is proportional to $E$, we find that the current is indeed uniformly distributed across the cross section. This result is true only for uniform cross section and uniform material, in the steady state. The current is not uniformly distributed across the cross section in the case of high-frequency (non-steady-state) alternating currents, because time-varying currents can create non-Coulomb forces, as we will see in a later chapter on Faraday's law.

Say the field is non-uniform but completely longitudinal. That would result in the integral above being non-zero, which is impossible, so the field always has to be uniform if its longitudinal.

If the wire is made of different types of material, the different parts would act as dielectrics, which would cause the material to polarize and dilute the electric field at different points, would cause the electric field to be non-uniform but still longitudinal, which contradicts the point made in the paragraph above.

How can I resolve this apparent contradiction?

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  • $\begingroup$ Please, reread the sentence after the one you are focusing right now with the word group "in the case of high-frequency (non-steady-state)". This sentence states that the uniform material distribution over the cross-section is not relevant for zero E-field circulation but the steady state is. A non-steady state current density causes a varying H-field which causes a varying B-field which causes a time-varying magnetic flux with causes a non-zero E-field circulation. I think the real explanation comes later in the book. Sorry, you have to wait a bit. $\endgroup$ – Tobias Feb 18 '14 at 5:39
  • $\begingroup$ @Tobias Sorry, your comment was a bit over my head since I don't know a lot about magnetism. From what I understand (correct me If I'm wrong), your saying that a non-steady state generates a magnetic field, which results in the integral being non-zero. But a point has to be equipotential with itself, and since the integral is just subtracting the voltage at the point with itself, doesn't it have to be zero? How could a magnetic field make the integral non-zero? $\endgroup$ – dfg Feb 18 '14 at 5:54
  • $\begingroup$ The E-field can only have a potential if it has zero circulation. For changing B-field the E-field does not have a potential anymore. I think the reasoning is the other way around. For a stationary current density you have zero E-field circulation. If only have a longitudinal E-field component that implies that you have a uniform E-field distribution. If the material is non-uniform over the cross-section you have non-uniform current density. $\endgroup$ – Tobias Feb 18 '14 at 5:55
  • $\begingroup$ @Tobias I made my question clearer, I hadn't explained it very well. $\endgroup$ – dfg Feb 18 '14 at 6:12
  • $\begingroup$ Regarding your edit: We do not need to take the dielectric properties into account here. Please, let us stay with a uniform permittivity to keep the example simple. The variation in conductivity over the cross-section would cause the non-uniform current density distribution. $\endgroup$ – Tobias Feb 18 '14 at 6:19
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If the electric and magnetic fields are steady, $\oint \mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0$ is a law of nature. Even if you change the material across the wire, the electric field should still respect $\oint \mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0$ and therefore must be uniform. In fact, this is one of the boundary conditions at dielectric interfaces, if I recall correctly.

However, by Ohm's law, $\mathbf{J} = \sigma \mathbf{E}$, the current density is no-longer uniform. Total current must be conserved, so if I suddenly create an interface with a different conductor of equal thickness, the electric field $\mathbf{E}$ will have to change, but it will do so uniformly, so that $\oint \mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0$ remains.

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