Electric field in wire's cross section?

Question



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

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?

Answer 1

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.