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

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.

I understand why the electric field could be non-uniform across a cross section of non-uniform material. But exactly what part of the above argument (the voltage across a circular path must be zero) fails if the cross section is of non-uniform material?



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

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?



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

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.

I understand why the electric field could be non-uniform across a cross section of non-uniform material. But exactly what part of the above argument (the voltage across a circular path must be zero) fails if the cross section is of non-uniform material?



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

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.

I understand why the electric field could be non-uniform across a cross section of non-uniform material. But exactly what part of the above argument (the voltage across a circular path must be zero) fails if the cross section is of non-uniform material?