How electron transport in conductors

[chenhon5]

Say we have a solid wire made of Cu. It seems most electrons transport at the surface of the wire. Right? So is there any difference of the resistance if we have a hollow Cu wire (same diameter with the solid wire). I think the answer shoule be "yes". Right? Becasuse, the resistance R=p*L/A (p is the resistivity).

But if the electron transport are only in the surface of the wires at both cases(solid and hollow wires) , why there is difference of the resistance?

 

[ImAnEngineer]

Say we have a solid wire made of Cu. It seems most electrons transport at the surface of the wire. Right? So is there any difference of the resistance if we have a hollow Cu wire (same diameter with the solid wire). I think the answer shoule be "yes". Right? Becasuse, the resistance R=p*L/A (p is the resistivity).

But if the electron transport are only in the surface of the wires at both cases(solid and hollow wires) , why there is difference of the resistance?

Why do you think electrons are only (or mainly) moving at the surface?

 

[Bob S]

For DC currents, the current density is uniform throughout the entire wire, so the resistance per unit length scales as 1/(pi R²). Thus any plane perpendicular to the length of the wire is an equipotential. For high frequency AC, the effective conduction is only near the surface, with a skin depth d, so the effectve area is only 2 pi R d and the wire resistance per unit length scales as 1/(2 pi R d).

The skin depth d is defined as d= sqrt(2/(w sigma mu)), and
w = 2 pi frequency
sigma = conductance (amps per meter²/volts per meter)
mu = permeability of conductor (copper = 4 pi 10^-7 henrys per meter)

 

[chenhon5]

Why do you think electrons are only (or mainly) moving at the surface?

I think I made a mistake and misunderstood something. And Bob give me a very clear and good explanation. Electron mostly transport in the surface with high frequency AC, not in all cases. Thanks a lot, Bob.

 

[ImAnEngineer]

For high frequency AC, the effective conduction is only near the surface, with a skin depth d, so the effectve area is only 2 pi R d and the wire resistance per unit length scales as 1/(2 pi R d).

Is there a simple explanation for this phenomenon or will it be quantum mechanical? I've never heard about it and it sounds rather peculiar to me. I'm just curious :) .

 

[Bob S]

We know that at very high requencies, the energy of a propagating electromagnetic signal is contained in transverse E and H waves, and that the Poynting vector S = E x H represents both the power and direction. When confined between resistive conductors (waveguide or coax cable) the waves penetrate the metal, and the waves attenuate because the Maxwell equation for curl H has a resistive component (where sigma is conductiviry and epsilon is permittivity).

del x H = sigma E + epsilon dE/dt

This resistive component leads to resistive losses (sigma E) and attenuation as the wave penetrates the metal conductor. The depth of penetration is called the skin depth mentioned above.

 

[ImAnEngineer]

We know that at very high requencies, the energy of a propagating electromagnetic signal is contained in transverse E and H waves, and that the Poynting vector S = E x H represents both the power and direction. When confined between resistive conductors (waveguide or coax cable) the waves penetrate the metal, and the waves attenuate because the Maxwell equation for curl H has a resistive component (where sigma is conductiviry and epsilon is permittivity).

del x H = sigma E + epsilon dE/dt

This resistive component leads to resistive losses (sigma E) and attenuation as the wave penetrates the metal conductor. The depth of penetration is called the skin depth mentioned above.

Thanks for the explanation!