Skin depth of zero thickness in DC circuits

[fisico30]

Hello Forum,

the formula for skin depth is both inversely proportional to conductivity and frequency omega.
There seem to be 3 cases when the skin depth of a metal is exactly (or tends to) zero:

1) if an EM wave of frequency f (any frequency) is incident on a perfect metal wall (infinite conductivity sigma). The current is a surface current that exists only on the surface of the ideal conductor;

2) If a AC battery is connected to a load via wires with finite resistivity. The AC current only distributes itself on the periphery of the conductor with the current density being largest near the surface of the conductor;

3) In the case of a DC battery connected to a load via ideal wires (sigma=infinity). We read in books that DC current flows uniformly across the cross-section of the conductor. However if the wires are perfect conductors the electric field E inside is equal to zero! How is that DC current generated then if E=0 and there is no F=q*E force to push charges and make a current...
I read that even in the DC case, if the wires have infinity conductivity (as postulated in intro book) ,the current ends up flowing only on the skin a not inside the conductor as most book say...
What is the true story as far as 3) goes?

I thought that in the DC case there was no skin effect...

thanks
fisico30

 

[Antiphon]

For 3), when you close the circuit there is a wave of electric and magnetic fields that starts going down the wire (in the space around the wire.) The current is pushed by the inductance of the wire behind the wavefront and pulled by the natural capacitance just ahead of the wavefront. Resistance of the wire is zero and has nothing to do with the current flowing.

Finite resistance *leads to* an electric field along the wire. It's not the cause of the current but an effect of it.

 

[fisico30]

Hi Antiphon,
so you say that "...The current is pushed by the inductance of the wire behind the wavefront and pulled by the natural capacitance just ahead of the wavefront. Resistance of the wire is zero and has nothing to do with the current flowing..."

That seems to be equivalent (correct me if I am wrong) to say that there are surface charges on the conductors caused by battery...They causes the capacitance and their motion the inductance that then cause the current.

"...Finite resistance *leads to* an electric field along the wire. It's not the cause of the current but an effect of it..."

Not sure I understand or agree with it: charges move if there is a force on them. The electric force is F=q*E. If there is finite resistance a force is needed and that is caused by an electric field being present. So I don't see how the E field is an effect and not the cause...

In the theoretical case 3) do you agree that a DC current flows uniformly across an ideal conductor or do you believe it flows on the skin instead?

thanks
fisico30

 

[Antiphon]

Yes to your first remark about the motion along the wire.

For the second remark, F=qE + qvXB. The electric term is not the only one. Think of an ideal conducting ring with a current flowing in it. Now if you introduce a finite but small resistance in the ring, a tangential electric field will appear suddenly which was not there before. No battery. The electric field is created by forcing current through the resistance. If you quickly take away the resistance, it's the voltage that drops to zero with it, not the current.

For an ideal conductor, the current flows in an infinitely thin sheet at the surface if it is AC current. To answer the DC question you must know the history of the fields. It is possible to cause DC currents to circulate indefinitely inside a perfect conductor.