Density of Electrons inside cylinder wire

[Mahbod|Druid]

Hi

assume J = I/A = equal in a circular segment of wire

then after a while those electrons forces other electrons to come closer to center but not all of them
(some thing like Haal effect (spell?) )


I want to find out the function of Density of electrons--Distance from center of wire

in books they assume all the electrons move in a line but actually they don't in my Question


https://www.physicsforums.com/showthread.php?t=313514"

 

[Born2bwire]

They will be more or less uniformily distributed on the surface.

 

[Mahbod|Druid]

on surface ?

but Wire has not any Extra charge so why should electrons go on surface ?

 

[Born2bwire]

Sorry, the currents will be confined to the surface. The electrons on the interior will more or less be unaffected.

 

[Bob S]

The resistivity of copper is about p = 17 nano-ohm-meters. Therefore the resistance of a copper wire L= 100 meters long and .001 meter diameter (A= 7.85 x 10^-7 m²) is
R = pL/A = 2.17 ohms.
This assumes the current is flowing uniformly throughout entire cross section of wire. This value agrees with my table of wire resistance for 18 gauge wire. The resistance would be much higher if the current flowed only on surface. In fact, the depth of the current penetration toward the center of the wire depends on the inverse square root of frequency, and is called the skin depth. See
http://en.wikipedia.org/wiki/Skin_depth

 

[Born2bwire]

The resistivity of copper is about p = 17 nano-ohm-meters. Therefore the resistance of a copper wire L= 100 meters long and .001 meter diameter (A= 7.85 x 10^-7 m²) is
R = pL/A = 2.17 ohms.
This assumes the current is flowing uniformly throughout entire cross section of wire. This value agrees with my table of wire resistance for 18 gauge wire. The resistance would be much higher if the current flowed only on surface. In fact, the depth of the current penetration toward the center of the wire depends on the inverse square root of frequency, and is called the skin depth. See
http://en.wikipedia.org/wiki/Skin_depth

If it was a perfect conductor (zero resistance/infinite conductance) then the current would flow only on the surface. The conductivity of copper is on the order of 10 M-mhos. So I would say you need to get down to low frequencies, like on the order of the 10 kilohertz before you hit a skin depth on the order of 1 mm.

If he is talking about DC currents then that would be an interesting thought exercise. With the voltage source sourcing and sinking the charges then there wouldn't be a charge differential setup to cancel out the applied electric field like we normally would assume in electrostatics. I would assume that the current will then be more or less distributed evenly throughout the wire. But under those conditions then the electron density will be uniform, just moving.

So I think the answer remains the same. With AC currents, the current is confined to surface of a perfect conductor and for a good conductor until you hit around the very low frequencies. And for DC the current is uniform I think (provided you can apply a uniform voltage across the cross-section of the wire) so in both cases the electron density is mostly uniform. Between DC and VLF things get sketchy, it would depend upon the frequency for the most part. The currents will not be uniform or symmetric through the cross-section of the wire. I remember seeing an applet that showed the EM wave for a twisted pair at a given frequency. It would be nice if I could find that again as it would give an idea of how the currents are.

But as for the charge distribution, I'm not sure. The voltage source is sourcing and sinking the charges so I'm not sure how to relate the EM waves of a transmission line and the charge distribution inside the conductors. We generally assume that the medium is source free when we solve the wave equations and so analysis of the electric field will not yield a charge distribution, I don't think.

 

[Bob S]

So I think the answer remains the same. With AC currents, the current is confined to surface of a perfect conductor and for a good conductor until you hit around the very low frequencies. And for DC the current is uniform I think (provided you can apply a uniform voltage across the cross-section of the wire) so in both cases the electron density is mostly uniform. Between DC and VLF things get sketchy, it would depend upon the frequency for the most part. The currents will not be uniform or symmetric through the cross-section of the wire. I remember seeing an applet that showed the EM wave for a twisted pair at a given frequency. It would be nice if I could find that again as it would give an idea of how the currents are.

But as for the charge distribution, I'm not sure. The voltage source is sourcing and sinking the charges so I'm not sure how to relate the EM waves of a transmission line and the charge distribution inside the conductors. We generally assume that the medium is source free when we solve the wave equations and so analysis of the electric field will not yield a charge distribution, I don't think.

I remember measuring the inductance per unit length of a coaxial transmission line, specifically RG-8. Below about 100 kHz the inductance was higher than the inductance above about 1 MHz. The inductance at low frequencies includes magnetic fields inside the center conductor, which are excluded at high frequencies. Because capacitance is a surface property, the charactistic impedance of RG-8 (Z = sqrt(L/C) ) dropped by roughly an ohm or so.

Note: The characteristic impedance of a transmission line is Z = sqrt(L/C) where L is the inductance per unit length, and C is the capacitance per unit length.

 

[Born2bwire]

I remember measuring the inductance per unit length of a coaxial transmission line, specifically RG-8. Below about 100 kHz the inductance was higher than the inductance above about 1 MHz. The inductance at low frequencies includes magnetic fields inside the center conductor, which are excluded at high frequencies. Because capacitance is a surface property, the charactistic impedance of RG-8 (Z = sqrt(L/C) ) dropped by roughly an ohm or so.

Note: The characteristic impedance of a transmission line is Z = sqrt(L/C) where L is the inductance per unit length, and C is the capacitance per unit length.

What's your point?