Does electricity flow on the surface of a wire or in the interior?

I was having a conversation with my father and father-in-law, both of whom are in electric related work, and we came to a point where none of us knew how to proceed. I was under the impression that electricity travels on the surface while they thought it traveled through the interior. I said that traveling over the surface would make the fact that they regularly use stranded wire instead of a single large wire to transport electricity make sense.

If anyone could please explain this for some non-physics but electricly incline people, I would be very appreciated.

• The dominant path for conductors is through the conductor and not on the surface. – Brandon Enright Apr 25 '14 at 1:51
• View a wire as a collection of many thin cylindrical shells. The outer shells have more cross sectional area compared to inner ones. All have same length. Hence, resistance is less outwards. View this as a parallel combination of these and you will see that current is more on the outer part of wire. – evil999man Apr 26 '14 at 7:12
• @Awesome The current density is the same in all shells (i.e. the current per unit cross-sectional area). – David Knipe Apr 27 '14 at 10:08
• @Awesome I'm pretty sure that's not what OP was asking. The same current goes through all regions that have the same area. (your shells do not have the same area) – Navin Apr 30 '14 at 3:11
• @Navin Doesn't the outer region has more area?$A=2\pi x dx$ – evil999man Apr 30 '14 at 3:14

It depends on the frequency. DC electricity travels through the bulk cross section of the wire.

A changing electrical current (AC) experiences the skin-effect where the electricity flows more easily in the surface layers. The higher the frequency the thinner the surface layer that is usable in a wire. At normal household AC (50/60hz) the skin depth is about 8-10mm but at microwave frequencies the depth of the metal that the current flows in is about the same as a wavelength of visible light

edit: Interesting point from Navin - the individual strands have to be insulated from each other for the skin effect to apply to each individually. That is the reason for the widely separated pairs of wires in this question What are all the lines on a double circuit tower?

• Pretty sure all power generation systems in the US run at 60 Hz — the "high voltage" just refers to the amplitude of the signal. For microwaves, where skin depth effect means that most of the metal volume is not conducting, you have to use waveguides to carry signals around. – rob Apr 25 '14 at 4:27
• It means that even very high power cables will be made of a number of thinner wires because once they are more than 1/2" thick the center is not being used efficiently. – Martin Beckett Apr 25 '14 at 5:31
• Note that normal stranded wire will not improve the situation since the current still sees it as one large wire. Litz wire prevents this by alternating the "inside" and "outside" wire. – Navin Apr 25 '14 at 8:03
• Very long distance power transmission can actually be DC rather than AC so no skin effect there. But I think most transmission is AC. As others said, the frequency, not the voltage, is the key here – Adam Apr 25 '14 at 22:33
• @Navin Fascinating, I've never seen such wires. I'm wondering whether the shape of the cross-section could also be used to make these high-frequency cables more efficient by using something that has a higher perimeter-to-area ratio than a circle. Triangles would even pack better than circles. Heck, there are even fractals that tessellate ;) – Christian Apr 28 '14 at 21:41

Stranded wire is used because it bends more easily, but it has essentially the same conductive properties.

Current flows throughout the entire wire. This is easily tested by measuring the resistance of round wires - the resistance will fall quadratically with the radius, indicating that it's the cross-sectional area that matters.

Amendment: this answer is only correct for direct current - see Beckett's below for AC. The changing magnetic fields introduce eddy currents which yield the skin effect, where current tends to be carried only within the "skin depth" of the wire, which is not proportional to the radius.

• You are assuming the resistance $R=\rho L/S$, where $S$ is the area of the wire that current flow(not necessarily the entire cross section of the wire) , I wonder if this holds for the AC frequency current, for $\rho$ is also change with the frequency. – an offer can't refuse Apr 25 '14 at 4:23
• @C4stor that's right, it doesn't verify that there isn't some $r$-dependence of the amount of current flowing. It does verify, however, that current isn't simply a "skin" property, where the current flow is limited to a fixed distance from the edge (or similarly, the center). In other words, while there might be some variation, it's fundamentally an area thing, not a circumference thing. The exact details of where the current flows are less interesting :P – Scott Lawrence Apr 25 '14 at 7:00
• It seems wrong to ignore the AC effects. See Wikipedia, it doesn't play a role in household power distribution but it's significant when radius exceeds 1 cm. – Blackbody Blacklight Apr 25 '14 at 9:02
• Another reason for the wire to be stranded is so that if there is a defect at any point and breaks, the breach is contained to a very small portion: that single fibre. – Davidmh Apr 27 '14 at 22:35
• The fact that, for certain types of AC, current runs only skin-deep is also why powerlines have a less-conductive steel core (for strength) with a more-conductive shell that runs the bulk of the current. – Vincent Vancalbergh Apr 28 '14 at 8:59

This is a bit unrelated to the original question, but it's worth mentioning that this can arise as a common misconception due to the fact that static electricity accumulates on the surface of a conductor. While this is true, it's correct that current tends to flow through the bulk of a conductor, and current density is measured in units of $\text{A}/\text{m}^2$.

Also, Martin's answer makes a good point, the skin effect is relevant for AC currents, but unless you're dealing with inch-thick wire, it won't really make a difference. At higher frequencies, stranded wire might help a little bit, but it would still be susceptible. There are special ways to strand wire (like the litz wire to mitigate/negate the effect, but that wouldn't be needed for mains electricity.

• Great example of litz wire! – user12029 Apr 25 '14 at 20:50

As already mentioned, the conductivity is both theoretically and empirically proportional to the cross-sectional area, not the circumference. An intuitive explanation (for DC or low frequency AC) results from the forces between moving electrons as opposed to static ones. Think of it as Ampere's Law, Maxwell's Equations, or the relativistic nature of electromagnetics -- either way, electrons moving in parallel directions attract. So, the actual cross-sectional current distribution would result from the net forces (both attractive and repulsive) of electrons as they course through the wire. I'm not about to calculate that distribution, and a quick search did not find it. Might check J. D. Jackson -- I don't have my copy any more. Anyway, the force of attraction between parallel moving electrons is the key to why electricity flows through the bulk of the wire as opposed to just on the surface (where static charges would reside).

I'd rather have just commented, but since I got an account here just because of this, I will attempt an answer, but cannot help but try to redirect some of the commentary here.

Simple answer: Yes, in an ideal case. If you construct the model you will see that that current density shrinks to zero at the centerline of the conductor, where E vector is zero. This takes some work beyond the statement of Maxwell's Equations.

Reality is of course not so cut and dried. But the gradient of current density is still very significant. Do you want to know why Nikolai Tesla could demonstrate the phenomenon using his own body? Well, here you have it.

So, use stranded wire for speaker cables, ipod jacks, etc. It's total current capacity (due to heat) is lower, so don't wire your house with it.

Finally, the separation of power transmission lines is to reduce losses due to capacitive coupling. But while we're on the subject, check out Hoover Dam. There you can buy a section of the original transmission line from the dam to the grid. It's copper, made of interlocking radial cross-section parts. And yes, it's hollow. For 60Hz.

There you go.

To abatter: Please try to understand the concept of current density in a conductor.

• Stranded cable is used for speakers because it is more flexible. Solid cable is used for permenent installs because it isn't flexible so won't move and potentially fray. Solid cable is used even for low current data cables in buildings. Capacitive coupling isn't an issue in the power cables in the link because the sep strands are on the same phase and at the same potential. Finally the hollow conductors at the hoover dam are probably to allow cooling rather than to provide a 2nd surface to reduce skin losses. – Martin Beckett Nov 1 '14 at 15:16

In the case of alternating current, the current density drops exponentially with distance from the outer surface of the wire (the "skin effect"), as explained by Martin Beckett. This can be shown analytically from the quasistatic approximation to Maxwell's equations, as is done in Jackson chapter 5.

The case of direct current is more interesting. First, you need to specify the external electric field ${\bf E}_0$ that "pushes" the current. This is usually taken to be uniform and parallel to the wire. The currents through the wire tend to attract each other and therefore cluster together (known as the "pinch effect"). The DC pinch effect is discussed in http://aapt.scitation.org/doi/abs/10.1119/1.1974305, http://aapt.scitation.org/doi/abs/10.1119/1.14075, and http://aapt.scitation.org/doi/abs/10.1119/1.17271. It turns out that Maxwell's equations are not enough to uniquely determine the current density distribution through the wire's cross-section; you also need to specify a microscopic model for the charge carriers.

At one extreme, you can treat both the positive and negative charge carriers as completely mobile and with equal charge-to-mass ratios. This is a good description of current conduction through plasmas, and plasma pinches can be strong enough to crush metal.

At the other extreme, you can treat the positive charges as completely stationary in the lab frame, at fixed density, and "immune" to the electromagnetic fields, with the current due entirely to the motion of the mobile negative charge carriers. This is a more realistic model for a metal wire, since the interatomic and Fermi exchange forces between copper atoms are much, much stronger than those induced by typical applied fields and electron currents. It turns out that in the lab frame, the wire's total linear charge density must be zero at equilibrium (otherwise it would exchange electrons with the fixed sources and sinks at the battery until it neutralized), but in the rest frame of the moving electrons, the bulk volume charge density must be zero (otherwise the electrons would experience a radial electric force drawing them toward or away from the wire's axis).

Combining these requirements, you get the following picture: define $R$ to be the wire's radius, $\rho_0$ to be the density of positive ions in the lab frame (in which they are at rest), $\beta = v/c$, where $v$ is the electron's drift velocity as seen in the lab frame, and $\gamma = 1/\sqrt{1-\beta^2}$. In the lab frame, the bulk positive volume charge density is $\rho_0$ and the bulk negative volume charge density is $-\gamma^2 \rho_0$, which is greater in magnitude. So the bulk net volume charge density $(1 - \gamma^2)\rho_0 = -\beta^2 \gamma^2 \rho_0$ is negative, and there is a radially inward electric field whose magnitude increases linearly with radius. (The internal generation of this radial electric field is sometimes called the "self-induced Hall effect.") The electric field balances out the radially inward attraction between electrons due to the current flow. There is a compensating positive surface charge density $\sigma = (R / 2) \beta^2 \gamma^2 \rho_0$ around the surface of the wire which balances out the negative bulk volume charge, so the radial electric field vanishes outside the wire. This surface charge is at rest in the lab frame, so it does not contribute to the current.

In the electrons' frame, there is no bulk volume charge density or radial electric field inside the wire. (There is a magnetic field from the motion of the positive ions, but the electrons don't feel it since they are at rest in this frame.) The surface charge in this frame is $\sigma' = (R / 2) \beta^2 \gamma^3 \rho_0$, and the total linear density in this frame is $\lambda' = 2 \pi R \sigma' = \pi R^2 \beta^2 \gamma^3 \rho_0$. In this frame, there is a radial electric field outside the wire, which does not effect the electons, but does attract or repel charged particles outside the wire.

But in a copper wire with typical currents, the electrons are extremely nonrelativistic ($\beta \ll 1$), so the net negative bulk charge and positive surface charge are extremely tiny.

Both in the interior (bulk) and at the surface, depending on the source voltage and frequencies. Surface charge is always required on a conducting wire, in order to establish powerflow over the wire. There are two types of current density $\boldsymbol J$: $\operatorname{div}\boldsymbol J = 0$ or $\operatorname{div} \boldsymbol J \lessgtr 0$, depending on the surface charge dynamics: $\operatorname{div} \boldsymbol J + \frac{\partial\rho}{\partial t} = 0$.

In most systems $\frac{\partial\rho}{\partial t}$ is so small that conducted current is free of divergence (typical drift current in wires). There are exceptional systems however, such that all the current is used to alternate the sign of surface charge on the wire, then the current is basically surface current. In principle, such a system might transport power. Thanks for sharing the good question, and for out of the box thinking.

The short answer is the surface. Being in a car during a lightening strike or high voltage line drop would kill you. Also think of the Tesla videos where someone is wearing a suit of armor and doesn't die from the arcs of electricity hitting him in the head; the difference in potential from his head to his feet, although only for a moment, is enough to kill him otherwise.

• This doesn't really answer the question. – Paŭlo Ebermann Apr 26 '14 at 17:08
• You're talking about the behavior of a Faraday cage, which isn't the same as a current carrying wire. – Robert Apr 26 '14 at 23:32

I'll try to keep it short and sweet; Stranded wire is capable of delivering high amperage without overheating because the strands devide the load..I.E. battery cables on your car. stranded wire is superior to solid but to expensive for long runs, so solid wire is used for long runs like for your house (easy to snake or bend) solid but flexible electric company supply line. Yes it mabe true that on a solid conductor there will be less resistance in the center, it would be nomimnal. Take your home appliances for instance, 120v is supplied to your home as a wave length (keeps voltage constant & helps keep line from overheating) Now examine everything you plug into the wall, if it has an elecrtic motor it usually runs A/C ah! but everything else runs on DC. most devices transform A/C to DC because DC can handle short runs with high (Ampherage, Current, Resistance, or Load) To be a little technical solid wire carrying A/C as a wave means there is space between the waves where electricity isn't flowing which help in delivery and cooling however you would need a scope to observe it......GOOD LUCK RAD3

• This doesn't answer the OP's question of whether the current flow is spread uniformly over a wire's cross section. – Selene Routley Nov 4 '14 at 11:04