# How is the current flow perpendicular to the wire?

This answer gives a great explanation of how surface charge builds up to force the current to move perpendicular to the wire: http://physics.stackexchange.com/a/102936/41086

However, it fails to address the magnetic field. Moving charges generate a magnetic field, so in addition to the electric field creating a force on the charges, so does the magnetic field.

The answer doesn't explain what effect (if any) the magnetic field generated by the moving charges in the transient state has on the charge build up or the effect of the magnetic field in the steady state.

So could someone give me an explanation of how the current flows perpendicular to the wire, by taking the generated magnetic field into account? Essentially, I'm looking for an extension to the answer I linked to above by including information on the steady state and taking the magnetic field into account.

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Have you tried to estimate the impact of the magnetic field in this case? What I mean by this is to compare $|\vec E|$ with $|v|.|\vec B|$ (SI units) for the electrons in the wire. If the second is much less and the first, there shouldn't be any important effect of the magnetic field. –  user23873 Mar 16 at 22:41
@user23873 That seems like a good approach, but I have no idea how to even get a ballpark estimate. –  user41086 Mar 16 at 23:29
If I'm to say, I would use something like the fermi velocity ($\alpha c\approx 1/137 c$) for the velocity of the particle, and $c|\vec B|=|\vec E|$. After next monday I'll probably be able to make a more decent answer. –  user23873 Mar 18 at 14:50

Let's model the real wire as two infinitely thin wires carrying 1 mA of current 1 mm apart.

$$B = \frac{\mu I}{2 \pi r}$$ $$F = q v B$$

Plugging this in, and considering the speed of the current being one third of the speed of light (close to the actual value, depending on the wire) we get that the acceleration exerted on each electron is of the order of $10^{12} m/s$ Calculation in Wolfram|Alpha. That is a quite high number, so my first model looks quirky.

We need something better. Take the section of the wire, the current flows on the surface. You want to see the magnetic field at the topmost point produced for each point in the surface. The first thing you notice is that, for each point, the one in the oposite side will have a magnetic field that will partially cancel each other. If you look at the wire from the direction where the current is going to, only the component to the left will remain. NOTE: the closer to the top, the more intense this would be, but also the stronger the cancellation.

Now, where is the magnetic force pointing at? Towards the centre of the wire. So, as electrons flow, they will be pushed towards the middle, but also that would bring them closer, and therefore augmenting the repulsion.

So, what happens in the end? If the wire is small enough, the magnetic forces are strong enough so they can push the electrons in, and the current flows all across the body of the wire. If you make it thick enough, the stronger components in this force, the ones coming from the opposite side of the wire, will be further away, so the flow will remain in the surface. That is one of the reasons for thick wires to be made out of many small ones bundled together. The others are: it is easier to bend, and if there is one defect (a crack, for example), it will only break that fiber, but not the whole cable.

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A comment on the value of the acceleration. The value itself is not important, what is important is the displacement that it causes, during the time that it actuates –  user23873 Mar 18 at 14:49
That is actually a very good point. Assuming a nanosecond wire (as Grace Hopper taught us), $10^{12} \cdot 10^{-12}^2 = 10^{-12} m = 10^{-9} mm$. –  Davidmh Mar 18 at 14:56