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The terminals of the batteries set up an electric field in the wire. Surface charges build up to ensure the field is perpendicular to the wire. This allows the electrons to move through the wire.

But moving charges generate magnetic fields. So wouldn't the magnetic fields generated disrupt the flow of electrons causing charge buildup that disrupts the electric field which causes a change in the magnetic field which then causes a change in the electric field and so on?

I guess what I'm asking is why doesn't the magnetic field generated by moving charges disrupt the flow of electrons which would otherwise flow perfectly because of the surface charge buildup?

I am not talking about the Hall effect. I'm asking about the magnetic field generated by the moving charges themselves not an external magnetic field.

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Are you familiar with the force law $\vec{F}_B=q\vec{v}\times\vec{B}$? Your answer may help frame others' answers. –  BMS Mar 4 at 4:04
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@BMS Yes, I am. –  Vimzy Mar 4 at 4:05
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The actual drift velocity of individual electrons in a circuit is quite small. Since the magnitude of magnetic force is proportional to the charge velocity, the magnetic component of the Lorentz force is much, much smaller than the electrostatic component. –  user27578 Mar 4 at 5:01
    
@dgh Sure, but the amount of charge is relatively high. Shouldn't this make up for the small velocity? –  Vimzy Mar 4 at 19:26
    
@dfg, the amount of charge on a single electron is not "relatively high" relative to anything. We're talking about the total force on each individual electron, so the effect is small. –  user27578 Mar 4 at 21:31

3 Answers 3

You are right in that a magnetic field is build up, which generates a electric field opposing the given potential. But the consequence is not an oscillation of current, but only a damping of the increase of the current. Therefore, if you have a Heaviside step function for the voltage, it'll result in an "exponential" increase of your current ($I(t) = I_0 \cdot(1-e^{\theta t})$). In general $\theta$ is big enough so that a $I_0$ is reached very fast, so no one takes notice, but if you look on an oscilloscope you'll see

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But if the voltage is low enough, shouldn't the dampening be enough to prevent current? –  Vimzy Mar 4 at 19:27
    
Lower voltage means proportional lower counter electric field. Same temporal behaviour, lower $I_0$ –  Lord_Gestalter Mar 4 at 20:56

This is an interesting question. Let us consider a current carrying wire and see what's going on inside it.

enter image description here

Consider two streams of electrons as shown above. They move with a drift velocity $\vec{v_d}$. $These$ $streams$ $can$ $be$ $thought$ $of$ $two$ $current$ $carrying$ $wires$ $each$ $with$ $the$ $thickness$ $of$ $an$ $electron$. Each carries $equal$ amount of current. Now, let us see how these affect each other.

Look at the diagram below,

enter image description here

Consider the stream at the top. We shall call it stream A. This stream creates a magnetic field $\vec{B}$ which affects the stream below( say stream B ). Remember, stream B also creates a magnetic field and I have not shown that in the diagram to avoid confusion. Don't worry, I'll get there soon. The dots in the diagram tell us that the magnetic field due to stream A is directed $out$ of the screen( I have used R.H.R to determine the direction of $\vec{B}$ ).

Once again using R.H.R to determine the direction of force on stream B we get,

enter image description here

Stream B feels a force $\vec{F}$ directed towards stream A.

Now, let us see how stream B affects stream A.

enter image description here

Stream B creates a magnetic field which goes into the screen. As a result, stream A experiences a force directed towards stream A as shown

enter image description here

Once again, I have used R.H.R to determine the force on stream A.

Since the separation is constant, force on stream A due to stream B is equal to the force on stream B due to stream A.

Now, let us analyze the forces

enter image description here

The forces tend to bring the two streams together. What keeps them in position is the Electric force of repulsion $\vec{F_e}$( the two streams are same in terms of their nature )

enter image description here

So, the net force on each stream is $zero$. Now, the wire can be thought to be made up of billions of such streams and all such streams come in pairs. This explanation holds good for ideal conducting wires that have uniform charge density. In the real world, this is close to perfect.

I have not included the effects produced by the lattice because for a conducting wire with uniform charge density, these effects cancel out one another.

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The magnetic field is created by the electrons; thus it would not disrupt the movement of the electrons.

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Sorry, I don't follow. Why wouldn't the field from other electrons affect an electron? –  Vimzy Mar 27 at 0:26

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