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There is something I am confused about when it comes to the force between two parallel wires carrying current, specifically why when they carry current in the same direction the wires are always attracted to each other, irrespective of the magnitude of the current they are carrying. I understand the explanation using Flemming's left hand rule and considering the electrons in each wire separately, but I come into problems when I think of the magnetic force as a relativistic effect...

If I were to consider the current in both wires being of the same magnitude and in the same direction, the electrons in each wire see the electrons in the other wire being stationary relative to them, but they see the positive ions in the other wire moving backwards. This means that they would see the space between the ions to be contracted, so the positive ion density is greater and the wire has an overall positive charge, and so the negative electrons are attracted to the other wire. Also, if I were to consider this from the case of the positive ions, they see the ions in the other wire as stationary and the electrons as moving, and so these are length contracted s the wire has an overall negative charge- againn the wires are attracted. So it seems to work out for this case.

However I am confused about the case when the current in one wire is more than twice the current in the other wire. If this were the case, then when considering the electrons in the wire with a smaller current, they would see the electrons in the other wire as having a greater speed relative to them than the positive ions in that wire, so theoretically the electrons should now see that the other wire has a greater negative charge density than positive charge density, and the wire should be repelled even thought the currents flow in the same direction. On the other hand, in the wire with the greater current the electrons would see the positive ions in the other wire as having a greater negative velocity than the electrons, so the wire with the larger current would see the wire with the smaller current as having an overall positive charge density, and therefore would be attracted to that wire. This makes no sense because the wires should experience equal and opposite forces- both should be either be attracted to each other or repelled from each other- by Newton's third law...

Am I thinking about this in the wrong way?

Thank you in advance :)

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  • $\begingroup$ A very long thread of comments deleted in its entirety because the comments were not directed at making this (or any) post better. $\endgroup$ – dmckee Nov 3 '14 at 4:54
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Short answer: If you want to analyze things in the electrons' rest frame in wire A (whose electrons move at less than half the speed, relative to the wire, as the electrons in wire B), you have to take into account not just the electric force on the electrons in the wire A, which will be repulsive as you say, but also the electric force on the positive charges (the ions that have lost electrons) in wire A, along with the magnetic force on the positive charges in wire A, which are moving in this frame. When I did this in a numerical example below, I did find that the net force on wire A was attractive despite the fact that the electric force on the electrons was repulsive.

Long answer: The average speed of electrons in conductors is actually quite slow, but let's look at an example where the speeds are large to make the numbers a little easer--say in wire A the electrons are moving at $\sqrt{0.0199}c$ = 0.14106736c relative to the wire rest frame (the number chosen so the spacing will shrink by exactly 0.99 due to Lorentz contraction, and in wire B they are moving at $\frac{(0.6 + \sqrt{0.0199})c}{1 + 0.6\sqrt{0.0199}}$ = 0.68323783c (this number chosen to make addition of velocities work out nicely below). So the wire itself is moving at v = -0.14106736c in the frame of the electrons in wire A, and the electrons in wire B are moving at u = 0.68323783c relative to the wire itself, so we can use the addition of velocities formula to find the speed of the electrons in wire B in the frame of wire A: $(v + u)/(1 + vu/c^2)$ = (-0.14106736c + 0.68323783c)/(1 - 0.14106736*0.68323783) = 0.6c. This means the spacing of electrons in wire B will be reduced by a factor of $\sqrt{1 - 0.6^2}$ = 0.8 in wire A's frame.

If the linear charge density of the positive charges in each wire's rest frame is $\lambda$, then in the frame of the electrons in wire A, the positive charge density in wire B is $\lambda_{B+} = \lambda / 0.99$ and the negative charge density in wire B is $\lambda_{B-} = -\lambda / 0.8$. Meanwhile, in this same frame, the linear charge density of electrons in wire A is just $ \lambda_{A-} = -\lambda$ and the density of positive charges in wire A is $\lambda_{A+} = \lambda / 0.99$. The attractive force per unit distance on one line charge due to another line charge at distance d is $- \lambda_1 \lambda_2 / 2 \pi \epsilon_0 d $, so the total attractive force/length on the negative charges in wire A, due to both the positive and negative charges in wire B, is $-(\lambda_{A-}\lambda_{B-} + \lambda_{A-}\lambda_{B+})/2 \pi \epsilon_0 d $ = $-((\lambda^2/0.8) + (-\lambda^2/0.99))/2 \pi \epsilon_0 d $ = $ - (0.19/0.792) \lambda^2 / 2 \pi \epsilon_0 d $ = $ - 0.2399 \lambda^2 / 2 \pi \epsilon_0 d $. And the total attractive force/length on the positive charges in wire A, due to both the positive and negative charges in wire B, is $-(\lambda_{A+}\lambda_{B-} + \lambda_{A+}\lambda_{B+})/2 \pi \epsilon_0 d $ = $-(\lambda_{A+}\lambda_{B-} + \lambda_{A+}\lambda_{B+})/2 \pi \epsilon_0 d $ = $-((-\lambda^2/0.792) + (\lambda^2/0.9801)) /2 \pi \epsilon_0 d $ = $ (0.1881/0.7762392) \lambda^2 / 2 \pi \epsilon_0 d $ = $ 0.2423 \lambda^2 / 2 \pi \epsilon_0 d $. So you can see here that the net attractive force on the positive charges in wire A is actually slightly larger than the net repulsive force on the negative charges in wire A.

Then to be sure about whether wire A will be attracted or repulsed by wire B when we analyze things in this frame, we also need to compute the Lorentz force per unit length on the positive charges in wire A. For a line current, current = (charge per unit length) * (speed of charges), so the current due to positive charges in wire A, which are moving at -0.14106736c in this frame, is $ \lambda_{A+} * -0.14106736c $ = $-0.14249228 \lambda c $. This is also the current due to positive charges in wire B in this frame. And the current due to negative charges in wire B, which are moving at +0.6c in this frame, would be $ \lambda_{B-} * 0.6c $ = $-0.75 \lambda c $. So in this frame the total current in wire A is $-0.14249228 \lambda c $ and the total current in wire B is $(-0.14249228 - 0.75) \lambda c $ = $-0.89249228 \lambda c $. So without even figuring out the force/length, we can see the magnetic force will be attractive, since currents in the same direction in parallel wires cause the wires to be attracted magnetically. To work out the math, the attractive force per unit length for two parallel straight wires at distance d with currents $I_1$ and $I_2$ is $ \mu_0 I_1 I_2 / (2 \pi d)$, so in this case the attractive force would be $ \mu_0 0.12717326 \lambda^2 c^2 / (2 \pi d)$.

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