# Magnetic force between moving charges Given two infinite parallel charged rods with equal charge d

Magnetic force between moving charges
Given two infinite parallel charged rods with equal charge density $\lambda$. They are moving with same constant velocity $\stackrel{\to }{v}$ parallel to the rods. Find the speed $v$ for which the magnetic attraction is equal to the electrostatic repulsion.
Well, I know how to solve this problem: we first find out the magnetic field created by one rod on the other using Biot's and Savart's law, then we use the definition of $\stackrel{\to }{B}$ ( $d\stackrel{\to }{F}=\stackrel{\to }{v}dq×\stackrel{\to }{B}$ ) to find the magnetic force, then equate magnetic and electrostatic forces to find $v$, which will be greater than or equal to $c$, thus conclude it is impossible for the forces to be equal.
However, one can argue as the following:
We all know that "same laws of physics apply in all inertial frames". With a constant velocity $\stackrel{\to }{v}$,the rest frame of the rods is an inertial frame. Therefore, if Biot-Savart law applies in our frame, it has to apply in the rest frame. If so, none of the rods will feel a magnetic field from the other one because their relative speed is zero, and there will be no magnetic force between the rods.
I've seen this question several times before in references, exams, exercise sheets,and in many different forms (parallel planes, beam of electrons ...),but no one ever used this argument.What is the problem in it ? Is it something related to Maxwell's equations or special relativity ? Or what else ?
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Eliezer Olson
I think about this problem the same way that you do, but I phrase it slightly differently.
First the intuitive solution. An observer in the rest frame of the two line charges would observe them accelerating away from each other due to electrostatic repulsion. Relativity demands, then, that no inertial frame exists where the two line charges accelerate towards each other due to magnetic attraction.
Now the mathematical solution. The Biot-Savart and Lorentz-force laws (which are already fully relativistic) tell you that in the limit that $v\to c$, you find that ${\stackrel{\to }{F}}_{B}$ approaches $-{\stackrel{\to }{F}}_{E}$. You know from your prior experience with relativity that $v\to c$ is physically an unreachable limit, so just because you can plug $v=c$ into some expression about forces doesn't mean you should get sloppy and say such things.
The reason for continuing to the mathematical solution is mainly to confirm that the Lorentz force is consistent with your special-relativistic intuition. It's possible to imagine a wrong expression for the Lorentz force which would predict that boost to a frame traveling at $c/2$ would cause the direction of the total force, and thus the net acceleration, to change sign. That's the kind of killer error you have to look for when you are coming up with solutions to new problems or trying to model new phenomena.
Here's a mathematically precise way to phrase your relativistic argument if we make the minor change that the two lines are oppositely charged:
In the rest frame of the line charges, electrostatic attraction will cause them to touch after some finite time $t$. In reference frames where the line charges have velocity $v$ the time for them to touch is magnified by time dilation to $t/\sqrt{1-{v}^{2}/{c}^{2}}$; an observer in this reference frame would ascribe the reduced acceleration to an additional force ("magnetism"). The time before contact can made arbitrary large in the limit $v\to c$, so the magnetic force can be made arbitrarily close to, but not equal to or larger than, the electric force.
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Aedan Tyler
As far as I can tell that seems perfectly valid. As long as the net force is equal in each reference frame then that should tell you that you did it right.
When you say "No one has ever used this argument" what did you mean? I believe it's pretty fundamental to understand that you can change your reference frame so that the force from a magnetic field is zero. The force from the electric field will change however because of relativistic effects on the charge density of the rods.