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Given two infinite parallel charged rods with equal charge density $\lambda$. They are moving with same constant velocity $\vec v$ 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 $\vec B$ ( $d\vec F=\vec vdq\times\vec B$ ) 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 $\vec v$,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 ?

I know a similar question was asked before, but the answers weren't satisfying. Please provide your answers with necessary mathematics.

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  • $\begingroup$ It is a completely valid argument. $\endgroup$ – Ruben Verresen Mar 18 '16 at 17:41
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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 $\vec F_B$ approaches $-\vec F_E$. 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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    $\begingroup$ thank you for this answer, it looks useful. I've figured out two deductions out it: 1- The velocities appearing in Biot-Savart and Lorentz laws are not the relative velocities of the charges, but rather the velocities in the considered frame of reference. 2-Magnetic force is relative, it can be zero in the rods' frame and non-zero in the observer's frame. Is this right or did I just understand nothing? ...*I think you meant "F_B approaches -F_E" $\endgroup$ – Tofi Mar 19 '16 at 4:10
  • $\begingroup$ @H.Tofaili Thanks for pointing out my typo (#2). On the other hand, #1 is the crux of the argument. $\endgroup$ – rob Mar 19 '16 at 18:25
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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.

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Imagine the two wires connected every meter by identical insulating springs. The springs stretch by an amount to balance the electrostatic repulsion. If you begin moving in the direction of the wires, there is no reason to think the springs would contract. Thus, the force on each spring is the same regardless of your speed. Conversely, if the wires are moving, the force on each spring is the same regardless of their speed.

To demonstrate this consider the following.

In the rest frame of the wires the force per unit length between the wires is the force on each spring $F_i$. This is given by $$F_i=\frac{\lambda^2}{2\pi\epsilon_0r}$$ This will take the same form in every frame however, due to length contraction, there is a different charge per unit length, $\lambda'$. For a frame moving with speed $v$ the force is $$F'_i=\frac{\lambda'^2}{2\pi\epsilon_0r}\left(1-\frac{v^2}{c^2}\right)$$ Where the second term in the parentheses arises from the magnetic force.

To relate $\lambda$ and $\lambda'$ use the fact that $\lambda=dq/dx$ and $\lambda'=dq/dx'$. Then, since charge is invariant, $dq$ is the same in either case, but $dx'=dx\sqrt{1-(v/c)^2}$ by length contraction, ($dx$ is the rest length). Combining these gives, $$\lambda'=\frac{\lambda}{\sqrt{1-v/c}}$$ Substituting this into the expression for $F_i'$ above reveals $F_i'=F_i$.

Note that this is the force on a single spring. Except in the frame of the wires, this is not the force per unit length. Since the spring would appear closer in a frame where the wires are moving, the force per unit length would increase as there appears more spring per unit length exerting the same force. Yet, in every frame, the total number of charges between any two springs is the same. Thus, since the force on each spring is equal in every frame, so is the force on each charge. This result is true in general. The force between two charges, stationary with respect to one another, is the same in every rest frame that observes them. To show this in the case of point charges requires much more work. Showing the result in this case is simplified greatly by symmetry.

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How about to start with a postulate that an induced magnetic field is the result of the alignment of the magnetic dipole moments of the involved particles?

Please hold in a moment and think about this. Recapitulate, that it is easy to divide electric charges since a long time and the intrinsic property of electric charges is obvious. The intrinsic magnetic dipole property of particles was found long after the formulation that a moving charge induces a magnetic field. This is right but not precise.

A moving charge is a charge that was accelerated before. The acceleration of a charge is the reason for the following alignment of the intrinsic spin of this charge and by this for the alignment of the magnetic dipole moment of this charge. The acceleration and not the movement of the charge is important. Follow the historical experiments and you will see that always acceleration of charges is involved.

One has to to check (to prove) the postulate that any magnetic field is the result of an alignment (the induction) of the magnetic fields of the involved particles. If there will be no contradiction then it is easy to explain that any current carrying wire has an magnetic field which was induces than switching on the current and which after the initial induction holds due to self induction. This postulate implies that single moving and undisturbed charges have equally distributed magnetic dipole moments and this will be a non provable postulate because any measurement process will align this magnetic dipole moment.

Reply to Robs comment

Rob commented:"The field due to a stack of magnetic dipoles will be everywhere parallel or antiparallel to the direction of the stack. You can't produce a Biot-Savart field from a superposition of dipoles."

Here the self inductance came into play. It is a self-reinforcing process. As you know, switching off the current the magnetic field will switch of it opposites field and due to this a opposite current will flow and so on. Of course this process is damped but every time one switch off or on a current this self-inductance phenomenon exists. I'm not sure is thsi process due to the bumping of the electrons in the wire and will the measurement of a beam of electrons show a different magnetic field. For example "parallel or antiparallel to the direction of the stack".

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    $\begingroup$ You can't do this because the magnetic field of a line current has a different set of symmetries than the magnetic field due to a combination of dipole moments. The field due to a current has curl parallel to the current, so that the field lines encircle the wire. The field due to a stack 0f magnetic dipoles will be everywhere parallel or antiparallel to the direction of the stack. You can't produce a Biot-Savart field from a superposition of dipoles. $\endgroup$ – rob Mar 18 '16 at 21:14
  • $\begingroup$ @rob Please see the reply to your serious objection in my edited answer. $\endgroup$ – HolgerFiedler Mar 19 '16 at 7:14
  • $\begingroup$ Self-inductance doesn't change my argument about symmetries at all. $\endgroup$ – rob Mar 19 '16 at 18:21
  • $\begingroup$ @rob At the moment the potential difference get lost (the current was switched off) the electrons get accelerated (stopped) and a magnetic field will be induced. This led to a current in the opposite direction to the original direction. This changes the magnetic field again. Acceleration of electrons are the reason for the alignment of the magnetic dipole moments and a resulting magnetic field. Not to draw this in recital is strange. $\endgroup$ – HolgerFiedler Mar 19 '16 at 18:41

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