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This question already has an answer here:

I think they shouldn't since they are relatively at rest to each other. The actual answer to this question (it was in an exam I took) is that there will actually be a magnetic force acting on both of them.

Here's my reasoning: Since both the charges are unaccelerated, I can view them from an inertial frame which is at rest relatively to both the charges. Now, these two charges would appear to be at rest and as such they should only affect each other by electrostatic force. What is wrong in my reasoning?

Thank you

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marked as duplicate by John Rennie, Kyle Kanos, Jon Custer, heather, AccidentalFourierTransform Feb 1 '17 at 12:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Related: physics.stackexchange.com/questions/24010/… $\endgroup$ – BowlOfRed Jan 31 '17 at 7:43
  • $\begingroup$ Independent from their movement electrons have a magnetic dipole moment and this leads to the alignment between this charges and an attracting force. For free electrons the repealing force from their electric fields will be stronger and the distance between them increases. In wires the electric net charge will be zero BUT the moving electrons of each wire induce their own magnetic field in dependence of the direction of the current and by this the wires would or attract or repeal each over. $\endgroup$ – HolgerFiedler Jan 31 '17 at 8:42
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If you look at it from stationary reference frame, the electrostatic force between the charges will actualy increase, but so will the magnetic force (according to stationary observer). It will change in such a way that the repulsive force gets just a bit smaller, which is consistent with special relativity which predicts that moving objects appear slower to stationary observer. But if we move the reference frame with the moving charges, those charges would only see the electrostatic force, but that repulsion would be just a bit greater because of no time dilation, since you're moving with the speed of charges.In different reference frames the same force (electromagnetism) can appear as electrostatic, magnetic or electromagnetic force.

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The answer to this question, in fact, requires knowledge of special relativity!

Imagine you have a wire with a current going through it and the electrons are going from left to right. Now let's place a negative (electron) charge underneath this wire. your point of view is from this negative charge.

The negative charge is moving at the same velocity as the electrons in the wire. Notice that when you move at the same velocity as the electrons, you will feel like the electrons are stationary. However, the protons in the wire are moving from the left to right (backward). These protons will undergo a length contraction (special theory of relativity). As a result, there will be more protons than electrons per unit length. The wire is now positively charged. You (the negatively charged particle) is now attracted to the wire via a magnetic force.

What was wrong with your reasoning? You didn't factor in length contraction of the positive particles, otherwise, you would've got it!

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  • $\begingroup$ Length contraction matters if there are multiple charges around (so the distances between them appear to shrink and the charge density increases). In the OP, on the other hand, each charge only sees a single charge. $\endgroup$ – Emilio Pisanty Jan 31 '17 at 8:36
  • $\begingroup$ This answer is not correct ,see this physics.stackexchange.com/questions/145721/… $\endgroup$ – Lapmid Jan 31 '17 at 9:01
  • $\begingroup$ I get what you're saying, but in my case, I don't have a current carrying wire. I have two isolated identical charges. $\endgroup$ – Anindya Mahajan Jan 31 '17 at 9:38
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Well there will be a magnetic force with respect to ground frame given by $F=q\,v\,B\sin(\theta)$. The field $B$ is due to the other charge. In my opinion nothing is wrong with your reasoning. The magnetic force in the ground frame will manifest as the electrostatic force due to $$\text{Lorentz force} =q\,E+q\,v\,B\sin(\theta)$$ in your frame.

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