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Case 1 ($O_1$):

We have wire which has current $I$, in this condition the wire is neutral, and we have a charged particle $+q$

I am put them as following. enter image description here

There will be no attraction since wire is neutral and particle is not moving relative to wire. (and for simplicity we can say wire and $+q$ is stationary to the ground reference system (let say $O_1$))


Case 2($O_2$): Now we are taking this system inside a train, and the train then moves with constant velocity $v$. $O_2$(observer 2) sees this motion outside the train stationary relatively to the ground.

enter image description here

$O_2$ observes that there must be magnetic force between wire and particle $+q$ since particle is moving and there is current in the wire.

enter image description here

Since there is nothing different for $O_1$, there must be no net force between wire and particle, and since $O_2$ outside train sees magnetic force there must be a force equalize the magnetic force for $O_2$.


Question: I do not understand what this equalizer force should be and why this occurs? I thought it must be electric force, but how electric force occurs although it was a neutral wire. In addition, the wire and the whole train will contracted due to Lorentz Transformations. However, how this contraction causes a breaking of a symmetry in current density (or something else related to it because I do not know ).

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  • $\begingroup$ Write down the effect of a Lorentz transformation on a current and you will see. $\endgroup$
    – my2cts
    Commented May 6, 2018 at 14:07

1 Answer 1

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There will not be any force on the q with respect to O2 also, because O2 will observe both magnetic and electric forces on the charge which cancel each other.

This is what you might be looking for : What would happen to a charge in uniform magetic field when observed from a moving reference frame?

According to this source

enter image description here

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