1
$\begingroup$

suppose we take a charged particle and a magnet and place them at some particular distance apart .Now let's take 2 frame of reference.

[the charged particle and the magnet are in rest with respect to each other through out the whole event]{both of the frames are inertial}

frame(a): this frame is in rest with respect to the charged particle

frame(b): this frame is in motion with respect to the charged particle

now in frame(a) there is no change in the magnetic field due to the charged particle

now in frame(b);; As in this frame the charged particle is in motion.so it will generate a magnetic field,and if we assume the velocity is high enough to attract(let's say it is attraction in this case,if it will be repulsion then the result will be the opposite,but that will still come in this question) the magnet then after time(T) the both object will touch each other.

now after time(T) from the start of the event;

1.in frame(a) the charged particle and the magnet are at the same distance from each other as they were from the start of the event.

2.in frame(b) the charged particle and the magnet are touching each other.

now my doubt is ,is the result is true and if wrong what is the fault in my assumptions or in the theory?

thank you for answering.

$\endgroup$
1
$\begingroup$

In the moving reference frame there will be no net force on either the magnet or the charge. For instance, the moving magnet will create both, a magnetic field (due to the magnet itself) and an electric field (this last due to the maxwell Faraday equation), and the net force on the charge will be zero. Same as viwed by the magnet, the moving charge will create no magnetic field because the charge generates two contributions to the magnetic field that cancel each other, one due to the current it creates, the other due to the changing electric field it creates as it moves (ampere law with maxwell's addition)

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.