B Magnetic Force and Frame of Reference

1. Dec 2, 2015

UncertaintyAjay

So, a friend asked me this question in school and I've come up with an answer, I'm just not sure that it is right.
Say you have a charged particle moving perpendicular to a magnetic field. There will be a force acting on it given by F=qvB.
Now suppose your'e moving along at the same velocity as said particle. It's velocity respective to you is zero, so you should not observe any magnetic force even though there is one.

My answer is this:
From an inertial frame, the charged particle will be moving in a non-inertial reference frame. So if you move along at the same velocity as the particle, yours is an accelerated reference frame too and you don't observe any force on the particle because of pseudo forces.
Is this right?

2. Dec 2, 2015

Orodruin

Staff Emeritus
No. You can always place yourself in an inertial frame where the particle is momentarily at rest. The solution to your conundrum has nothing to do with accelerated frames. It is based on the fact that the electromagnetic fields transform under changes of inertial frames. What might have been a pure magnetic field in your original frame will be a combination of electric and magnetic fields in the instantaneous rest frame of the particle. What appeared as a magnetic force in your original frame is an electric force in the particle rest frame.

3. Dec 2, 2015

UncertaintyAjay

Thanks a lot!!

4. Jul 1, 2016

Ajmal Hassan

No. Magnetic force is not a real force, Neither is electric force. Only sum of magnetic and electric force (electromagnetic force) is a real force. What your friend said was absolutely correct. Force is frame independent and velocity is frame dependent. Then how the hell can we relate these two? It completely violates the rules of inertial frame and non inertial frames.
Also seeing your friends statement, we can conclude that magnetic force is not really a force. And by the way your statement is wrong. The charged particle will move in an inertial frame, as it is moving with a constant velocity.

Last edited: Jul 1, 2016