Current / Magnetism Questions: Proton near magnetic field of moving electrons

AI Thread Summary
The discussion revolves around the behavior of a proton in the magnetic field created by two beams of charged particles, one of electrons moving left and one of positive particles moving right, both at 1x10^8 m/s. The current flows to the right, opposite to the electron flow. The force on the proton, as observed by a stationary person, can be calculated using the equation F=qvB, but the interactions between the proton and both beams must be considered, as both attraction and repulsion are at play. For the running observer, the proton appears stationary, leading to the conclusion that no magnetic force acts on it, but an electric force due to the electric field must be accounted for. The discussion also touches on relativistic effects and the perceived velocities of the beams from different reference frames.
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Homework Statement



Ok guys, this problem has a few steps to it, and id appreciate any help I can get.

qu.jpg


Ok, so beam of electrons moving left, positive particles with same but opposite charge moving right, both moving at 1x10^8 m/s. Then there is a proton moving left on top of the beam nearby AT ALSO 1x10^8 m/s. A person standing nearby is observing, and so is a guy running ALSO 1x10^8 m/s nearby.

Questions:
1) Direction of current?
2) Force on the proton as seen by the person standing, and the person running seperatly.
3) Velocity/direction of the two beams from the running persons point of view?
4) What current will the man running measure and in which direction?

Homework Equations


1) none
2) F=qvBsin(theta) (or in this case just F=qvB since the direction of the proton is perpendicular to the magnetic field..?)
3) none
4) ?

The Attempt at a Solution



1) Current flows is opposite of current flow, so therefore in this case current flows to the right. Is this correct?

2) Person standing still: There is a magnetic field flowing out of the screen from the view we have, if the current is indeed traveling to the right. Therefore, I simply use the equation F=qvB to solve. Is this true, or would I need to factor in the fact that the electrons are also drawing the particle towards them?

Person running: since he is moving at the same speed and direction as the proton, it will seem to him that it is not moving.

3) The beam moving in the same direction will seem as though it doesn't move, but the beam moving in the opposite direction will seem to move twice as fast.

4) will he measure no current? since he is moving as fast as the electrons?Thanks for any help guys! I appreciate it.
 
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1) I assume you meant to say that current flows in the opposite direction from electron flow? Anyway, to the right is correct.

2) In the reference frame of the person standing, yes, use F = qvB. Think about this: if the electrons are attracting the proton towards them, why would the positive particles (in the current) not also be repelling the lone proton away from them? How do the strengths of these two effects compare?

In the reference frame of the running person: yes, he sees it as not moving and therefore sees no magnetic force. But he sees another force acting on the proton in place of the magnetic force. Can you identify it?

3) The beam moving along with the runner (that is, the electrons) will seem to him like it's not moving, true. The other beam will seem like it's moving almost twice as fast. (Why not exactly twice as fast?)

4) Aren't you forgetting the beam of positive particles?
 
Have you studied relativity? I am wondering because the speeds involved are (1/3)c. If one were not to consider relativistic effects, why pick this particular speed for charge distributions and observer?
 
Diazona-

You brought up some amazing points! Thank you so much for your help.

For number 2, for the person running, you are referring to the force of the electric field, am I correct?

And one other question, what direction would the electron seem to be traveling from the point of the running person?
 
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