# Moving charge and a magnet

## Homework Statement

A magnet is moving with constant velocity vector ##\vec V##. A charged particle fly through it. At first, the objects are place very far from each other, and the charged particle velocity vector is ##\vec V0##, after the charged particle cross over and fly away from the magnet, it velocity vector is ##\vec Vc##. We know all the velocity ##\vec V0=110m/s## and ##\vec Vc=134m/s##. The angles are ##|\vec V;\vec V0|=120°## and ##|\vec V;\vec Vc|=45°##. Find ##|\vec V|##
Note: Assume that the charged particle do not affect the motion and magnetic properties of the magnet.

## Homework Equations

3. The Attempt at a Solution [/B]
It must be some kind of momentum and angular momentum conservation, since we have no information about the magnetic field created by the magnet and the charge of the charged particles, nor do it mass. I assume a elastic collision between the charged particle and the magnet, but that do not hold since we cannot explain the velocity change in the direction perpendicular to ##\vec V##. I have no idea how to apply angular momentum conservation.

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Orodruin
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I have no idea how to apply angular momentum conservation.
What makes you think angular momentum is conserved?

There is, however, one conservation law that does apply. If we were in the rest frame of the magnet, how would the velocity of the particle change?

HotFurnace
I don't know, but if only momentum is conserved, then how would the velocity of the charged particle change in the direction perpendicular to ##\vec V##?? And from what I know, if a magnet is moving, it will generate a magnetic field and a electric field (note there's no relativity involved, this is a classical problem). If we consider the effect of the magnetic field then we would need to conserve angular momentum.
I just tried to do so, but I didn't realize anything special, more hint is needed.

Orodruin
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Momentum is not conserved.

And from what I know, if a magnet is moving, it will generate a magnetic field and a electric field (note there's no relativity involved, this is a classical problem).
This is why I asked you about the rest frame of the magnet!

If we consider the effect of the magnetic field then we would need to conserve angular momentum.
Why?

HotFurnace
Yes, momentum is not conserved, what i mean with post #1 and #3 is that we can consider the interaction between the particles as an elastic collision, imagine a mass with velocity v collide with a moving wall with velocity u, the wall will just be moving on with u and the ball would return with velocity v+2u. This is not conservation after all, but that's what i mean. Sorry if i confused you.
I don't know why i think i need to conserve angular momentum , but after solving lots of problem related to charge moving in magnetic field, i had the feeling so. I'm still in high school and preparing for the National Physics Olympiad.
Maybe this is what you want to say, you wanted me to take the frame of the magnet so that the electric field generated by the magnet is zero. In this case then the charged particle energy is conserved. Maybe I got what you mean, let's try again.

Orodruin
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is that we can consider the interaction between the particles as an elastic collision,
If by "particles" you mean the particle and the magnet, then no. If you send the magnet mass to infinity, then yes.

Maybe this is what you want to say, you wanted me to take the frame of the magnet so that the electric field generated by the magnet is zero. In this case then the charged particle energy is conserved.
Right, because the magnetic force (and hence the acceleration) is perpendicular to the velocity of the particle. How do you express the velocity of the particle in the rest frame of the magnet in terms of the velocities of the particle and magnet in the lab frame?

HotFurnace
I solved it yesterday, but had to sleep and take class. The key information is that energy is conserved in the magnet frame, so a little trigonometry will do the job: $$Vc^2+V^2+2Vc*V*cos(180-45)=Vo^2+V^2+2Vo*V*cos(180-120)$$, solve it for V we will get the answer $$V=19.55m/s$$
Thanks a lot for your help!

Orodruin
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Great that you solved it. On a LaTeX note: Instead of writing Vc or Vo to get ##Vc## or ##Vo##, you can write V_c or V_0 to obtain ##V_c## or ##V_0##. Things will become much easier to read.