# Pos. et neg. particle in circ. motion

1. Jan 11, 2008

### henxan

1. The problem statement, all variables and given/known data

You have two particles, both mass m.
One particle negative (-q) and other (+q).
Thay are going in a circular motion, velocity v or angular velocity (omega).
What is the difference in scale of the electric and magnetic attraction?

2. Jan 11, 2008

### Shooting Star

This is the Homework forum. Why don't you tell us your thoughts first?

3. Jan 11, 2008

### henxan

Well, straightforward you've got the square decreasing electric field. But, concerning the magnetic field:
This is actually a parallel problem to a two wire problem. Two wires separated with a distance D, carrying a current in the same direction. This is, because, if you take a snapshot of the system, it is two particles, traveling in opposite direction, with opposite charge.

Using vector potential you would also get a B which decreases with the square of r. So, my conclusion is, the order of magnitude of difference would be related to myu_0 and epsilon_0

Is this correct? Or have I messed around and gotten a factor difference somewhere? :/

4. Jan 11, 2008

### Shooting Star

> So, my conclusion is, the order of magnitude of difference would be related to myu_0 and epsilon_0

Very good. But the magnetic force will also depend on the speeds of the particles. For low speeds, the magnetic force is very low compared to the electric force. You can get a rough idea by treating the moving charges as tiny currents and using Biot-Savart’s law. The speeds will then come in via the current.

5. Jan 11, 2008

### henxan

LOL.. I actually forget to mention that that the speed of the particle also mattered. So it is correct to say that epsilon_0 vs speed*myu_0?.. Thanks for the reply.. I have to admit this question came as a result to the fact that i didnt bother to actually calculate the forces.. Ill set ut the equations and figure this out later.. And it ought to be straight forward unless one steps into the relative roam v>10% of lightspeed :)..

6. Jan 11, 2008

### Shooting Star

Post your calculations. It'll involve $v^2$. Don't go relative for now.

7. Jan 11, 2008

### henxan

LOL.. I wont.. Ill do it tomorrow or sunday.. I think ill set the mass as unknown, and calculate the two forces first.. Thanks..