# Homework Help: Bring a particle to rest within a field

1. Apr 3, 2013

### Octavius1287

1. The problem statement, all variables and given/known data
A charged particle is moving in a magnetic field. what is the electric field E needed to bring the particle to a rest at 100 us, and v is 1 km/s

2. Relevant equations
I know the magnetic force is q*vxB
and Fc=(mv^2)/r

I have all the parts I think to solve this, I'm just not sure about what method to use

2. Apr 3, 2013

### Staff: Mentor

What is the relative orientation of the fields and the velocity?
The magnetic field will not influence the velocity directly, but it can lead to a curved path.

3. Apr 3, 2013

### Octavius1287

v is perpendicular to the field. I know B, m, and q and r

4. Apr 3, 2013

### Staff: Mentor

Please post all values you know - ideally, copy the exact problem statement. It is pointless to try to help based on some fraction of the problem statement.

5. Apr 3, 2013

### rude man

Use a cyclotron in reverse!

6. Apr 3, 2013

### Octavius1287

A particle with a charge 10^-24 C and a mass 10^-24 kg has a linear velocity v of 1 km/s.
Part a of the problem was finding the radius of curvature if it was entering a magnetic field B perpendicular to v
part B was if v and b was parallel. But stated B=2T

Part c is what i posted.
Find the breaking electric field E needed to bring the particle to a rest at 100 us, and a linear v is 1 km/s

7. Apr 3, 2013

### rude man

So in part c are the B and E fields still in parallel?

8. Apr 3, 2013

### Octavius1287

in part B it just says find R if v is parallel to B, which is 0 because sin(180) is 0.

C is worded exactly as i typed it. Part C is the first time the electric field is mentioned. I assume V is perpendicular to B, because if it was parallel everything would be 0

9. Apr 3, 2013

### rude man

Actually, R = ∞ if B and v are parallel since there would be no bending of the beam path.
No. v would be a straight line. If v were still parallel to B then the solution to c) would be easy. If v is perpendicular to B then I would revert to "get a cyclotron".

Hopefully someone brighter will give us a better idea ... you could set up an alternating E field with a square or sine or some other wave I suppose, but the math ...

10. Apr 3, 2013

### Octavius1287

ugh this has been killing me for hours

11. Apr 3, 2013

### rude man

EDITed:

It will keep killing you until you assume that B and v are still parallel (as in part b).

Last edited: Apr 3, 2013
12. Apr 3, 2013

### Octavius1287

my brains so tired now im not seeing it even if they are parallel

13. Apr 3, 2013

### rude man

If B and v are parallel what does B do to the path of the charged particle?

14. Apr 4, 2013

### Octavius1287

i think nothing

15. Apr 4, 2013

### Staff: Mentor

Right. I think you should assume that B is parallel to v, as in b, so you can ignore the magnetic field for part c. The electric field is aligned with the velocity as well (otherwise it cannot completely stop the particle).
Therefore, all motion is constrained to one direction, and the equations get easy.

16. Apr 4, 2013

### Octavius1287

the force of the lectric field need to be equal to the force of the moving particle right

Last edited: Apr 4, 2013
17. Apr 4, 2013

### Octavius1287

i guess i dont see the equation where time would play a factor

18. Apr 4, 2013

### rude man

The electric field applies a force to the moving particle to slow it down to a stop.

Just like throwing a ball up. It goes up a certain distance, then stops before returning.

19. Apr 4, 2013

### rude man

If you threw a ball up with a certain velocity v0 and the ball stopped going up after time T, could you compute g? It's exactly the same problem.

20. Apr 4, 2013

### Octavius1287

i mean in equation form, i see how it would but reading my book i dont see how to set up the equation

21. Apr 4, 2013

### Staff: Mentor

There is no "force of the moving particle". You want to decelerate the particle (can you calculate numbers for that? Using the velocity different and the stopping time...) with the force of the electric field (which formula can you use to calculate this?).

22. Apr 4, 2013

### Octavius1287

F=ma, but with a - acceleration...damn it i didnt see that