Bring a particle to rest within a field

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Discussion Overview

The discussion revolves around the problem of determining the electric field required to bring a charged particle to rest within a magnetic field. Participants explore the relationships between the magnetic and electric fields, the particle's velocity, and the time frame for stopping the particle. The conversation includes theoretical considerations and mathematical reasoning related to the forces acting on the particle.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the need to find the electric field E to stop a charged particle moving at 1 km/s in a magnetic field.
  • Another participant questions the orientation of the velocity relative to the magnetic field, noting that the magnetic field will not directly influence the velocity but can cause a curved path.
  • Some participants clarify that the velocity is perpendicular to the magnetic field and discuss the implications for calculating the radius of curvature.
  • There is a suggestion to use a cyclotron in reverse to conceptualize the problem.
  • Participants express confusion about the role of the electric field in the context of the problem and the conditions under which it is applied.
  • One participant emphasizes that if the velocity and magnetic field are parallel, the magnetic field does not affect the particle's path.
  • Another participant proposes that the electric field must be aligned with the velocity to effectively stop the particle.
  • Discussions arise about the equations needed to relate force, mass, and acceleration, with some participants struggling to connect time to the equations involved.
  • There is a mention of the analogy of throwing a ball to illustrate the relationship between velocity, stopping time, and force.

Areas of Agreement / Disagreement

Participants express varying views on the orientation of the fields and the implications for the problem. While some suggest assuming parallel conditions to simplify calculations, others highlight the complexities introduced by the perpendicular orientation. The discussion remains unresolved regarding the exact approach to take for calculating the required electric field.

Contextual Notes

Participants note the importance of clearly defining the orientations of the velocity and fields, as well as the assumptions made in the problem. There are unresolved mathematical steps and dependencies on the definitions of forces and fields.

Octavius1287
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Homework Statement


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

Homework 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
 
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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.
 
v is perpendicular to the field. I know B, m, and q and r
 
What about the electric field?
I know B, m, and q and r
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.
 
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
 
So in part c are the B and E fields still in parallel?
 
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
 
Octavius1287 said:
in part B it just says find R if v is parallel to B, which is 0 because sin(180) is 0.
Actually, R = ∞ if B and v are parallel since there would be no bending of the beam path.
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

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 ... :eek:
 
  • #10
ugh this has been killing me for hours
 
  • #11
Octavius1287 said:
ugh this has been killing me for hours

EDITed:

It will keep killing you until you assume that B and v are still parallel (as in part b). :smile:
 
Last edited:
  • #12
my brains so tired now I am not seeing it even if they are parallel
 
  • #13
Octavius1287 said:
my brains so tired now I am not seeing it even if they are parallel

If B and v are parallel what does B do to the path of the charged particle?
 
  • #14
i think nothing
 
  • #15
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
the force of the lectric field need to be equal to the force of the moving particle right
 
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  • #17
i guess i don't see the equation where time would play a factor
 
  • #18
Octavius1287 said:
the force of the lectric field need to be equal to the force of the moving particle right

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
Octavius1287 said:
i guess i don't see the equation where time would play a factor

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
i mean in equation form, i see how it would but reading my book i don't see how to set up the equation
 
  • #21
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
F=ma, but with a - acceleration...damn it i didnt see that
 

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