Bring a particle to rest within a field

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SUMMARY

The discussion centers on calculating the electric field (E) required to bring a charged particle to rest within a magnetic field. The particle has a charge of 10-24 C, a mass of 10-24 kg, and an initial velocity (v) of 1 km/s. The magnetic field (B) is given as 2 T, and the stopping time is specified as 100 microseconds. Participants emphasize that the electric field must be aligned with the particle's velocity to effectively decelerate it, and they suggest using the relationship between force, mass, and acceleration to derive the necessary equations.

PREREQUISITES
  • Understanding of Lorentz force and its components: magnetic force (qvxB) and electric force (qE).
  • Familiarity with kinematic equations and concepts of acceleration and deceleration.
  • Knowledge of the relationship between electric fields and forces on charged particles.
  • Basic principles of cyclotron motion and the effects of magnetic fields on charged particles.
NEXT STEPS
  • Calculate the required electric field E using the formula F = ma, where F is the force exerted by the electric field.
  • Explore the relationship between stopping time and acceleration in the context of charged particle motion.
  • Investigate the implications of magnetic field orientation on particle trajectory and motion.
  • Review the principles of cyclotron motion to understand the effects of magnetic fields on charged particles in motion.
USEFUL FOR

Physics students, electrical engineers, and anyone involved in particle physics or electromagnetic field studies will benefit from this discussion.

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
 
Last edited:
  • #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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