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Bring a particle to rest within a field

  1. Apr 3, 2013 #1
    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. jcsd
  3. Apr 3, 2013 #2

    mfb

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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.
     
  4. Apr 3, 2013 #3
    v is perpendicular to the field. I know B, m, and q and r
     
  5. Apr 3, 2013 #4

    mfb

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    What about the electric field?
    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.
     
  6. Apr 3, 2013 #5

    rude man

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    Use a cyclotron in reverse!
     
  7. Apr 3, 2013 #6
    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
     
  8. Apr 3, 2013 #7

    rude man

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    So in part c are the B and E fields still in parallel?
     
  9. Apr 3, 2013 #8
    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
     
  10. Apr 3, 2013 #9

    rude man

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    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 ... :eek:
     
  11. Apr 3, 2013 #10
    ugh this has been killing me for hours
     
  12. Apr 3, 2013 #11

    rude man

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    EDITed:

    It will keep killing you until you assume that B and v are still parallel (as in part b). :smile:
     
    Last edited: Apr 3, 2013
  13. Apr 3, 2013 #12
    my brains so tired now im not seeing it even if they are parallel
     
  14. Apr 3, 2013 #13

    rude man

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    If B and v are parallel what does B do to the path of the charged particle?
     
  15. Apr 4, 2013 #14
    i think nothing
     
  16. Apr 4, 2013 #15

    mfb

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    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.
     
  17. Apr 4, 2013 #16
    the force of the lectric field need to be equal to the force of the moving particle right
     
    Last edited: Apr 4, 2013
  18. Apr 4, 2013 #17
    i guess i dont see the equation where time would play a factor
     
  19. Apr 4, 2013 #18

    rude man

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    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.
     
  20. Apr 4, 2013 #19

    rude man

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    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.
     
  21. Apr 4, 2013 #20
    i mean in equation form, i see how it would but reading my book i dont see how to set up the equation
     
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