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Motion of charge in electric and magnetic fields

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A proton is accelerated from a rest position into a uniform electric field and magnetic field that are perpindicular to each other, as shown, The proton passes through the parralel plates without being deflected, at a constant velocity. When the proton leaves the plates, it only experiences a magnetic force. The magnitude of the electric field is 1.2 x 10^4 n/c and the magnetic field is 0.20T.

    IMG_3457.JPG

    a) Find the speed of the proton.
    b) Find the radius of the path of the proton as it leaves the plates.

    2. Relevant equations
    FM = qvB
    FE = Eq
    Fc = (mv^2)/r

    3. The attempt at a solution

    a) Fnet = FM + FE
    0 = FM - FE
    FM = FE
    qvB = Eq
    v = E/B
    = (1.2 x 10^4 N/C) / (0.20 T)
    = 6.0 x 10^4 m/s

    b) FM = FC
    qvB = mv^2/r
    qB = mv/r
    r = mv/qB
    = (1.67e-27 kg)(6.0e4 m/s)/(1.60e-19 C)(0.20T)
    = 3.1 x 10^-3 m

    I'm not certain how to think about this situation since the velocity associated with magnetic force is perpendicular to the force.


     
  2. jcsd
  3. Mar 2, 2016 #2

    haruspex

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    Why is that a problem?
     
  4. Mar 2, 2016 #3
    It's not a problem so much as a source of uncertainty/confusion for me. I'm used to working with net force in one dimension at a time.

    The electric force between the plates is pushing down on the proton but not causing it to move from it's horizontal path. I assume that's because there's an equal and opposite magnetic force pointing up? And the proton is moving sideways because of the magnetic force? So there's no horizontal force. It is moving at a constant velocity, so no acceleration.
     
  5. Mar 2, 2016 #4

    haruspex

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    Not sure what you mean by that. Sideways with respect to its initial motion or sideways as viewed in the image? The magnetic force will be vertical. By 'moving' do you mean accelerating?
    Other than that, what you wrote looks right.
     
  6. Mar 2, 2016 #5
    Oh, when it's between the plates it's moving at a constant velocity rightwards, in the diagram, perpendicular to the magnetic force, because of the magnetic force?
     
  7. Mar 2, 2016 #6

    haruspex

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    Depends what you mean by "because of the magnetic force". It moves at constant velocity because there is no net force, and that is because the electric and magnetic forces cancel (which you used in your calculation).
     
  8. Mar 2, 2016 #7
    Okay, the proton was accelerated outside the fields and already had that speed upon entering them. The magnetic force is not actually giving it speed. Absent the electric field, the uniform magnetic field would cause the proton to curve off into uniform circular motion, but instead the electric force pushes back down on it so it stays in a straight line.

    Do my steps look about right?
     
  9. Mar 2, 2016 #8

    haruspex

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    Yes.
     
  10. Mar 2, 2016 #9
    Alright, thanks for your help
     
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