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Alpha particle magnetism

  1. Jul 16, 2006 #1
    "An alpha particle is accelerated in the +x direction through a voltage of 1000V. The particle then moves in an undeflected path between two oppositely charged parallel plates in a uniform magnetic field of 50 mT in the +y direction. (a) If the plates are parallel to the xy plane, what is the magnitude of the electric field between them?"

    Since the particle's path is undeflected, the electric force and the magnetic force on the particle must be equal in magnitude. So
    Ee = qE = Fm = qvB
    E = vB

    However, I don't have v, the velocity. How do I solve this?
     
  2. jcsd
  3. Jul 16, 2006 #2

    Doc Al

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    But you do have the voltage. Make use of it.
     
  4. Jul 16, 2006 #3
    I know that V = Ed, but then I don't have d...

    I'm still not sure what to do.
     
  5. Jul 17, 2006 #4

    Hootenanny

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    HINT: Voltage can be defined as work done per unit charge, thus;

    [tex]V = \frac{W}{Q}[/tex]

    Think kinetic energy :wink:
     
  6. Jul 17, 2006 #5
    [tex]W = Vq = K = \frac{1}{2}mv^2[/tex]
    [tex] v = \sqrt{\frac{2Vq}{m}} [/tex]
    [tex] E = vB = 1.6 \times 10^4 V/m[/tex]
    ...that seems correct, but doesn't [tex]W = \Delta K[/tex]? If it does, isn't the change in kinetic energy zero, because there's only 2 forces working on the particle, and both of the forces are not acting in the direction of the particle's displacement...?
     
  7. Jul 17, 2006 #6

    Doc Al

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    Once the particle gets up to speed and enters the region of crossed magnetic & electric fields, then its KE remains constant. But so what? (The purpose of considering the change in KE is to find the particle's final speed before it enters that second region.)
     
  8. Jul 17, 2006 #7
    ah ic. thanks
     
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