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Momentum in a magnetic field

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A particla of charge q enters a region of uniform magnetic field B (pointing into the page). The field deflects the particle a distance d above the original line of flight. In terms of a, d, B, and q, find the momentum of the particle.


    2. Relevant equations
    F=dp/dt
    Fmag=(vXB)q
    I put v=vy and B=-Bx


    3. The attempt at a solution
    I started by setting the two above equations equal to eachother.
    dp/dt=(vXB)q
    I took the cross product and got
    dp=Bvqzdt
    and since v=ds/dt
    dp=Bq(ds)z
    and since the displacement was from when the particle entered the field to when it would be leaving, I said it went from 0 to a and
    p=Bqaz

    This is the answer I got but I was just concerned because it didn't include the value that the particle was vertically displaced by the magnetic force, d. Is this answer right, or does the vertical displacement appear somewhere that I've overlooked?
    EDIT: I just remembered that the direction of the velocity changes because of the magnetic force. So instead of the displacement being just a wouldn't it now be √(a2+d2 ?
     
    Last edited: Nov 17, 2009
  2. jcsd
  3. Nov 17, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    It is not possible to figure out what is going on without a diagram defining quantities a and d with respect to the particle's trajectory. Whatever the case may be, the magnetic field changes the direction of the particle's momentum but not its magnitude, so if the question asks you to find the momentum. All you need to provide is the new direction and that you can probably figure out from the diagram.
     
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