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Direction of a particle in a uniform magnetic field.

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    a beam of particles with velocity v^vector enters a region that has a uniform magnetic field B^vector in the +x direction.

    show that when the x component of the displacement of one of the particles is 2 * Pi * (m / qB) * vcos(theta), where theta is the angle between v^vector and B^vector, the velocity of the particle is in the same direction as it was when the particle entered the field

    2. Relevant equations

    the period of circular motion is T = 2 * Pi * m / q * B

    3. The attempt at a solution

    if I divide v = d * t = x * t

    if I choose t to be T then I find v = v cos(theta).

    then thing is, I don't really undertstand the question, and I want to know how to proceed from there.
     
  2. jcsd
  3. Mar 31, 2009 #2

    tiny-tim

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    Hi nvictor! :smile:

    (have a theta: θ and a pi: π and use bold for vectors :wink:)
    The displacement in the x-direction is (vcosθ)t (because speed in that direction is constant), so the question is asking you to prove that the period is 2πm/qB :smile:
     
  4. Mar 31, 2009 #3
    thanks a lot for the reply.
     
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