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Magnetic Force on Moving Charges

  1. Dec 10, 2008 #1
    A proton high above the equator approaches the Earth moving straight downward with a speed of 355 m/s. Find the acceleration of the proton, given that the magnetic field at its altitude is 4.05 X 10^-5 T.


    2. Relevant equations

    F = MA
    =eVBsin?
    a = evB/m


    3. The attempt at a solution

    a = ((1.6x10^-9C)(355 m/s)(4.05x10^-5)/(1.673x10^-27)) = 1.38x10^16 m/s^2



    My attempt is pretty straight forward, but I think im missing something. There has to be some relevance to the magnet force near the equator and the angle that the particle is moving, I think its perpendicular to earths magnetic lines, but im not sure.

    Thanks.
     
  2. jcsd
  3. Dec 11, 2008 #2

    tiny-tim

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    Science Advisor
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    Hi AcidicVision! :smile:

    I think you're right … assuming they mean the magnetic equator :rolleyes:

    not a very good question, is it? :grumpy:
     
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