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Cyclotron Motion Due to Earth's Magnetic Field

  1. Jan 14, 2015 #1
    1. Say that an electron is heading towards the earth from the sun with an initial known velocity v. And we know that at earth's surface the magnetic field is given by B1. This B varies as (R1/R2)^2 where R1 is the radius of the earth. How can I find the location in space, R3/R1, where R3 is the distance where the electron's cyclotron radius equals the distance from earth?

    2. w = qB/m

    R = v/w ( - cos (wt) in x direction + sin (wt) in y direction )



    3. So I started by assuming that

    - since B varies with (R1/R2)^2 then
    B = B1 (R1/R2)^2

    I'm not sure if this is a right assumption...

    But anyhow,

    since we know B now then

    w = [ q B1 (R1/R2)^2 ] / m


    then I took this w and plugged it into R = v/w ( - cos (wt) in x direction + sin (wt) in y direction ) giving me

    R = (v R2^3 m)/(qB1R1^3) ( -cos (([ q B1 (R1/R2)^2 ] / m) t) + sin (([ q B1 (R1/R2)^2 ] / m)t) )

    Now I don't know what to do because of the t in the cos and sin function...


    I thought of squaring both sides to get cos^2+sin^1 = 1 but I don't think that would work out...
     
  2. jcsd
  3. Jan 14, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    You can't get the angle out, because the actual radius will always depend on the direction. I guess you are supposed to assume the motion is perpendicular to the magnetic field.
     
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