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Radius and Pitch of an electron's spiral trajectory

  1. Jul 21, 2014 #1
    1. The problem statement, all variables and given/known data
    The uniform 45.6 mT magnetic field in the picture below points in the positive z-direction. An electron enters the region of magnetic field with a speed of 5.29E+6 m/s and at an angle of θ = 30.1° above the xy-plane. Find the radius r and the pitch p of the electron's spiral trajectory. (Enter the radius r first and the pitch p second.)
    Picture attached! Please let me know if you cant see it and I'll attempt to try attaching again :D

    Where,
    B= 45.8mT = 0.0456T
    v=5.29*10^6m/s
    Θ=30.1
    m= 9.11*10^-32kg
    q=1.60*10^-19C


    2. The attempt at a solution
    F=qvBsinΘ=mv^2/R
    qBsinΘ=mv/R
    R=mv/(qBsinΘ) = (9.11*10^-32kg)(5.29*10^6m/s)/[(1.60*10^-19C)( 0.0456T)*sin(90-30.1)]
    =7.63*10-4m

    For the second part of the question, could someone please explain to me the concept of pitch and possibly how the equation has been derived (d=vsinΘ*2Pi*m/[qB]), because I have searched google and I don't understand it much other than that it is the measurement of the height that the electron has travelled in one revolution~ Or is this equation something I should just memorize?

    Thank you sooo much!
     
  2. jcsd
  3. Jul 21, 2014 #2
    It's not something you should need to memorize, it's just the distance between points on a helix.

    You should be able to derive the equation you need quite simply if you think a little bit about the components of the electrons velocity.
    If it were travelling in a straight line without any magnetic field you have x- and y-componentnets, then a magnetic field is turned on, how are the components effected?

    The equation you gave:
    d=vsinΘ*2Pi*m/[qB]
    isn't completely necessary here you can derive a much simpler equation for it since you already know radius. (hint: or rather, you already know how far the electron will travel in one of its components as it completes a full rotation)

    However, if you're really curious about deriving that equation exactly just be careful with your angles. Remember the one you used was 90-30.1, and not the actual angle shown in the picture.
    The angle used in that equation you found should be the angle shown in the pic.
     
    Last edited: Jul 21, 2014
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