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Cyclotronic motion problem.

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the required magnetic field to give an electron deflection angle of θ=90deg. The other variables (such as Rd, Re, and the electron energy in keV) are currently dummy variables such that this problem will initially be solved symbolically.


    2. Relevant equations
    (1) F=q(v x B)
    (2) F = qvbsinθ
    (3) QBR = p

    3. The attempt at a solution
    Using the following figure (apologies for poor paint skills), I've taken the cross product of (v x B), recognizing that vx = 0, By = Bz = 0. This gives me that v x B = (vzBx)y - (vyBx)z where y and z are unit vectors.

    Edit: Apologies; I seem to be awful at remembering to attach photos, files in e-mail, etc. Here's the figure: http://i.imgur.com/V5TN9.png

    I am having trouble fitting the cross product to another equation to yield any kind of results. Let me know if I should do more work (I've been reading through Griffith's Intro to Electrodynamics for about two hours now trying to make any headway).
     
    Last edited: Oct 7, 2011
  2. jcsd
  3. Oct 7, 2011 #2

    rude man

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    The figure shows x as + into page. That is not a right-handed coordinate system. Either the B field is directed into the page or into the +x direction.

    If you intended a left-hand system we would like to know about it. It's very rare, although my high school textbook of the '50's actually used it.


    Is the B field understood to be confined to within the circle of radius RB?
     
  4. Oct 7, 2011 #3

    SammyS

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    The one I saw in the figure is right handed.

    (+)x is into the page, y is up, z is to the right. --- That's RIGHT handed. Sitting in the 1st octant looking towards the origin, x to y to z is counter-clock-wise.
     
  5. Oct 7, 2011 #4
    Yes, B is to be confined within the circle radius RB (accidentally noted RB as Rd in the description; apologies). As I understand it, we have to take into account the relativistic and classical momentum to find the more realistic values of B (assuming all electrons are uniform, originating from the same source traveling in the same direction). I also may be wrong with taking the cross product of the force, and I may just need (2) as well as the following equations:

    (4) pv/r = F
    (5) E^2 = m^2 * c^4 + p^2 * c^2.
     
  6. Oct 7, 2011 #5

    rude man

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    I beg everyone's pardon, it is right-handed. How embarrassing for me.
     
  7. Oct 7, 2011 #6

    diazona

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    Consider this: since you've already got the directions of all relevant quantities specified in the figure, you can work with equations that only involve magnitudes. So you don't need to calculate out the components of the cross product.
     
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