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Magnetic Fields of ion source

  1. Jul 14, 2009 #1
    1. The problem statement, all variables and given/known data

    The ion source is a spectrometer produces both singly and doubly ionized species, X+ and X2+. The difference in mass between these species is too small to be detected. Both species are accelerated through the same electric potential difference, and both experience the same magnetic field, which causes them to move on circular paths. The radius of the path for the species X+ is r1 while the radius for species X2+ is r2. Find the ratio r1/r2 of the radii.

    So this means:

    2. Relevant equations

    3. The attempt at a solution
    We know that the charge of the X2+ is twice the X. I am just so confused about where to go from here. My teacher has written out the solution but it still makes no sense.

    Please help :(

  2. jcsd
  3. Jul 14, 2009 #2


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    You just about have it with your equation don't you?

    Consider the equation for each particle.

    Now take the ratio of the equations.

    If the masses are virtually identical, the acceleration through the electrical potential and hence the velocities are also the same then aren't they?

    By the time you get through canceling things out it looks to me like you can determine r1/r2 by inspection.
  4. Jul 14, 2009 #3
    so heres where i am at:


    I then divide r1 and r2 getting

    then with canceling out I get r1/r2=2

    I am confused, the teachers notes say its the square root of 2, I am not sure where this comes in
  5. Jul 14, 2009 #4


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    Oops. Sorry. The acceleration through the electrical potential will result in a velocity that is not identical. It will be different how?
  6. Jul 14, 2009 #5
    EPE= Vq
    so if the force goes up... and the mass is the same the acceleration will have to go up with the second one by 2 right?
  7. Jul 14, 2009 #6


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    I think they want you to accelerate that from rest. So ...

    W = q*ΔV = 1/2*m*v2

    Which means that the ratio of v2 = 2:1, since v2 is proportional to the charge.
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