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Rolling Rod in a Magnetic Field

  1. Oct 17, 2008 #1
    Edit: The title is misleading because I initially was going to ask a different question.

    1. The problem statement, all variables and given/known data
    A singly charged ion of mass m is accelerated from rest by a potential difference ΔV. It is then deflected by a uniform magnetic field (perpendicular to the ion's velocity) into a semicircle of radius R. Now a triply charged ion of mass m' is accelerated through the same potential difference and deflected by the same magnetic field into a semicircle of radius R' = 6R. What is the ratio of the masses of the ions?


    2. Relevant equations
    W = qV
    L = mvr


    3. The attempt at a solution
    I know that V stays the same in each instance, but I am not sure how exactly to relate the masses. Thanks for any help.
     
  2. jcsd
  3. Oct 17, 2008 #2

    Defennder

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    You need to derive an expression of circular motion caused by the centripetal force being equivalent to the force exerted by B-field. Then find the ratio of the respective radii.
     
  4. Oct 18, 2008 #3
    Well F = mv^2/r = qvB

    W = qV

    If I use the above I get something where the variables depend on one another...so I get stuck.
     
  5. Oct 18, 2008 #4

    Defennder

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    How did you get stuck? From the above, you can easily get r = mv/Bq. You then use the work energy theorem to get the final speed of the charged ions after they have been accelerated through ΔV. Then find the ratio of the two with r = (formula). The variables will all cancel out leaving only m' and m and.
     
  6. Oct 20, 2008 #5
    I decided to go to bed that night and the next day I successfully solved the problem. Thanks for your help.
     
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