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Equating mass and charge ratio of cathode rays

  1. Aug 14, 2003 #1

    i was approached by this problem,

    prove q/m = (2V)/(B^2r^2) [2V over B-squared r-squared]

    so far i have,

    F[electric] = qE

    F[magnetic] = Qvb

    F[electric] = F[magnetic]

    therefore, v = E/B

    now i am a bit confused/wrong

    E[kinetic] = (mv^2)/2

    therefore, q x V = (mv^2)/2

    therefore, q/m = E^2/(2VB^2)

    jus by looking at the result, i have to get E= 2V/r.

    any insights will be greatly appreciated....
    if my method even correct?
  2. jcsd
  3. Aug 14, 2003 #2
    F[magnetic] = qvB

    Elecrons will be deflected in a circular path, so

    F[centripital] = mv^2/r

    Equate these, and get v = Bqr/m

    Electrons accelerate through a potential, so

    E[kinetic] = mv^2/2 = qV

    Substitute v = Bqr/m into this, and solve for q/m.
  4. Aug 15, 2003 #3
    thanks a lot..

    works out perfectly
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