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Radius of curvature atom path

  1. Apr 7, 2009 #1
    A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?


    the equation i was using was
    r = mV/|q|B

    m = 6.68 x 10^-27
    V = 4.00 x 10^3
    B = 0.450T
    q = 1.6 x 10^-19

    I dont think this is right. any help pr tips?
    Thanks :)
     
  2. jcsd
  3. Apr 7, 2009 #2

    LowlyPion

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    Homework Helper

    First figure the kinetic energy, from the work done on the charge.

    W = q*ΔV

    From ½mv² you can derive a value for v .

    Then you can use your second equation derived from

    F = mv² /R = qv*B

    R = mv/qB
     
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