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Mass of charged particle in magnetic field

  1. Apr 6, 2009 #1
    1. A 1.56024 μC charged particle with a kinetic
    energy of 0.115905 J is placed in a uniform
    magnetic field of magnitude 0.150267 T.
    If the particle moves in a circular path of
    radius 3.13685 m, find its mass. Answer in
    units of kg.

    2. Relevant equations
    Fe= Fm
    Fe= qE
    KE=mv^2/2
    Fm= qvB= mv^2/r




    3. I decided to first find E: E =Ke (q/r^2)= 8.98e10 (1.5602e-6 C/3.13685^2)=1425.455

    Then I solved for Fe= qE= 1425.455 (1.5602e-6)= 2.2239e-3= Fm

    Then I used Fm= qvB and solved for v = 9503.84615 m/s
    Then I used KE = mv^2/2 to sove for m = KE(2)/ v^2= (0.115905)(2)/ (9503.84615 m/s )^2=3e-9 kg=m
    But this isn't correct. What am I missing here? Thank you in advance for any help!
     
  2. jcsd
  3. Apr 6, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper



    What is the E field? I don't see that.
    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c2

    Don't you have a charge that is moving in a circle in a B field.

    |qV X B| = mV2/R

    And you know that 1/2mV2 = .115 J
     
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