1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Nabeshin

    User Avatar
    Science Advisor

    Your work is a little tough to follow, but it is worth noting that this problem has nothing to do with electric fields, only magnetic ones. So, I'll help you out by saying that only equations 3 and 4 under your list of relevant equations are actually relevant.
     
  4. Apr 6, 2009 #3
    So would I be able to solve eqautions 3 and 4 somehow to find mass and velocity? Those are my two unknowns. I need velocity to solve for mass.
    Is there another relationship I am missing to help me solve for velocity?
     
  5. Apr 6, 2009 #4

    Nabeshin

    User Avatar
    Science Advisor

    Well you have equations 3 and 4, and only two unknowns like you say, m and v. 2 equations 2 unknowns sounds like a solvable system to me. (Try substitution!)
     
  6. Apr 6, 2009 #5
    Substitution worked wonderfully! Thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook