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Magnetic Field and Moving Charge

  1. Jul 29, 2009 #1
    1. The problem statement, all variables and given/known data
    The drawing shows a charged particle (q=2.80x10^-6C) moving along the +y axis with a speed of 4.80X10^6 m/s. A magnetic field of magnitude 3.35x10^-5 T is directed along the +z axis, and an electric field magnitude 123 N/C points along the -x axis. Determine the (a) magnitude and (b) direction of the net force that acts on the particle.

    q=2.80x10^-6C V=4.80X10^6 m/s B=3.35x10^-5 T E= 123 N/C

    2. Relevant equations



    3. The attempt at a solution
    2.80x10^-6C / 123 N/C = 2.28X10^-8 N

    F=2.28X10^-8 N

    Vsin@ = F/Bq 4.80X10^6sin@ = (2.28X10^-8)/ (3.35x10^-5)( 2.80x10^-6)

    @= .0029deg in the positive direction

    I think I am doing the problem right, but I think my algebra might be bad?

    Attached Files:

  2. jcsd
  3. Jul 29, 2009 #2


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    Homework Helper

    Force due to magnetic field F = qVBsinθ.
    Find the direction of the magnetic filed by Flemming's left hand rule.
    Find the direction of the electric field. Then find the resultant.
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