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Magnetic force of a proton

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A proton with the speed 3.7*10^6 m/s is moving in a magnetic field, parallel to the field.
    The magnetic B-force is 0.45T.

    a)Draw a figure that shows the proton in the magnetic field. Indicate the direction of the proton's velocity, the direction of the magnetic field and the direction of the magnetic force.
    b) Determine the magnetic force.
    c) Determine the radius of the circular motion that the proton will make.
    2. Relevant equations
    a) Using the right hand rule
    b) F= qvB
    c) r= mv/qB

    3. The attempt at a solution
    a) The picture I drew looked like this
    [​IMG]

    where v is the direction of velocity, Fm the magnetic force and the B-force is the cross going into the paper.
    b) F = (1.602*10^-19)*3.7*10^6*0.45=2.67*10^-13 N
    c) r=mv/qB=[(1.673*10^-27)*(3.7*10^6)]/[(1.602*10^-19)*0.45]
    r=0.086 m

    Could anyone tell me if this is correct? I would really appreciate some help!
     
  2. jcsd
  3. Sep 2, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    A proton with the speed 3.7*10^6 m/s is moving in a magnetic field, parallel to the field.The magnetic B-force is 0.45T.

    It should be perpendicular to the field.
    Rest of the calculation is correct
     
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