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Kinetic Energy and Magnetic Fields

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data[/b
    An electron has a kinetic energy of 5.80 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.10 10-5 T. Determine the radius of the path?

    I know

    KE = 1/2mv^2

    Using KE = 1/2mv^2 and saying KE = 5.8 x 10^-17, and m = 9.10938 x 10^-31 KG

    I get that v= 11284559 m/s

    Then using F = mv^2/r, I end up with an answer of 2.27 x 10^-12 m.

    It tells me this answer is wrong.

    Any help would be great.
  2. jcsd
  3. Feb 24, 2010 #2

    I obtained the same v as you, but what did you use for the force?
    I used F=q(vxB), in this case v cross B is vB because they are perpendicular.
    F= qvB = mv^2/r
  4. Feb 24, 2010 #3
    Ahh thanks, I was using B for force, instead of multiplying by the charge of the electron.

    Correct answer is 1.26 m.
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