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What is the radius of it's path?

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data
    A 5.6 MeV (kinetic energy) proton enters a 0.18 T field, in a plane perpendicular to the field.
    What is the radius of its path?

    2. Relevant equations
    kinetic energy: KE = 1/2 mv^2
    velocity: v = sqrt(2KE / m)
    radius: r = mV/qB

    3. The attempt at a solution

    KE = 5.6 MeV = 8.9722x10^-13 J
    m(p) = 1.67x10^-27 kg
    so v = 3.3x10^7 m/s

    r = (m(p))(v) / (q(e))(B)

    where m(p) is mass of proton
    v is velocity as determined by manipulating the KE equation
    q(e) is the charge of the proton which is the same as that of an electron (except positive)
    B is the given magnetic field

    the homework system I'm on, keeps telling me that my answer is wrong. if you see what i'm doing wrong or where i've made a mistake, please let me know.
  2. jcsd
  3. Aug 6, 2011 #2
    If anyone was in the process of helping me on this one, thank you; but i've figured it out.

    I was probably making some sort of calculating error.

    r = 1.9 m
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