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Magnetism Problem

  1. Apr 25, 2005 #1
    A proton moves in a circular path perpendicular to a 1.15-T magnetic field. The radius of its path is 8.40 mm. Calculate the energy of the proton in eV.

    So, I calculated the velocity with:
    [tex]qvB = \frac{mv^2}{r}[/tex]
    [tex](1.60 \times 10^-19)(1.15) = \frac{(6.27 \times 10^-27)v}{0.0084}[/tex]
    [tex]v = 2.47 \times 10^5 m/s[/tex]
    Then calculated the energy:
    [tex]E = qv[/tex]
    [tex]E = (1.60 \times 10^-19)(2.47 \times 10^5)[/tex]
    [tex]E = 3.95 \times 10^-14 J[/tex]
    Then divided that by eV (1.60 x 10^-19) but I get something like 247000. The answer says 4.47 keV though. What am I doing wrong?
  2. jcsd
  3. Apr 25, 2005 #2


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    Homework Helper

    Mass of a proton is:[tex]1.67\times10^{-27}kg[/tex] not the number you used.

    Also you have the wrong equation for energy. I think you took E=qV...where V is voltage not velocity... however that equation does not apply here.

    The energy of the proton is kinetic energy... so use [tex]E=\frac{1}{2}mv^2[/tex]
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