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Homework Help: Radius of the path of an electron in a magnetic field?

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data
    15MeV electrons, enter magnetic field with strength of 0.7T. What is the radius of the path of the electrons? What energies could the electrons have and be confined to orbits with radii within 5% of the 15MeV electrons?

    2. Relevant equations

    [itex]E = \frac{m_{0}c^{2}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}[/itex]

    [itex] r = \frac{mv}{qB} [/itex]

    3. The attempt at a solution

    I solved for v using relativistic energy equation. I got:

    [itex] v = c \sqrt{\left(1-\frac{m_{0}c^{2}}{E}\right)^{2}}[/itex]

    [itex]v = 2.898x10^{8} m/s [/itex]

    and plugged that value into formula for radius

    [itex] r = \frac{\left(9.11x10^{-31}kg\right)\left(2.898x10^{8} m/s\right)}{\left(1.602x10^{-19}C\right)\left(0.7T\right)} [/itex]

    I get r = 0.002354m

    I then take [itex]\pm[/itex]5% of the radius and get 0.0022363m to 0.0024717m

    When I take the radius r=0.0022363 and work backwards by using it to find the velocity then energy I get a scant 1.3 MeV

    and when I take r = 0.0024717m and work backwards, I end up with a velocity greater than c!

    Is my approach correct?

    Thank you
  2. jcsd
  3. Sep 22, 2011 #2


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    Homework Helper

    [tex]r = \frac{p}{qB} = \frac{\gamma mv}{qB}[/tex]
    for a relativistic electron. (I'd encourage you to try to prove this yourself, it's not that complicated)

    Also, you can save some work by not solving for [itex]v[/itex], if you can directly relate energy to momentum.
  4. Sep 23, 2011 #3
    Thank you for the reply. I corrected the formula for [itex]r[/itex], and I took your advice and related energy to momentum.


    [itex]r = \frac{p}{qB} = \frac{\gamma mv}{qB}[/itex]

    [itex]E^{2} = \left(pc\right)^{2} + \left(m_{0}c^{2}\right)^{2}[/itex]

    [itex]p = \frac{\sqrt{E^{2} - \left(m_{0}c^{2}\right)^{2}}}{c}[/itex]

    [itex]p = \frac{\sqrt{\left(2.4x10^{-12}J\right)^{2} - \left(8.187x10^{-14}J\right)^{2}}}{3x10^{8}\frac{m}{s}} = 7.995x10^{-21} kg\frac{m}{s}[/itex]

    [itex]r = \frac{7.995x10^{-21} kg\frac{m}{s}}{\left(1.602x10^{-19}\right)\left(0.7T\right)} = 0.071m[/itex]

    [itex]\pm[/itex] 5% of 0.071m = 0.0677m[itex]\rightarrow[/itex] 0.0749m

    [itex]p = rqB[/itex]

    [itex] r = 0.0677 m[/itex]

    [itex] p = \left(0.0677 m\right)\left(1.1214x10^{-19}\right)\frac{kg}{s} = 7.596x10^{-21} kg\frac{m}{s}[/itex]

    [itex]E = \sqrt{\left[\left(7.596x10^{-21} kg\frac{m}{s}\right)\left(3x10^8 \frac{m}{s}\right)\right]^{2} + \left(8.187x10^{-14}kg \frac{m^{2}}{s^{2}}\right)^{2}} = 2.28x10^{-12}kg \frac{m^{2}}{s^{2}} = 14.2MeV[/itex]

    Then do the same for [itex]r = 0.0749m[/itex], and [itex]E = 15.7MeV[/itex]

    I believe this looks right, any thoughts?

    Thank you
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