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Electron moving in a betatron

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data

    The magnetic flux through the orbit of an electron increases by 5Wb every second. The electron is accelerated to the point where its energy is 25MeV. Electron's orbit has a radius of 25cm. How many orbits does the electron have to complete in order to gain that much energy?

    2. Relevant equations

    [tex] F=qvB [/tex]
    [tex] B=\frac{\Phi}{S} [/tex]

    [tex] a=\frac{v^2}{r} [/tex]
    [tex] F=ma [/tex]
    [tex] m=\frac{m_{e}}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

    [tex] E=\frac{mv^2}{2} [/tex]

    3. The attempt at a solution

    [tex] \frac{d\Phi}{dt}=5 \Rightarrow \Phi (t)=5t [/tex]

    [tex] F=q_{e}vB [/tex]
    [tex] F=ma=\frac{m_{e}}{\sqrt{1-\frac{v^2}{c^2}}}\frac{v^2}{r} [/tex]

    The Lorentz force and the centripetal force should be equal and in opposite directions, therefore [tex] q_{e}vB(t)=ma=\frac{m_{e}}{\sqrt{1-\frac{v^2}{c^2}}}\frac{v^2}{r} [/tex] where [tex] B(t)=\frac{\Phi(t)}{S}=\frac{5t}{\pi r^2} [/tex]

    I get [tex] q_{e}v5t=\frac{m_{e}}{\sqrt{1-\frac{v^2}{c^2}}}\frac{v^2}{r} [/tex]
    and therefore [tex] v=\frac{1}{\sqrt{\frac{m_{e}^2 \pi^2 r^2}{25q_{e}^2 t^2}+\frac{1}{c^2}}} [/tex]

    Since I now have v(t) and I know the final speed of the electron (since I know the final energy) I could also derive t(v) and see how long it would take for an electron to accelerate to this point.
    Eventually I would get the answer by solving this equation:

    [tex] s=\int_{0}^{t_{final}}v(t)dt [/tex]

    I know that the answer should be about 8000km which is about 41 million orbits. So far I haven't got even close to that. I can't seem to find any mistakes in my equations nor have I found any conceptual flaws.

    Any ideas?
  2. jcsd
  3. Dec 10, 2007 #2


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    Homework Helper

    Here's a thought. When you gave the rate of magnetic flux change, I considered using Faraday's Law. You still would do this, but it's not so fancy. (It helps that the orbit radius is held fixed.)

    Consider that Wb/sec = T·(m^2)/sec ; since T = N/(A·m) , then Wb/sec = J/(A·sec) = J/C = V . (Actually, Faraday's Law also says as much directly). So the electron is effectively starting "from rest" and experiences a 5 V emf on each orbit (from the line integral of E·ds along the circle). How many orbits does it take to gain 25 MeV of kinetic energy and what total distance is that? (BTW, for the radius you cite, 41·10^6 circuits of the fixed circle is not 8000 km.)

    Oh, and you should be all right solving this classically: 25 MeV of kinetic energy barely brings the Lorentz factor to 1.05...
    Last edited: Dec 10, 2007
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