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B Does an electron beam bending lose energy by photon emission

  1. Apr 3, 2016 #1
    A beam of electron in vacuum with velocity v enter a region of spa e with a electric field E. The field is such the electrons circle with radius r. The electrons are now accelerating at constant tangential speed.

    Because this is not an atomic orbital then by classical physics the electrons radiate photons of frequency f.

    Does this radiation make the electrons lose energy and spiral inward.

    The answer is no because such circular beams can be created and are stable, so what compensates for the radiated energy loss?
     
  2. jcsd
  3. Apr 3, 2016 #2
    The 2.45 Ghz frequency generated by the electron beam circling(bending) in the Magnetron in a microwave oven produces no light and therefore no photons.
    Visable light is associated with photons is within the frequency range of ~405 - 790 THz. It is not usually associated with electron beams.
     
  4. Apr 3, 2016 #3

    Drakkith

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    That is incorrect. ALL electromagnetic radiation is composed of photons, not just visible light.

    The radiation would indeed lead to an in-spiraling of the electron beam. Stable electron beams, such as in circular particle accelerators, have to keep supplying energy to keep them in a steady path.
     
  5. Apr 3, 2016 #4
    I never implied visible photons.

    So in a lab bench set up where the beam of electrons is bent by the field from Helmholtz coils which I assume anyone reading are familiar with in most undergrad labs, is the energy removed from the magnetic field measurable. Shouldn't we see a fluctuations in the coil current?? Or should the coils draw more power when the beam is present?
     
  6. Apr 3, 2016 #5

    Drakkith

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    I don't think energy is removed from the magnetic field. Those beams are probably not stable. The in-spiraling of the electron beam just probably isn't noticeable since the beam isn't moving relativistically or being curved hard enough.
     
  7. Apr 3, 2016 #6
    Not wanting to mention manufacturers names but these beams are clearly stable in bench top devices. You can visually measure the radius. The coil currents are only a few amps.
     
  8. Apr 4, 2016 #7

    Drakkith

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    I'd need a link to such a device to see it and help answer your question.
     
  9. Apr 4, 2016 #8
    Both Teltron and Pasco Scientific sell such devices. I used a Pasco its called e/m experiment.

    Will link if that's not spamming.
     
  10. Apr 4, 2016 #9

    Drakkith

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    Lol, no. It's not spamming. :biggrin:
     
  11. Apr 4, 2016 #10
  12. Apr 4, 2016 #11

    Drakkith

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    Okay, we just had one of those in my E&M class the other week. I can't guarantee anything, but I was under the impression that the electrons are emitted at a low velocity, in a low-strength magnetic field, and the reaction force from the radiation they release isn't measurable at this scale.
     
  13. Apr 4, 2016 #12

    davenn

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  14. Apr 4, 2016 #13

    jtbell

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    Look up "synchrotron radiation" and try calculating the rate of energy loss for electrons in that apparatus. The accelerating voltage is typically a few 1000 V (in our setup it was about 1500 to 5000 V), so the initial kinetic energy is a few keV. The radius of curvature is maybe 5 cm.
     
  15. Apr 4, 2016 #14
    I think it is a (non relativistic) cyclotron not a synchrotron, unless a sync and cyc are the same.

    Formula is here;

    https://en.m.wikipedia.org/wiki/Cyclotron


    and

    0e3315fcbd7a6dea6f4bace9051212f2.png

    If this equals hf then

    f= (q^2 B^2 R^2)/(h 2m)

    I'm not sure it is valid to assume the kinetic energy of the electrons all turns into photons and can be equated to hf though??
     
    Last edited: Apr 4, 2016
  16. Apr 4, 2016 #15

    ZapperZ

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    1. There IS energy loss when an electron beam moves in a circular path.

    2. For a tabletop experiment shown in a typical e/m experiment, the loss is VERY SMALL, because the charge on each electron is very small, and the speed of these electrons is "small".

    3. Evidence for such energy loss can seen in when the electron beam are in MeV and GeV energy range. Synchrotron radiation is a clear example of this. It is also why the LHC is so large in diameter, so as to minimize the bending radius.

    Zz.
     
  17. Apr 4, 2016 #16
    Yes, it is called Bremsstrahlung radiation. It is how X-Rays are generated in your doctor's X-ray machine.
     
  18. Apr 4, 2016 #17

    sophiecentaur

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    Ifaik, Bremsstrahlung radiation happens due to close interaction of high speed electrons with atoms in a target. X rays are very high energy but the photons produced by larger structures are surely a lot lower energy because the acceleration of the charges is lower. Would they be classed as Bremsstrahlung radiation?
     
  19. Apr 4, 2016 #18
    Quite not, doctors xray is by high energy collision knocking k-shell electrons out of atoms in a solid target. Not even close to this phenomena.
     
  20. Apr 4, 2016 #19
    I stand corrected. I was told by the X-ray tech the X-rays were from Bremsstrahlung. I guess I was misinformed. Thanks Houlahound.
     
  21. Apr 4, 2016 #20
    It could be called bremss. in Xrays. I don't know the names I just happen to know the process of medical xray production in dentist and doctor surgeries.
     
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