# why dosen't a charged particle radiate energy in circular motion in a uniform magnetic field?

I have studied in my Physics course that one of the drawbacks of Rutherford's atomic model was that when an electron will revolve around the nucleus, it is undergoing acceleration and so it should radiate energy and consequentially fall into the nucleus.

Similarly when a charged particle is projected in the plane perpendicular to a uniform magnetic field it executes uniform circular motion withradius $r=mv/qB$.

My question is why isn't the charged particle radiating energy here? Even in this case the charged particle is accelerating, just as it was in Rutherford's model of the atom. So shouldn't the radius decrease in this case also?

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A charged particle circulating in a magnetic field does radiate energy, and it is called synchrotron radiation. All circular particle accelerators have energy losses due to this radiation.

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It's only synchrotron radiation if the particle velocity is relativistic. Otherwise it's cyclotron radiation. The emission dynamics are quite different for the two cases, hence the nitpick. :) –  Kitchi Nov 8 '12 at 13:37
so does that mean that the radius will keep on decreasing with time? –  Shantanu Nov 8 '12 at 15:03
Yes. The radius decreases, that is why accelerators have to feed pulses to beams to keep them at the same momentum/ radius –  anna v Nov 8 '12 at 15:08
@Kitchi both cyclotron and Synchrotron are at relativistic energies. When one talks of particles one is at fractions of c. It is at the ultra high energies, when the harmonics of the cyclotron radiation become so dense it becomes a continuum. astro.umd.edu/~miller/teaching/astr601/lecture16.pdf –  anna v Nov 8 '12 at 15:11
@annav - Synchrotron radiation is only for relativistic motion, because then you will see relativistic beaming of the radiation in the direction of motion. Cyclotron radiation is in the non-relativistic limit, and as long as $\lambda$ >> size of the radiating system, we can approximate a dipole radiation field. These two links explain that in much more detail. (pdf warning). –  Kitchi Nov 8 '12 at 16:17