Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

When an electron travels in circles in a uniform magnetic field, it must lose energy because all accelerated charges radiate, and must therefore spiral down to the center. Is this energy compensated by the magnetic field? Or where does this energy go?

share|improve this question
    
Spiral? how exactly? Is it doing a helical motion?If not, please post a diagram. –  udiboy1209 Jul 23 '13 at 10:34
    
spiral motion. spiral motion or helical motion whatever may be, it is accelerated and release energy. –  albedo Jul 23 '13 at 10:38
    
If its a helical motion it will not release energy, even if it is accelerated, just like electrons revolving around a nucleus do not release energy. –  udiboy1209 Jul 23 '13 at 10:40
    
electron moving in a magnetic field is not like the electron in atoms. According to the quantum mechanics the electrons that are bound to an atom are standing waves that completely engulf and surround the nucleus. It won't release the energy... –  albedo Jul 23 '13 at 10:53

2 Answers 2

up vote 1 down vote accepted

You are right. An electron in a uniform magnetic field will travel in circles (or in a helix, up to a change in frame of reference), but this means that it is an accelerated charge and it must therefore radiate and lose energy. This radiation is known as synchrotron radiation, and it is a major design issue for particle accelerators. (In fact, it is the reason for a recent trend to go back to linear accelerators, which are less efficient as each accelerating stage only works once per particle, but are not subject to this.) It can also be harnessed to make synchrotron light sources, and with some extra work one can build a free-electron laser using that principle.

In short, then, the electron will spiral down to the centre and lose all its kinetic energy as electromagnetic radiation.


(For the more quantum-mechanical minded, now that Landau eigenstates have joined the fray, this means that all excited Landau states will have to decay through radiative coupling to the ground state with zero angular momentum. Once there, though, the uncertainty principle kicks in and stops the electron getting localized to radii smaller than the characteristic harmonic oscillator length $$x_0=\sqrt\frac{\hbar\omega_c}{m}=\frac{\sqrt{\hbar eB}}{m}$$ corresponding to the cyclotron frequency $\omega_c=eB/m$.)

share|improve this answer
    
Thanks, but I din't understand what prevents the electron from stopping after a long time, if the electron is circulating in a magnetic field for long time? –  albedo Jul 23 '13 at 13:24
    
@albedo The electron will spiral into the centre of the circle (slowly if it's nonrelativistic). However, at the very end it will not be perfectly localized at the centre, since that is forbidden by the Uncertainty Principle. Instead it will have a gaussian wavefunction of characteristic size $\sigma_x=x_0$. (This $x_0$ is chosen so that the minimum momentum uncertainty $\sigma_p=p_0=\hbar/\sigma_x$ will make the electron circle with a radius of order $x_0$.) –  Emilio Pisanty Jul 23 '13 at 14:13
    
So, if we somehow inject an electron bunch perpendicular to the magnetic field, this buch of electron will lose energy continuously. It will spiral and eventually will we get a buch of electron almost concentrated at some point in the magnetic field? –  albedo Jul 23 '13 at 14:27
    
Yes. For a real electron bunch, though, space charge (repulsion between the different electrons) will prevent this. –  Emilio Pisanty Jul 23 '13 at 15:18

I think the question is somewhat related to landau energy level(one electron in uniform magnetic field).

share|improve this answer
    
Could you clarify and elaborate? –  Emilio Pisanty Jul 23 '13 at 12:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.