# Acceleration of orbiting electrons in magnetic field.

1. Mar 27, 2012

### tasnim rahman

Suppose we have an electron orbiting a nucleus,in an anti-clockwise direction, in a magnetic field, with field lines coming out of the paper/plane. If it has a particular velocity v1,it will experience a Lorentz force due to the velocity, towards the center of the orbit. We assume the extra force only changes the velocity of the electron, keeping the radius of orbit constant. Now it has a new velocity v2, where v2>v1. In a purely classical approach, wouldn't the electron feel a greater Lorentz force due to the higher velocity? This would increase the velocity even more for a fixed orbit radius. Thus, shouldn't the velocity and thus the Lorentz force on the electron keep increasing infinitely? Urgent help would be appreciated.

2. Mar 27, 2012

### K^2

It doesn't work this way because electron doesn't orbit nucleus as a point particle. This is a quantum mechanics problem, and cannot be adequately resolved in classical physics.

3. Mar 27, 2012

### tasnim rahman

Approaching it in a classical way, based on the classical explanation of diamagnetism, for a single electron orbiting a nucleus, what will be the answer? Thanks anyway K^2.

4. Mar 27, 2012

### K^2

Classically, if you have a charged object orbiting another, much heavier, charged object, and you add magnetic field, velocity will increase, radius will decrease, and angular momentum will remain constant. You should be able to use this to solve for new orbit.

5. Mar 27, 2012

### tasnim rahman

Thanks for the help. But I was talking about the equation FL=m(v2)/r, where m, mass of the electron;and r,radius of the orbit remains constant. v, the velocity of the electron changes. FL represents the extra centripetal force provided by the Lorentz force, in addition to the Coulomb force, FC which is also acting as the centripetal force.

6. Mar 27, 2012

### K^2

Radius doesn't remain constant, though. That's the whole point. If the charged object is already in orbit, and you apply magnetic field, the orbit diameter will either shrink or grow, depending on direction of the field you apply.

If you mean that you try launching the charged particle to force it to the same orbit each time, then yes, it will either require higher velocity or lower, again, depending on direction of the field. Notice that Lorentz force increases linearly with particle's speed, while the centripetal acceleration needed grows as a square of velocity. So you'll be able to find a new velocity that gives you the same orbit.

7. Mar 27, 2012

### Philip Wood

Using the interpretation of k^2, I think you'll find that
$$\frac{mv^2}{r} = \frac{mv_{0}^2}{r} + Bev$$
in which $v_{0}$ is the speed without B.

So we have a quadratic in $v$.

8. Mar 27, 2012

### tasnim rahman

Thanks for all your help everyone, but I believe this is not what I meant.Here's a diagram for aid. In the first diagram, the electron is orbiting the nucleus, with speed v1, due to the centripetal force provided by the Coulomb attractive force between the proton and the electron. In the second diagram, a new force FL (Lorentz force) acts on the electron, due to the presence of the magnetic field (direction: out of plane of screen), which is directed towards the center of orbit, using Flemming's left-hand rule. This force manifests itself in the form of increasing the centripetal force(Centripetal force without magnetic field:FC. Centripetal force with magnetic field: FC + FL). As the total centripetal force on the electron increases, the speed of the electron increases from v1 to v2. From F=mv2/r, as F increases, v increases; m (mass of electron) and r(radius of orbit) remains constant.

My question is that: is not the electron which is now at a higher velocity (v1>v2) supposed to experience a greater Lorentz force due to the higher velocity? If so, now it will experience a new even greater centripetal force. This centripetal force will also increase the velocity of the electron; if m (mass of electron) and r(radius of orbit) remains constant. If the velocity increases it will experience a new greater Lorentz force,..................and so on.Thus the velocity of the electron will keep increasing.

Is this what will happen? This is what I could think of. Not sure whether I am correct or not.
Any help/verification is welcome.

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9. Mar 27, 2012

### Philip Wood

Thought my equation did address your problem. It shows there is a v at which the Lorentz force (including the Bev) provides the centripetal force. In other words the v doesn't go on increasing indefinitely. The fundamental reason why it doesn't was pointed out by k^2 a few posts ago:

"Notice that Lorentz force increases linearly with particle's speed, while the centripetal acceleration needed grows as a square of velocity."

Last edited: Mar 27, 2012
10. Mar 27, 2012

### tasnim rahman

Thanks a lot, Philip Wood and K^2.Probably didn't understand it the first time.I think I understand it now.Thanks, again.

11. Mar 27, 2012

### Bob S

Orbiting electrons in the magnetic field of a betatron are accelerated but the orbit is unchanged as the magnetic field is increased, as long as the magnetic field on the orbit is half the average magnetic field inside (linking) the orbit. Electrons have been accelerated to over 300 million volts using this principle. See http://teachers.web.cern.ch/teachers/archiv/HST2001/accelerators/teachers notes/betatron.htm

and

http://storage.lib.uchicago.edu/apf/apf2/images/derivatives/apf2-00056r.jpg [Broken]

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