Radius for the path of an electron in a magnetic field

[KarenRei]

Just double checking this - I found this formula describing the radius of the path of an ion in a cyclotron:

R = sqrt(2*E*m)/(e*B)

.. where R is the radius in meters, E is the energy of the particle in joules, m is the mass in kilograms, c is the charge in coulombs, and B is the strength of the magnetic field in Teslas.

Is this correct, and would this be the maximal radius required for the collimation of electrons emitted from a flat surface inside an axial magnetic field? So for example, for an electron - charge -1.60e−19C, rest mass 9.11e-31kg, energy 16MeV=2.56e-12J - emitted inside say a 1T field - would curve parallel to the field lines in sqrt(2*9.11e-31*2.56⋅10e-12)/(-1.6e−19C*1) = 0.013m = 1.3cm if 100% perpendicular to the field, less otherwise?

Just making sure here :)

 

[ChrisVer]

it will start circling at this radius...you can think the straight line as circle of infinite radius (which comes from $B=0$ /non magnetic field )...
The more you raise the magnetic field's strength, the radius will start getting less and less until it'll be able to create a circle in your lab...of course ever after B=0, the trajectory is a circle but it can be really huge (not small enough to go around in your accelerator or device)
If not perpendicular to the field, then it's going to move into a spiral motion (you can decompose its momentum/velocity in a component perpendicular and parallel to the field). Again as a limit (when the velocity is exactly parallel to the magnetic field) this spiral motion tends to a straight line.

It's not like a maximal radius though, it's the radius that all particles with mass $m$ and momentum $p$ (or energy $E$) will have once they go through that magnetic field.

 

[KarenRei]

Thanks for the reply ChrisVer. :) But how would that not be a maximal radius, if it's the radius that all particles with mass m and momentum p (or energy E) will have once they go through the magnetic field? If the field is fixed then what would be changing the momentum / energy (or mass)?

 

[ChrisVer]

I'm saying it's not maximal because there is no smaller radius for the particles of momentum $p$ and mass $m$.
I'm sorry I didn't understand the last question.

 

[KarenRei]

Ah, I think I see the difference here - you're saying that the angle that the particle was emitted doesn't make a difference. And that makes sense, given that Lorentz force is proportional to the sine of the angle between the path of travel and the field, so a more parallel-emitted particle may have longer to travel to reach that radius but the force on it will be correspondingly less during that time.

Thanks :)

 

[ChrisVer]

However about your formula, where did you find it?

 

[KarenRei]

I don't recall, it was in some online set of physics problems with associated solutions. Is something wrong with the formula?

 

[ChrisVer]

It looks weird...
r = \frac{\sqrt{2Em}}{qB}

Whereas the relativistic formula (I looked) is:

r= \frac{ \gamma u m_0}{qB}= \frac{p}{qB}

 

[KarenRei]

Found it: The second half of solution B (starting at "The radius R...") to the cyclotron problem here:

http://physicstasks.eu/uloha.php?uloha=551

Am I misinterpreting the applicability of this formula?

Also, your formula doesn't take into account rest mass? Hmm.. I guess that might make sense, a lighter particle would move faster at a given energy level than a heavier one but react to the force more quickly; relativistically, it's effectively heavier by virtue of its energy.

 

[KarenRei]

Hmm, but your formula is giving some weird results: 2.56e-12 / (-1.60e−19 * 1) = 16000000 meters? I'm guessing I'm using the wrong units here? I know for a fact that it doesn't take 16 thousand kilometers to curve the path of a 16MeV electron with a 1T field ;)

The numbers in the formula I used seem much more realistic. For example, the LHC does 13TeV with a radius of 2800 meters, that would correspond to a radius of 0,34 centimeters for a 16 MeV particle - versus my calculated 1,3, but of course they use higher field strengths. Although that could be just coincidental.

What's going on here?

 

[ChrisVer]

Obviously different units.
eB \approx 9 \times 10^{10} \frac{MeV}{c^2 sec} For B=1T

If you say that the energy is E \sim 16MeV whereas the mass of the particle is just half MeV, then I'd say that it's really relativistic and I can write E \approx p =16~MeV/c. For comparing: the exact momentum would be \sqrt{E^2 - m^2} = \sqrt{16^2 - 0.5^2} MeV/c = 15.99~MeV/c (so I'm going less than 1% off with my assumption).

So:

r = \frac{p}{qB} \approx \frac{16}{9 \times 10^{10} s^{-1}}c \approx 5.33 \times 10^{-2} ~m
or 5.33~cm

 

[ChrisVer]

As for the problem OK... it uses that the particle gets boosts everytime it passes from the dees up to a maximum value...

 

[KarenRei]

Thank you very much. But where is that 9e10 MeV/c²sec coming from? And what's the meaning of those units - energy over (c² times time)? My mind is trying to change them to other forms to make sense of them (replacing energy over time with power, replacing c^2 with E/m, etc) but so far nothing is helping. And the second time you write it I don't get the unit conversion. So we're using p = 16 MeV/c, and eB = 9e10 MeV/c²sec, but then you're writing eB in the place for qB as 9e10s¯¹, with the whole MeV/c² disappearing?

Also, why is the problem different? Yes, on the big picture the electron has been boosted up by the dees, but between the dees it's just being curved by the magnetic field, is it not?

 

[ChrisVer]

I'm sorry about the units... these are the most common units when you are dealing with particle physics.
The 9 e10 MeV/c^2 sec comes from the fact that ( brackets is units ):
[qB] = C T = C \frac{kg}{C ~sec} = \frac{kg}{sec}
1 Tesla= 1 kg/(Cb sec)
And I used q=e = 1.6 e-19 Cb...
Then I turned kilograms into MeV/c^2 ... 1kg= 5.60958885 e29 MeV / c^2
One easy way to do this conversion is to remember that the electron's mass in MeV is 0.511MeV/c^2 (and you know its value in kg). Easy way also to remember the energy/c^2 is the famous (half-written) formula : E=mc^2.

So eB= 5.6*1.6 e10 MeV/(c^2 sec) ~ 9 e10 MeV/(c^2 sec)

I also wrote the momentum in MeV/c (easy to remember from special relativity formula E=pc).

so I had:
\frac{p}{qB} = \frac{16 \frac{MeV}{c} } {9 \times 10^{10} \frac{MeV}{c^2 sec}} = \frac{16}{9} \frac{c}{10^{10}s^{-1}}

The MeV/c cancels out and I bring the other 1/c in the denominator as c in the nominator.

 

[ChrisVer]

Also, why is the problem different? Yes, on the big picture the electron has been boosted up by the dees, but between the dees it's just being curved by the magnetic field, is it not?

The magnetic field is everywhere. When the particle gets boosted, it will have a different energy/momentum after that, and so its radius at which it circles (which is derived in Solution A in your link) will get larger and larger... at some point it will stop getting boosted (having some certain energy/if it won't radiate it away) and the circle radius will be fixed.

 

[KarenRei]

Got it, now I see how the unit conversions got to that point. Thanks :)