# Homework Help: Maximum producible energy (electromagnetism; cyclotron)

1. Jun 25, 2009

### ephedyn

This isn't exactly homework, but I felt that I had too much to ask that it wouldn't be right to put this in the regular section. Anyway, I've been reading the original paper on the design of a cyclotron (Lawrence & Livingstone) and there were two parts which didn't agree with what I knew... and the numbers didn't tally with what they had claimed. It's unlikely that they're wrong, so could someone give some possible explanations as to where I'm wrong?

1. The problem statement, all variables and given/known data

1a. I would understand non-relativistic Lorentz force on a charged particle is $Hev$. What is c doing in $\frac{Hev}{c}$?

2. This is untrue!(?) I substituted the values and got 59.0m.

3. Is there a formula to determine the number of semicircular arcs completed by a proton in a cyclotron with magnetic field $H$ and radius $r$? Although I'm guessing ab initiio, '300' was determined from maximum V empirically found, divided by the known parameter $\frac{H^2 r^2}{/c^2} \frac{e}{m}$...

2. Relevant equations

If high frequency oscillations having peak values of 4000V are applied to the electrodes and protons are caused to spiral around in this way 150 times, they will receive 300 increments of energy, acquiring thereby a speed corresponding to 1,200,000V.

1. Along the circular paths within the electrodes, the centrifugal force of an ion is balanced by the magnetic force on it,

$\frac{mv^2}{r} = \frac{Hev}{c}$

It follows that the time for traversal of a semi-circular path is

$t = \frac{\pi r}{v} = \frac{\pi mc}{He}$

which is independent of the radius of the path and velocity of the ion. The relationship between the wavelength $\lambda$ of the oscillations and the corresponding synchronizing magnetic field $H$ is in consequence

$\lambda = \frac{2 \pi mc^2}{He}$

2. Thus for protons and a magnetic field of 10,000 gauss, the corresponding wavelength is 19.4m; for heavier particles the proper wave-length is proportionately longer.

3. The energy $V$ in eV of the charged particles arriving at the periphery of the apparatus on a circle of radius $r$

$V = 150 \frac{H^2 r^2}{/c^2} \frac{e}{m}$

Thus the theoretical maximum producible energy varies as the square of the radius and the square of the magnetic field.

3. The attempt at a solution

1. I couldn't interpret the factor $1/c$. I guessed c should've been referring to the speed of light - but I did a search and couldn't find this factor used anywhere else to correct $Hev$.

I also encountered another strange problem if I were to assume

$\frac{mv^2}{r} = Hev$

instead. Taking a magnetic field of 10,000G = 108T, mp=1.67x10-27, r = 0.12m, I found $v$ VERY large, ~1015 m s-1. I am guessing you'll have to reduce it somehow but throwing in $1/c$ seems too arbitrary.

So instead, I thought it was most probable that it was the Lorentz factor. But this is again contradictory, since the paper goes on to say that $\frac{\pi mc}{He}$ is independent of the velocity of the ion. Or is there a way to get the Lorentz factor based on rest mass of the proton so that it's still independent of the velocity as claimed?

2. Now, assuming there is nothing wrong with c referring to the speed of light and equation 1, I substituted and found

$\lambda = \frac{2 \pi \times 1.67 \times 10^{-27} \times 9 \times 10^{16}}{10^8 \times 1.6 \times 10^{-19}} \approx 59.0m$

Anyone manage to get 19.4?

3. I can derive

$V = \frac{1}{2} \frac{H^2 r^2}{c^2} \frac{e}{m}$

in eV, assuming equation 1 is correct, and then taking $\frac{1}{2} mv^2$. But again, is equation 1 correct? The next problem was, the maximum producible energy is still insanely inflated here... with a cyclotron of radius 0.57m, magnetic field = 14,000 gauss, the paper claims that we can produce 25 MeV protons with 500 semicircular arcs completed. But making the substitutions, I found V = 3.39GeV for protons. I know that we have to make relativistic corrections, but this is just too large? Most of all, the difference between the claim on the paper and the calculation is just too large...

Last edited: Jun 25, 2009
2. Jun 25, 2009

### ideasrule

I also don't know what "c" is doing in the first equation. However, for the second equation, 10 000 G is not 10^8 teslas; it's 1 tesla, as the tesla is much larger than the gauss. Fixing that and changing the "c^2" term to "c" (because one of the c's was introduced in the first equation) would give you a wavelength of 19.4 m.

3. Jun 25, 2009

### ephedyn

Ah, the conversion factor! And yeah, it makes sense to make the fix fix from c^2 to c, which would mean that I'd remove 1/c from the original equation. With this, I fixed the maximum producible energy to be

$V = \frac{1}{2} m_{p} ( \frac{Her}{m_{p}} )^2$

I think there's another mistake on the paper here whereby they're multiplying the number of semicircular arcs completed to the maximum kinetic energy of protons at the cirucmference which 'resonate' with the magnetic field, which doesn't make sense. And so this fix yields about 30 MeV protons for the maximum producible energy, which agrees with their claim of >25 MeV.

Everything is correct now. Thanks a lot! I wish you a good weekend ahead of you. ;)

4. Oct 3, 2009

### tomliew

The equation is okay. It is just that he is using the cgs guassian unit and you are more familiar with the SI units. For example the Lorentz force in SI unit F = q v x B but in cgs it is q (v/c) x B.

Each time the charge particle crosses the gap between the Dees it gains an among of energy equal to qE (SI unit). So the more passes the more energy eventually the charge particles will gain and of course it will also corresponding be traveling in increasing larger radius.