# Is an alpha particle's curvature in a magnetic field visible with a homemade cloud chamber?

I'm trying to do some calculations to see just how strong a magnet you'd have to have, in order for curvature to be noticeable in a rudimentary cloud chamber, with lead-210 as an alpha particle source. I'm guessing that this would have to be huge, but I'm not getting through my back of the envelope calculations successfully. Starting from the tensor expression: $$\frac{d^2 x^\mu}{d\tau^2}=\frac{q}{m}F^\mu_\nu \frac{dx^\nu}{d\tau}$$ and expanding using the relation $d\tau=\frac{dt}{\gamma}$, I find: $$\frac{q}{m}F^\mu_\nu \frac{dx^\nu}{dt}=\gamma\ddot{x}^\mu+\dot{x}^\mu \frac{d}{dt}\gamma$$ Assuming the electric field is zero, no work is done, so $\frac{d\gamma}{dt}$=0, this gives in vector form (zero $ct$ component): $$\gamma^{-1} \frac{q}{m}\dot{\vec{x}}\times \vec{B}=\ddot{\vec{x}}$$

Plugging numbers in, with the mass being about four times that of a proton, and the charge being twice that of a proton, and the energy released by lead 210 in alpha decay $3792 \text{KeV}$, I find that the emitted particles have a velocity around $\frac{1}{20}c$, and I get an acceleration on the order of $10^{15}$ in a 1 tesla field, and calculating $v^2/a$ for the radius of curvature gives me a radius of .28 meters.

I'm skeptical of my result because it seems great, but I haven't seen any images or stories of this being tried with a homemade cloud chamber and a neodymium magnet. Is this result correct? Is there something I'm not taking into account? It would be disappointing to set the experiment up and see no result!

(I can post the Mathematica code used to query these values and calculate everything, on request.)

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I'm not sure about the curvature, but you can definitely observe alpha tracks in homemade cloud chambers and it's definitely worth it to build one. After that, putting the magnet in is easy, and if they don't curve - well, they don't curve. –  Emilio Pisanty Sep 11 '13 at 17:45

it is prohibitively expensive to get a 1T magnet around the size comparable to the expected radius of curvature of the charged particles. assuming cost was not the issue, you would still have trouble finding one that size. a 6" Nd disc is \$800--with a magnet this "small" you just might detect some curvature in the cloud tracks with some luck.

if you tried to build a solenoid you would still need a few thousand turns of wire running 10 amps...

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Building a Tesla strength electromagnet is a mechanical engineering challenge, too. Compute the forces between the poles of that puppy. Those things have to be massive, and if one were to break under load it would be muchos dangerous. –  dmckee Sep 11 '13 at 19:25
I have a friend with such a neodymium magnet, and I was planning on using Mathematica to try to get a nice distribution after several thousand tracks. I'm still skeptical but I think the only thing to do now is try it! –  NeuroFuzzy Sep 11 '13 at 20:32
@NeuroFuzzy make sure you put it all on youtube! –  gregsan Sep 11 '13 at 21:52
@gregsan Of course! Another thing that is worrying, this linke: wiki.answers.com/Q/… says that Rutherford did this in a vacuum chamber. I can't find online whether alpha particles pick up electrons in atmosphere or not, before they stop. –  NeuroFuzzy Sep 11 '13 at 22:48
@NeuroFuzzy They just stop in very fairly short distances. A few centimeters in most cases. Pulling a partial vacuum can stretch that out. It goes basically as the areal mass density (volumetric density times path-length). So if you pump down to circa 100 torr you can get a few tens of cm. 10 tor gets you a few meters path-length. As I imagine you are planning a chamber only a few tens of cm on a side, you don't need a really strong vacuum. –  dmckee Sep 12 '13 at 3:03