# 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.)

-
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