Charge to mass ratio inversely proportional to curved path radius? In a cloud or bubble chamber, charged particles follow circular paths. I learned that charge to mass ratio of the particles is inversely proportional to the radius of the path. Thus, a particle following a circular path with a large radius means that charge to mass ratio of that particle is smaller.
However, I want to know from where this relationship came from.
 A: Well, the particles won't always follow circular paths (for instance, the particles in this video).
But, if you apply a constant magnetic field across the chamber, charged particles moving in the field will be deflected according to the Lorentz Force Law.
The centripetal acceleration for a particle moving in a circle is $a=\frac{v^2}{r}$, where $v$ is the particles tangential velocity and $r$ is the radius of the circle.
Plugging this into Newton's Second Law we get
$$
m\frac{v^2}{r}=qvB
$$
Rearranging:
$$
\Big(\frac{q}{m}\Big)\Big(\frac{B}{v}\Big)=\frac{1}{r}
$$
Or, in other words, the charge-to-mass ratio is inversely proportional to radius of the particles circular orbit by proportionality constant $\frac{B}{v}$. 
Edit: I can't comment yet otherwise I would answer your second question in the comments. No, $v$ cannot be eliminated because it is integral to the magnitude of the Lorentz force. A light but slower-moving particle could have the same orbit as a heavier, faster particle assuming that they have the same charge, but you can calculate each particle's momentum in a creation event by using conservation of momentum and conservation of electric charge. So, no, I don't think you can say which particle is lighter if all you know are the two radii. Of course, things are different if you have a velocity selector.
A: It boils down to balancing the centripetal force,
$$\vec{F}=\frac{mv^2}{r}\hat{r}$$
with the magnetic force
$$\vec{F}=q\vec{v}\times\vec{B}$$
Equating these and considering the perpendicular velocity, we get
$$\frac{mv_\perp^2}{r}=qv_\perp B$$
Which can easily be solved for $q/m$:
$$\frac{q}{m}=\frac{v_\perp}{rB}$$
Thus, if you know the strength of the magnetic field, the velocity at which it travels, and the radius of the arc it travels, you know the charge-to-mass ratio.
