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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.

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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.

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  • $\begingroup$ is B/v always going to be a constant. I mean, can I use this relationship to compare the curved paths of two different particles (resulting from the breakdown of one particle in a bubble chamber)? $\endgroup$
    – Eliza
    Dec 7 '13 at 3:53
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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.

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  • $\begingroup$ Well what if the paths of two charged particles in a bubble chamber were given (for eg. a particle breaks down to form 2 other particles)and we do not know the velocities of the particles. Then, will the path with the larger radius necessarily be the one with the smaller charge to mass ratio? Is there any way to eliminate v from the equation? $\endgroup$
    – Eliza
    Dec 7 '13 at 3:45

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