# j j thomson's "previous" experiment

by gnome
Tags: experiment, thomson
P: 1,048
I thought I was going to like this book, but it's defying me.

I'm staring at pg 123 of Tipler's Modern Physics text (3rd ed.) This is talking about an experiment he says Thomson did before the "famous" J J Thomson experiment. It goes like this:

 In his first measurement, Thomson determined the velocity from measurements of the total charge and the temperature change occurring when the beam struck an insulated collector. For N particles, the total charge is Q = Ne, while the temperature rise is proportional to the energy loss W = N(½mu2). Eliminating N and u from these equations, we obtain $$\frac{e}{m} = \frac{2W}{B^2R^2Q}\hspace{200}eq.\hspace{5} 3-3$$
That's all he says about that experiment, except for a brief note later on that the measurements obtained from that one were actually more accurate than the measurements he obtained from the later, and more well-known, method.

Anyway, on the preceding page, he talks a bit about the circular path followed by charged particles in a magnetic field and gives

$$quB = \frac{mu^2}{R}$$

Obvious enough. (Here, little q is of course the charge on an individual particle, so it corresponds to e, not Q, in the other equations.)

But I don't see how that helps in getting to equation 3-3 above. All I get from the two preliminary equations is
$$\frac{e}{m} = \frac{Qu^2}{2W}$$
To get from that to his equation 3-3 I have to say
$$u^2 = \frac{4W^2}{B^2R^2Q^2}$$
and I don't see anything that lets me say that.

Any ideas?
 P: 991 I'm not familiar with Thomson's "previous" experiment, but if you look at the two equations: $$quB = \frac{mu^2}{R}$$ and $$\frac{e}{m} = \frac{Qu^2}{2W}$$ If you solve for u in the first equation you get $$u = \frac{eBR}{m}$$ which you can then substitute into the second equation to get the desired equation. cookiemonster
 P: 1,048 [:((] Hmmm...I hate to admit to how long I didn't see that. Thanks for providing the requisite smack in the head.

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