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Ratio of electron drift velocity?

Hey, just having a little bit of trouble with this homework, womdering if anyone could help out?

"A copper rod is stretched in a tensile testing machine so that its diameter at the widest section is three times that at the narrowest section. The rod is connected to a battery.

Calculate the ratio of the maximum to the minimum mean drift velocities of the free elections in the rod."

Right, so basically I have:

I = nAve

Because I, n and e are constant (same number density, elementry charge and current will be present) the change in A must make a change in v for I to stay the same, and it will be proportional. I guess.

The diameter changes by 3x, therefore so does the radius.

Therefore, the ratio must be (pi)r^2 : (pi)(3r)^2. (Not sure, but assuming..)

The answer in the book says that "The speed at the minimum diameter is nine times than the speed at the maximum".

How do I go about that?!

Thanks.

Reply 1

teachercol

YOure nearly there : you've got the ratio being 3^2 which is 9!

Reply 2

beesbees

Ah! I see..

Just punched some numbers in there, so if r = 2
r^2 = 4
3r^2 = 36
4:36 which is 1:9!

I've been stuck on this for about 15 minutes, is that it? :p: