Question

Electron drift speed?

Answer 1

A 1.7*10^-3 V/m electric field creates a 3.5*10^17 electron/second current, i(e), in a 1.1 mm diameter aluminum wire. What is the drift speed?

Answer 2

So, I know that drift velocity is\[v_d=\frac{e \tau }{m}E\]and that \[i_e=\frac{n_ee\tau A}{m}E\]so I can substitute v(d) in giving me\[i_e=n_ev_dA\]
I'm given i(e) and E in the equation. And A would just be the cross sectional area of a cylinder. So\[i_e=3.5*10^{17}\frac{electrons}{second}\]\[E=1.7*10^{-3}\frac{V}{m}\]and \[A = \pi r^2= \pi (.55*10^{-3}m)^2=.95033m^2\]

Answer 3

Solve i(e) for v(d) and I found\[v_d=\frac{i_e}{n_eA}\] n(e) (electron density) for aluminum is \[n_e=6*10^{28}m^{-3}\]Plug in numbers and I get\[v_d=\frac{3.5*10^{17}s^{-1}}{6*10^{28}m^{-3}*0.95033m^2}=6.138*10^{-12}\frac{s^{-1}}{m^{-1}}=6.138*10^{-12}\frac{m}{s}\]

Answer 4

This is apparently not the right answer. What am I doing wrong? My units check out and everything, and I've double checked my calculations making sure I didn't just punch it in wrong.

Answer 5

Your area is wrong.

Answer 6

How so?

Answer 7

Ah, okay I see. Should be 9.5*10^-7?

Answer 8

yep

Answer 9

Bleh..thanks