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Homework Statement
A 2.0-m length of silver wire has a uniform cross-sectional area of [tex]0.6mm^2[/tex]. If the drift velocity in this wire is 5.0 x [tex]10^{-5}[/tex] m/s and the electrons all move in the same direction along the wire. What is the total current carried by the wire?
Homework Equations
So, silvers atomic mass is 107.9 g/mol and the density is 10500 [tex]kg/m^2[/tex].
Also, conductivity of silver is 6.3 x [tex]10^7[/tex] (too lazy to add units)
So, I = JA, I have A, so I'll find J.
J=nq[tex]V_d[/tex]
where n = number density of atoms
q is charge constant (1.6 x [tex]10^{-19}[/tex])
and [tex]V_d[/tex] is the drift velocity.
The Attempt at a Solution
So, I think I have an answer, my instructor said it was wrong, but wouldn't tell me until after the homework is due (no free points, I guess it's fair).
n=10500/[tex]m^3[/tex]*(1mol/0.1079kg)*(6.02 x [tex]10^{23}[/tex]/1mol) = 5.85 x [tex]10^{28}[/tex]
J=nq[tex]V_d[/tex]
(5.85 x [tex]10^{28}[/tex])*(1.6 x [tex]10^{-19}[/tex])*(5.0 x [tex]10^{-5}[/tex])
= 468656.16
Area = [tex]0.6mm^2[/tex] = 0.6 x[tex]10^{-6}[/tex]m
Now, to find I. I=(468656.16 N/C)*(0.6 x[tex]10^{-6}[/tex]m) = 0.28Amps.
Does this look correct? Sorry if my work is a little sloppy.