What Factors Affect the Resistance and Drift Velocity of a Copper Wire?

AI Thread Summary
The discussion focuses on calculating the resistance and drift velocity of a copper wire formed from a 10-g piece at 50°C. The resistivity of copper is determined using the formula ρ=ρ0(1+α(T-T0), resulting in a value of 1.899*10^-8 Ωm. The length of the wire is calculated from its volume, derived from density and mass, leading to a length of approximately 35.605 cm. The resistance is then computed using R=ρ(L/A), yielding a very low resistance of 2.152*10^-4 ohms. This low resistance results in an extremely high current of 5577 amps, raising concerns about the wire's potential to overheat without adequate heat dissipation.
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Homework Statement



A 10-g piece of copper is to be formed into a wire of radius 1.0 mm at a temperature of 50°
C. What is the resistance of this wire? (Hint: you will need to look up the density of copper.) If a potential of 12 V is put across this wire, what is the drift velocity?

Homework Equations



J=I/A=nqvd
R=ρ(L/A)
ρ=ρ0(1+\alpha(T-T0)
ρ0=1.7*10^-8 at 20°C

The Attempt at a Solution



The resistivity of copper at 50°C is:
ρ=ρ0(1+\alpha(T-20) where \alpha=3.9*10^-3 1/°C
Substituting values, ρ=1.899*10^-8 Ωm

I hit a brick wall here, though. I don't know how to get from density of copper to length of the wire for R=ρ(L/A)

Any advice?
 
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What can you calculate from the density and mass of the piece of copper? How does that relate to A and L?
 
Hmm. Density of Cu=8.96 g/cm^3

(8.96g/cm^3 * 1 / 10g)^-1=1.119 cm^3 Ohh, that gives you a volume! D'oh!

V=1.119 cm^3
1.119=pi*r^2*L
L=1.119/(pi*.1^2)
L=35.605 cm

So, R=rho(L/A)
R=(1.899*10^-8)(.0356)/(3.142*10^-6)
R=2.152*10^-4 ohms

Does this sound right?
 
Except that gives a current of 5577 amps for the second part. What am I missing?
 
Everything looks ok to me except for your conversion of L from cm to m.
 
Yeah, I noticed that. I tried it with the correct conversion (.356 m) and 5577 amps was what I found for current. Isn't that really high though?
 
I think it's ok. The resistance of the wire is very small, so the current will be very high. In fact, the current would quickly heat the wire and it could melt if there's not sufficient heat transfer to the environment.
 
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