Find Drift Speed of Electrons in Copper Wire | Tom

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SUMMARY

The drift speed of electrons in a copper wire can be calculated using the formula Vd = I/nqA, where I is the current, n is the number of charge carriers, q is the charge of an electron, and A is the cross-sectional area of the wire. In this discussion, Tom calculated the drift speed to be 9.2 x 10^-5 m/s using a wire length of 10 meters, a temperature of 60°C, a current of 5 A, and a resistance of 0.05 ohms. The number of charge carriers was determined to be 8.5 x 10^28, and the resistivity at the given temperature was found to be 2.0 x 10^-8 ohm-meters.

PREREQUISITES
  • Understanding of electrical concepts such as current, resistance, and drift speed.
  • Familiarity with the formula for resistivity: p = po[1+α(T-To)].
  • Knowledge of the number of charge carriers calculation: n = density * 6.02 x 10^23/molar mass.
  • Basic algebra and unit conversions relevant to physics problems.
NEXT STEPS
  • Research the effects of temperature on resistivity in different materials.
  • Learn about the relationship between current density and drift velocity in conductors.
  • Explore the impact of wire diameter on the drift speed of electrons.
  • Investigate the significance of charge carrier density in various conductive materials.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and anyone interested in understanding the behavior of electrons in conductive materials like copper wire.

tom_paine
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Homework Statement



I’m trying to find the drift speed of electrons in a copper wire. The length of the wire is 10 meters, temp is 60°C, current is 5 A and the total resistance of the wire is 0.05 ohms.


Homework Equations



# of charge carriers (n) = density * 6.02 x 10^23/molar mass

R = pL/A, where p = resistivity of the wire, L = length and A = area

p = po[1+α(T-To)], where α = temp coefficient of resistivity at 20°C, po = resistivity at 20°C, and T= temp

I = nqAVd


The Attempt at a Solution



n = (8920 kg/m^3)(6.02 x 10^23)/(0.0635 kg/mol)
n= 8.5 x 10^28

p = 1.7 x 10^-8 [1 + 3.9 x 10^-3(60-20)]
p= 2.0 x 10^-8

Area = pL/R
= 2.0 x 10^-8 * 10/0.05
= 4 x 10^-6

Vd = I/nqA
= 5/(8.5 x 10^28 * 1.6 x 10^-19 * 4 x 10^-6)
= 9.2 x 10^-5 m/s

I just want to know if I’m on the right track.

Thanks,

Tom
 
Physics news on Phys.org
You did it well, nice work!

ehild
 
Thanks for the reply! I thought I got it right, but the book I got the problem from is missing the solutions page, so I wasn't 100% sure.

Tom
 

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