What Is the Average Velocity of Electrons in a Copper Wire Carrying 1.0A?

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Homework Help Overview

The discussion revolves around calculating the average velocity of electrons in a copper wire carrying a current of 1.0 A. The problem involves concepts from electromagnetism and material properties, specifically focusing on the drift velocity of charge carriers in conductive materials.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between current, electron density, charge, and drift velocity using the equation I = ρeAv. Some express confusion regarding unit conversions and the calculation of cross-sectional area. Others share their attempts to derive the average velocity and question discrepancies with the expected answer.

Discussion Status

Several participants have attempted calculations but report varying results, with some confirming the expected answer of 2 × 10^-5 m/s while others express frustration over not reaching this conclusion. There is an ongoing exploration of the formula and its components, with some participants offering guidance on the necessary calculations.

Contextual Notes

Participants note the importance of accurate unit conversions and the correct application of formulas. There is mention of a reference to a previous thread for additional context, indicating that some foundational knowledge may be assumed.

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Copper has approximately 10^29 free electrons per cubic meter. What is the approximate average velocity of electons in a 1.0 mm radius wire carrying 1.0A?

resisivity, p of copper = 1.68 * 10^-8 ohms meter

The answer in the book apparently is 2 * 10^-5 m/s.

Please help by showing all your work. Your attempts will be greatly appreciated.
 
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the answer is 2 * 10^-5 m/s.
please read this before posting...
https://www.physicsforums.com/showthread.php?t=28 :-p
 
i have tried it many times and it did not reach tat ans.
 
the current [itex]I[/itex] is equals to free electrons density [itex]\rho[/itex] times charge per electrons [itex]e[/itex] times drift velocity [itex]v[/itex] times cross section area [itex]A[/itex]
[tex]I=\rho e v A[/tex]
ring the bell?
since you have everything except v, this is one equation with one unknown... solving it is straight forward...
you must be careful on the unit conversion, it is a little bit tricky...
the area is not given directly, but you have radii, the area shold be piece of cake
charge of electron is on the back of your textbook... if not, look it up on internet
show me your calculation ...
 
Within 1 sec, e drifted for length L

L = v * 1 sec = v

total e drifted within 1 sec is 1 Ampere.

1.0 A = 6.24*10^18 e = 10^29 * v * (π * (0.001)^2)

v = 2 * 10^-6 m/s

therefore, does not match with the ans in the book.
 
Within 1 sec, e drifted for length L

L = v * 1 sec = v

total e drifted within 1 sec is 1 Ampere.

1.0 A = 6.24*10^18 e = 10^29 * v * (π * (0.001)^2)

v = 2 * 10^-6 m/s

therefore, does not match with the ans in the book.

what the hack are you doing here?
 
[tex]I=\rho e v A[/tex]
[tex]\rho[/tex] is the free electronss density
[tex]e[/tex] is the charge per electron...
[tex]A[/tex] is the cross section area
[tex]I[/tex] is the current
plug in all number and solve for v...
 
it appears that I have punched in the wrong notation the whole time. ths for ur help.
 
I don't know why, but i just can't seem to get the answer :(

the formula i use is

I = NaQv

so V = I / NAQ

A (cross sectional area) = (0.001)^2 multiplied by Pi
Q = 1.6 * 10^-19
N= 10^29
I = 1

is this right so far?

Regards,
Mo
 
  • #10
Yes,it's identical to what the OP did.You should get the same answer.[tex]2\times 10^{-5}m \ s^{-1}[/tex]

Daniel.
 

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