How Many Electrons Pass Through a Gold Wire Per Second?

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Homework Help Overview

The discussion revolves around a problem involving the calculation of the number of electrons passing through a gold wire carrying a current of 380 mA. The resistivity of gold and the dimensions of the wire are provided, but participants are exploring the relevance of the given information and the correct approach to the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of resistance using the formula R=ρL/A and question the appropriateness of dividing voltage by electron charge to find the number of electrons per second. There is also a focus on understanding the implications of current values and the relevance of the problem's parameters.

Discussion Status

Some participants have offered guidance on using the relationship between current and the number of electrons per second, while others express confusion about the calculations and the relevance of certain information. Multiple interpretations of the problem are being explored, particularly regarding the significance of the current value.

Contextual Notes

Participants note that the current is not 1 A, and there is discussion about the potential irrelevance of some information provided in the problem statement. There is also mention of a possible typo in the calculations presented.

victorializ
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Homework Statement



The resistivity of gold is 2.44 x 10-8 Ω m at a temperature of 20°C. A gold wire, 0.5 mm in diameter and 44 cm long, carries a current of 380 ma. The number of electrons per second passing a given cross section of the wire, is closest to:

A) 2.4 × 10^18
B) 1.2 × 10^22
C) 2.8 × 10^14
D) 2.4 × 10^17
E) 6.3 × 10^15

Homework Equations



R=ρL/A

The Attempt at a Solution



R=(2.44 x 10^-8) (.44)/∏(.00025^2)
R= .0547
then i multiplied the resistance by the current (.38) to get .021 for a voltage.
then i divided that by (1.6 x 10^-19) and got 1.3 x 10^17 which isn't right but is close to the answer which is A
 
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Why did you divide a voltage by the electron charge? Do you expect the right unit "per second" as result of that calculation?

You are given the current. What does a current of 1 A (example value) mean?
 
victorializ said:

Homework Statement



The resistivity of gold is 2.44 x 10-8 Ω m at a temperature of 20°C. A gold wire, 0.5 mm in diameter and 44 cm long, carries a current of 380 ma. The number of electrons per second passing a given cross section of the wire, is closest to:

A) 2.4 × 10^18
B) 1.2 × 10^22
C) 2.8 × 10^14
D) 2.4 × 10^17
E) 6.3 × 10^15

Homework Equations



R=ρL/A

The Attempt at a Solution



R=(2.44 x 10^-8) (.44)/∏(.00025^2)
R= .0547
then i multiplied the resistance by the current (.38) to get .021 for a voltage.
then i divided that by (1.6 x 10^-19) and got 1.3 x 10^17 which isn't right but is close to the answer which is A

Something's off by a factor of 18.5 and you call that 'close'?

Learn that people are going to give you problems with a lot of irrelevant information!
 
mfb said:
Why did you divide a voltage by the electron charge? Do you expect the right unit "per second" as result of that calculation?

You are given the current. What does a current of 1 A (example value) mean?

so i need to use the fact that 6.241 x 10^18 electrons/sec pass through a circuit which is 1 A 

when i use that number though i still have it wrong so I'm not really sure where else my calculations are messed up.
 
victorializ said:
so i need to use the fact that 6.241 x 10^18 electrons/sec pass through a circuit which is 1 A 

when i use that number though i still have it wrong so I'm not really sure where else my calculations are messed up.

But the current isn't 1 A.
 
dauto said:
But the current isn't 1 A.

oh my goodness so most of the information in this problem is irrelevant? all i need is to do a proportion between 1A and the 38 mA in the problem?
 
victorializ said:
oh my goodness so most of the information in this problem is irrelevant? all i need is to do a proportion between 1A and the 38 mA in the problem?
If that gets you closer to the right answer, go for it! :smile:

380 mA would be even better ... :wink:
 
victorializ said:
oh my goodness so most of the information in this problem is irrelevant? all i need is to do a proportion between 1A and the 38 mA in the problem?
Yes (apart from the typo (?) NascentOxygen pointed out).
 

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