How Many Electrons Pass Through a Gold Wire Per Second?

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SUMMARY

The discussion centers on calculating the number of electrons passing through a gold wire carrying a current of 380 mA. Using the resistivity of gold at 20°C (2.44 x 10-8 Ω m), participants derived the resistance and voltage but struggled with the final calculation of electron flow. The correct approach involves recognizing that 1 A corresponds to approximately 6.241 x 1018 electrons per second, allowing for a direct proportion to find the electron flow for 380 mA. The closest answer to the problem is A) 2.4 × 1018.

PREREQUISITES
  • Understanding of electrical current and its units (Amperes)
  • Knowledge of resistivity and resistance calculations
  • Familiarity with the charge of an electron (1.6 x 10-19 Coulombs)
  • Basic algebra for proportional calculations
NEXT STEPS
  • Study the relationship between current, charge, and electron flow in circuits
  • Learn about resistivity and its impact on electrical resistance in different materials
  • Explore Ohm's Law and its applications in circuit analysis
  • Investigate common mistakes in electrical calculations and how to avoid them
USEFUL FOR

Students in physics or electrical engineering, educators teaching circuit theory, and anyone interested in understanding electron flow in conductive materials.

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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