Find Electron Current Through Iron Wire

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SUMMARY

The discussion focuses on calculating the electron current through a 1.4-mm-diameter iron wire subjected to a 0.070 V/m electric field. The key equation used is ie = neAVd, where ne is the electron density (8.5×10^28 m^-3) and A is the cross-sectional area (1.5×10^-6 m^2). The initial attempt to calculate the drift speed (vd) resulted in an incorrect current of 4.6×10^7 electrons/s, while the correct current is 8.4×10^18 electrons/s. The correct drift speed must be derived accurately to achieve the correct current value.

PREREQUISITES
  • Understanding of electric fields and their effects on charge carriers
  • Familiarity with the equation for current density (ie = neAVd)
  • Knowledge of electron density in materials, specifically iron
  • Basic principles of drift velocity and collision times in conductors
NEXT STEPS
  • Review the derivation of drift velocity in conductors
  • Study the relationship between electric fields and current density in metals
  • Explore the Hyperphysics website for detailed explanations on Ohm's Law
  • Investigate the impact of temperature on electron mobility in iron
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the behavior of electrons in conductive materials.

bburke95
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Homework Statement


What electron current is driven through a 1.4-mm-diameter iron wire by a 0.070 V/m electric field? The mean time between collisions in iron is 5.2×10^−15s .
A=1.5*10^-6 m^2
ne=8.5*10^28 m^-3

Homework Equations


I know I need to use ie = neAVd, but I'm having trouble finding the drift speed.

The Attempt at a Solution


I set up ie = neAVd , as I currently know both ne and A, but the vd value is what's throwing me off. I multiplied the collision speed by the electric field strenght and got vd=3.64*10^-16m/s, which led to a final answer of ie=4.6*10^7 electrons/s. However, I was wrong and the answer is 8.4*10^18 electrons/s.
 
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