Electron drift speed, Current, and Electric field in hollow wire?

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SUMMARY

The discussion focuses on calculating the electron drift speed, current density, and electric field in a hollow copper wire with an inner diameter of 1.4 mm and an outer diameter of 2.6 mm carrying a current of 2.50 amps. The relevant equations include I=neeVdA for drift speed, J=I/A for current density, and J=σE for electric field calculations. The correct approach involves calculating the cross-sectional area by subtracting the inner area from the outer area, as highlighted in the feedback regarding algebraic errors. The conductivity of copper is noted as 6.0x107 (ohm m)-1.

PREREQUISITES
  • Understanding of electric current and drift speed
  • Familiarity with the concept of current density
  • Knowledge of electric field calculations
  • Basic algebra for area calculations
NEXT STEPS
  • Calculate electron drift speed using the formula I=neeVdA
  • Determine current density by applying J=I/A
  • Analyze the relationship between current density and electric field using J=σE
  • Explore the impact of wire dimensions on electrical properties
USEFUL FOR

Students in physics or electrical engineering, educators teaching electromagnetism, and anyone interested in the electrical properties of conductive materials.

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Homework Statement



A hollow copper wire with an inner diameter of 1.4 mm and an outer diameter of 2.6 mm carries a current of 2.50 amps. Copper has an electron density of 8.5x1028 m-3 and a conductivity of 6.0x107 (ohm m)-1.
a. What is the electron drift speed in the wire?
b. What is the current density in the wire?
c. What is the electric field in the wire?

Homework Equations



a) I=neeVdA
b)J=I/A
c)J=σE

The Attempt at a Solution


I believe I have the correct answers but was just curious if someone could double check me!
Thank you!
 

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Bra, your algebra is wrong. When you find the area, do it for each section separably and subtract them from each other you will see your error. X^2-Y^2 does not equal (X-Y)^2
 

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