Chapter 1: Problem 17
In typical residential wiring, the copper wire has a diameter of \(2.05 \mathrm{~mm}\) and carries a maximum current of \(15 \sqrt{2}\) A due solely to electrons, each of which has a charge of \(1.60 \times 10^{-19}\) C. Given that the free electron (these are the electrons capable of moving through the copper) concentration in copper is \(10^{29}\) electrons \(/ \mathrm{m}^{3}\), find the average velocity of the electrons in the wire when the maximum current is flowing.
Short Answer
Step by step solution
Identify Key Variables
Calculate the Wire's Cross-sectional Area
Use Current Density Formula
Calculate Current Density
Solve for Electron Drift Velocity
Final Interpretation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Current Density
- \( J = \frac{I}{A} \)
Free Electron Concentration
- These free electrons facilitate the flow of electric current.
- They act as carriers and turn the copper into an excellent conductor.
Copper Wire Properties
- High electrical conductivity: This allows electric current to flow with minimal resistance compared to other materials.
- Flexibility: Copper can be easily shaped into wires without losing its conductive properties.
- Durability: Copper is corrosion-resistant, which helps maintain its electrical properties over time.
Electric Current
- In copper wires: This flow is primarily due to the movement of free electrons.
- Equation for current: \( I = nAqv_d \) where \( n \) is the free electron concentration, \( A \) is the cross-sectional area, \( q \) is the charge of an electron, and \( v_d \) is the drift velocity.