Solution Found!
Answer: A current-carrying gold wire has diameter 0.84 mm.
Chapter 25, Problem 23E(choose chapter or problem)
Problem 23E
A current-carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire?
Questions & Answers
QUESTION:
Problem 23E
A current-carrying gold wire has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What are (a) the current carried by the wire; (b) the potential difference between two points in the wire 6.4 m apart; (c) the resistance of a 6.4-m length of this wire?
ANSWER:
Solution 23E
Introduction
We have to calculate the current in the given gold wire. Then we have to calculate the potential difference in the given length of the wire. And finally we have to calculate the resistance of te wire for a given length.
Step 1
The diameter of the wire is
Hence the area of the wire is