Answer: A current-carrying gold wire has diameter 0.84 mm.

Chapter 25, Problem 23E

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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?

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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

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