Exercise 25.23 Part A A current-carrying gold wire has a diameter of 0.85 mm. Th

Question

Exercise 25.23 Part A A current-carrying gold wire has a diameter of 0.85 mm. The electric field in the wire is 0.50 V/m What is the current carried by the wire? Express your answer using two significant figures. Submit My Answers Give Up Part B What is the potential difference between two points in the wire 6.9 m apart? Express your answer using two significant figures Submit My Answers Give Up Part C What is the resistance of a 6.9-m length of this wire? Express your answer using two significant figures. Submit My Answers Give Up

Explanation / Answer

a) resistivity of gold wire = 2.44 * 10-8 ohm.m

current density = I / A

I = j A = E A / p = (0.50 * pi * 0.4252 * 10-6) / (2.44 * 10-8)

= 11.63 A

b) Potential difference = E L = 0.50 * 6.9

= 3.45 V

c) Resistance = V / I = 3.45 / 11.63

= 0.297 ohm

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