A current-carrying gold wire has a diameter of 0.90 mm . The electric field in t
ID: 1456934 • Letter: A
Question
A current-carrying gold wire has a diameter of 0.90 mm . The electric field in the wire is 0.46 V/m .
Part A
What is the current carried by the wire?
Express your answer using two significant figures.
I = A
Part B
What is the potential difference between two points in the wire 6.0 m apart?
Express your answer using two significant figures.
V = V
Part C
What is the resistance of a 6.0-m length of this wire?
Express your answer using two significant figures.
R=
V = V
Part C
What is the resistance of a 6.0-m length of this wire?
Express your answer using two significant figures.
R=
Explanation / Answer
use the relation Current density J = E /rho = i /A
where E is Electric field = 0.46 V/m
rho is density of gold = 2.44 e-8
i is current = ?
so current i = E A/rho
i = (0.46 * 3.14 * 0.45 e -3* 0.45 e-3)/(2.44e -8)
i = 11.98 Amps or 12 Amps
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use the relation E = V/d
so Potential V = Ed
V = 6 * 0.46
V = 2.76 Volts
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Use resistance R = rho L/A
so
R = 2.44 e -8 * 6/(3.14 * 0.45 e -3* 0.45 e-3)
R = 0.23 ohms