A current-carrying gold wire has a diameter of 0.84 mm electric field in the wir
ID: 1881361 • Letter: A
Question
A current-carrying gold wire has a diameter of 0.84 mm electric field in the wire is 0.45 V/m You may want to review (Pages 820-823) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Electric field potential difference, Part A What is the current carried by the wire? Express your answer using two significant figures. ANSWER Part B What is the potential difference between two points in the wire 6.9 m apart? Express your answer using two significant figures. ANSWER Part C What is the resistance of a 6.9-m length of this wire? Express your answer using two significant figures. ANSWE:Explanation / Answer
1)
Let's take the resistivity of gold () = 2.44 * 10^-8 .m
Cross-sectional area of wire of radius (r) {Units alert. convert (r) to metres!}
A = r² = * (4.2 * 10^-4)² = 5.54 * 10^-7 m²
Resistance (R) of a wire of length (L) and cross-sectional area (A) is given by:
R = L/A
R/L = /A {The units of (R/L) will be /m}
R/L = (2.44 * 10^-8 .m) / (5.54 * 10^-7 m²) = 0.044 /m
(0.45 V/m) / (0.044 /m) = 10.23 V/ = 10.23 A
(b) (0.45 V/m * 6.9 m) = 3.105 V
(c) R = L/A = (2.44 * 10^-8 .m) * (6.9 m) / (5.54 * 10^-7 m²) = 0.304
2)
Electrons density = 8.50*10^28 el./m^3
a)
cable volume = 0.7854*0.205^2*67 = 2.21 cm^3 = 2.21*10^-6 m^3
8.50*10^28 el./m^3*2.21*10^-6 m^3 = 1.88*10^23 el.
El./Coulomb = 6.24*10^18 el/(A*sec)
6.24*10^18 el/(A*sec)*5 A = 3.12*10^19 el/sec
1.88*10^23 el./3.12*10^19 el/sec = 6025 sec = 6025/60 = 100.4 min
b)
cable volume = 0.7854*0.412^2*67 = 8.93 cm^3 = 8.93*10^-6 m^3
8.50*10^28 el./m^3*8.93*10^-6 m^3 = 7.6*10^23 el.
El./Coulomb = 6.24*10^18 el/(A*sec)
6.24*10^18 el/(A*sec)*5 A = 3.12*10^19 el/sec
7.6*10^23 el./3.12*10^19 el/sec = 24359 sec = 406 min