How long does it take for one electron to travel the length of the cable?
A 240-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,040 A. If the conductor is copper with a free charge density of 8.5 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 10^7 for the number of seconds in a year.)
So far this is what I got.
L = 240km = 2.4 x 10^5
I = 1040A
A = Pi*r^2 = Pi (1)^2
I = vneA
V = I / neA = 1040 / (8.5x10^28)(1.6x10^-19)A = ????
I am confuse on how to finish this problem because I keep getting the wrong answer. If anyone could help I would really appreciate it. 10 points to best answer.
- billrussell42Lv 77 years agoFavorite Answer
drift velocity Vd
Vd = I / nqA
Vd is drift velocity in m/s
I is current
n is the number of charge carriers per m³
for copper, 8.5e28 electrons per m³.
A = πr² is the cross sectional area in m²
q is the charge of the charge carriers (electrons)
q = 1.602e-19 Coulomb (charge on an electron)
8.5 1028 ?? I'll use my number
A = πr² = π(0.01)² = 0.000314 m²
Vd = (1040) / (8.5e28)(1.602e-19)(0.000314)
Vd = 0.0002432 m/s
240 km = 2.4e5 m
t = 2.4e5 m / 0.0002432 m/s = 9.87e8 sec
or 274000 hours
or 31 years
- Big DaddyLv 77 years ago
The drift velocity is found by: V = I / n A q
Current is given. n is the charge carrier density (given in the problem), A is the cross section of the cable (easy to calculate), and q is the charge on the charge carrier (one electron).
r = 1cm
A = pi (1cm)^2
A = 3.1416 x 10^-4 m^2
V = 1040A / (8.5x10^28 e/m^3 3.1416x10^-4 m^2 1.602x10^-19 C)
V = 1040A / (4.278x10^6 A s/m)
V = 2.43x10^-4 m/s
t = d / V
t = 240km / 2.43x10^-4m/s
t = 9.87x10^8s = 31.3 year