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# 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.

### 2 Answers

- 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