Question

The mean time between collisions in iron is 5.0×10?15 s .

What electron current is driven through a 1.9-mm-diameter iron wire by a 0.070 V/m electric field?

Express your answer using two significant figures.

Answer 1

Given mean time T = 5 * 10^-15 s

diameter d = 1.9 mm

radius r = 1.9 / 2 = 0.95 mm = 0.95 *10^-3 m^2

Area of cross section

A = pi * r^2

A = pi * (0.95 * 10^-3)^2

A = 2.83 m^2

electric field E = 0.07 V/m

Charge of electron   e = 1.6 * 10^-19 C

mass of electron m = 9.1 * 10^-31 kg

electron density in iron is n = 8.5 * 10^28 m^-3

The drift velocity is given by

vd = e * E * T / m

vd = (1.6 * 10^-19 * 0.07 * 5 * 10^-15) / (9.1 * 10^-31)

vd = 6.15 * 10^-5 m/s

The electric current   I = n * e * A * vd

I = 8.5 * 10^28 * 1.6 * 10^-19 * 2.83 * 10^-6 * 6.15 * 10^-5

I = 2.37 A