Question

A copper wire with diameter 1.00 mm carries a current of 0.50 A. The number density of conduction electrons in copper is 8.5 x 1028 m-3 and the resistivity is 1.7 x 10- 2m.

(a) What are the drift velocity and electric field in the wire?
(b) To estimate the mean time τ between collisions for a conduction electron in the wire. assume that, between collisions, an electron moves with a constant acceleration a due to the electric field. Under this assumption, explain why the drift velocity should be Ud ar. Then solve for the numerical value of τ.

Answer 1

Answer:

a)Drift velocity,  $v_{d} = 4.63 * 10^{-5} m/s$

Electric field in the wire, E = 1.083 * 10⁴ V/m

b) $\tau = 4.84 * 10^{-20} s$

Explanation:

a) Diameter of the wire, d = 1.0 mm = 0.001 m

Area of the wire, A = $\frac{\pi d^{2} }{4}$

A = $\frac{\pi 0.001^{2} }{4}$

A = 0.000000785 m² = 7.85 * 10^-7

Current carried by the wire, I = 0.50 A

Number density of the conduction electrons, n = 8.5 * 10²⁸ m⁻³

charge of an electron, e = 1.62 * 10⁻¹⁹C

Current, $I = neAv_{d}$ , where $v_{d} =$ Drift velocity

$0.5 = 8.5 * 10^{28} * 1.62 * 10^{-19} * 7.85 * 10^{-7} v_{d} \\v_{d} = \frac{0.5}{10809.45} \\v_{d} = 4.63 * 10^{-5} m/s$

Resistivity, $\rho = 1.7 * 10^{-2} ohm-meter$

Electric field, $E = \rho J$

$J = I/A = 0.5/(7.85 * 10^{-7}) \\J = 636942.68 A/m^{2}$

E = 1.7 * 10⁻² * 636942.68

E = 1.083 * 10⁴ V/m

b) If an electron moves with constant acceleration,

S = v t.............(1)

If the electron starts from rest

S = 0.5 at².........(2)

Equating (1) and (2) and making $t = \tau$ and v = $v_{d}$

$v_{d} \tau = 0.5a \tau^{2} \\\tau = \frac{v_{d} }{0.5a}$

To get the value of a

F = qE and F= ma

qE = ma

a = qE/m

Mass of an electron, m = 9.1 * 10⁻³¹kg

a = (1.602 * 10⁻¹⁹*1.083 * 10⁴)/(9.1*10⁻³¹)

a = 1.91 * 10¹⁵m/s²

$\tau = v_{d} /o.5a\\\tau = \frac{4.63 * 10^{-5} }{0.5 * 1.91 * 10^{15} } \\\tau = 4.84 * 10^{-20} s$