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

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`

Step-by-step explanation:

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

Area of the wire, A =
`(\pi d^(2) )/(4)`

A =
`(\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) = (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 = (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 = (4.63 * 10^(-5) )/(0.5 * 1.91 * 10^(15) ) \\\tau = 4.84 * 10^(-20) s`