Question

Using the concept of free electrons in a conductor derive the expression for the conductivity of a wire in terms of number density and relaxation time hence obtain the relation between current density and applied electric field E

Answer 1

J= IA
I= NeAVd
J= NeVd
Vd= eEt

J= Ne eEt/m
J= sigma. E

Answer 2

______?

Solution:

The drift velocity of electrons in a conductor is given as

V_d = e E r _________(1)

m

where,

E = electric field set up across a conductor

m = mass if the electron

average relaxation time

Now the current flowing in the conductor can be derived as [see attachment 2]☺

Electron drift to a small distance in a time

Δt , q = lΔt

or l = neAvd ________(2)

where n = no of free electrons per unit volume or

number density

Now from equation (1) & (2), we get

I = ne^2 ArE ________(3)

m

Since the resistivity of a conductor is given as;

ρ = m

ne^2

Now we know that conductivity of a conductor is mathematically defined as the reciprocal of resistivity of conductor.Thus

ρ = 1 /σ _________(4)

where σ = conductivity of the conductor.,Thus

from equation (3) & (4).,we get,

σ = ne^2r _________(5)

m

Now from equation (3)& (5) ., we have,

I = σE. _________(6)

A

and the current density is given as

J = I

A

Thus,

J = σE✅