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 the applied electric field E.

Explanation

The electric field E exerts an electric force or an electron , `vecF=-vecEe`

According to each electron `veca=(-evecE)/m`.........(1)
where m=mass of an electron and e=charge or an electron.
Drift velocity ,`vecv_d=(vecv_1+vecv_2+.......+vecv_n)/n`
`=((vecu_1+vecar_1)+(vecu_2+vecar_2)+.....+(vecu_n+vecar_n))/n`
Now, `vecu_1,vecu_2,........vecu_n` are the terminal velocities of the electrons,
`vecar_1,vecar_2,....vecar_n` are the velocities acquired by electrons
`r_1,r_2,.....r_n` are the time elapsed after the collision
`thereforevecv_d=((vecu_1+vecu_2+....+vecu_n))/n+(veca(r_1+r_2+.....+r_n))/n`
Since `(vecu_1+vecu_2+......+vecu_n)/n` =average thermal velocity =0, we get
`vecv_d=veca_r`.......(2)
Where `r=(r_1+r_2+r_3+.....+r_n)/n` is the average time elapsed between two successive collisions of electons which is known as relaxation time of electron.
From equation (1) and (2) we have,
`vecv_d=(-evecE)/mr`........(3)
Let N number of free electrons passing through the cross section (A) of the wire in time t, is confined within a cylinder lenght l, Then electric current flowing through the conductor,
`I=(-Ne)/t=(-nAl e)/(l/v_d)` [n=number density of free electron]
or,`I=-n eAv_d=(n e^2Ar)/mE`
or,`j=I/A=((n e^2r)/m)E=sigma E`