Question

(a) Define the term ‘conductivity’ of a metallic wire. Write its SI unit.
 (b) Using the concept of a free electron in a conductor, derive the expression for the conductivity of wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E.

Answer 1

Hint: The reciprocal of resistivity of a material is called its conductivity and denoted by σ.The relation between conductivity(σ) is found to be directly proportional to the number density, square of electric charge of the electron, relaxation time τ, and inversely proportional to the mass of the electron.
Formula used:
The relation between the conductivity of wire in terms of number density and relaxation time is given below:
$\sigma = \dfrac{{n{e^2}\tau }}{m}$
Where σ is conductivity, n is number density, τ is relaxation time, e is the charge of electron and m is the mass of the electron.

Complete step-by-step answer:
The conductivity of a metallic wire is its ability to conduct electricity. It is defined as the reciprocal of the resistivity of a metallic wire. Mathematically given below
$\sigma = \dfrac{1}{\rho }$
The S.I. unit if conductivity is${\Omega ^{ - 1}}{m^{ - 1}}$ or$S/m$ .

(i)Conductivity in terms of electron density and relaxation time is derived as follow-
We know that The resistance R of a conductor of length l, area of cross-section A and resistivity ρ is given by
$R = \rho \dfrac{l}{A}$
(We know,$\rho = \dfrac{1}{\sigma }$ )
$R = \dfrac{1}{\sigma }\dfrac{l}{A}$
$R = \dfrac{l}{{\sigma A}}$
$\sigma = \dfrac{l}{{RA}}$
But we also have $R = \dfrac{{ml}}{{n{e^2}\tau A}}$ ,
Putting this value of R in the above equation, we have
$\sigma = \dfrac{{n{e^2}\tau }}{m}$ (1)
From the above equation, we get that the conductivity of metallic wire is directly proportional to the electron density (n) and relaxation time (τ).

(ii) The relation between current density and electric field.

For an electron of charge (-e), electric field$\vec E$, the drift velocity ${\vec v_d}$ is given below.
${\vec v_d} = - \dfrac{{e\vec E\tau }}{m}$
And current density$\vec j$ is given as follow
$\vec j = n( - e){\vec \upsilon _d}$
Putting the value of ${\vec v_d}$ in the above equation we get the following relation
$\vec j = ( - e)( - \dfrac{{e\vec E\tau }}{m})$
Let us simplify the above equation.we get,
$\vec j = \dfrac{{n{e^2}\tau }}{m}\vec E$
Now we know, $\dfrac{{n{e^2}\tau }}{m} = \sigma $. Let us use this relation in the above expression.
$\vec j = \sigma \vec E$ (2)
Hence, the required relations are$\sigma = \dfrac{{n{e^2}\tau }}{m}$,$\vec j = \sigma \vec E$
Additional information:
Drift velocity: It is the average velocity that is attended by the moving particle, like an electron, etc. Drift velocity is directly proportional to the current applied, thus mathematically it can be written as follows,
${u}$ = ${\mu} E $
Where,$u$ = drift velocity,${\mu}$ electron mobility of material,$ E $ is the electrical field.

Note:
The electron density of a metallic conductor is defined as the number of free electrons per unit volume.
The relaxation time τ is defined as the average time between the two successive collisions of an electron.