Question

(a) Define the term 'conductivity' of a metallic wire. Write its SI unit.
(b) 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.

Answer 1

(a) Conductivity of a metallic wire is defined as its ability to allow electric charges or heat to pass through it.
Numerically, conductivity is the reciprocal of resistivity.
SI unit : ${\text{ohm}}^{-1}$ $m^{-1}$ or mho $m^{-1}$ or $S{m}^{-1}$.
(b) Consider a potential difference V applied across a conductor of length l and cross-section area A.
Electric field inside the conductor, E = $\frac{V}{l}$
Due to the external field the free electrons inside the conductor drift with velocity $v_d$.
Let, number of electrons per unit volume = n,
charge on an electron = e
$\therefore$ Total electrons in length l = nAl
And, total charge, q = neAl



Time taken by electrons to enter and leave the conductor,

t= $\frac{l}{v_d}$

Current, I = $\frac{q}{t}=\frac{neAl}{l/v_d}$
=$neA{v}_d$

Current density, J = $\frac{I}{A}=ne{v}_d$ ...(i)

We know, $v_d=\frac{eE\tau }{m}=\frac{eE\tau }{ml}$

$\therefore I=neA{v}_d=\frac{neAVe\tau }{ml}$

$\Rightarrow \frac{V}{I}=\frac{ml}{m{e}^2\tau A}[\because \frac{V}{I}=R;$ Ohm's law]

Resistivity, $\rho =\frac{RA}{l}=\frac{mlA}{n{e}^2\tau A}$

or $\rho =mln{e}^2\tau$

Since conducitivity, $\sigma =\frac{1}{\rho }$

$\therefore \sigma =\frac{n{e}^2\tau }{m}$ ...(ii)

Relation between current density and field:

For an electron, charge q=-e

And current density, $J=\frac{I}{A}=-ne{v}_d$ [from(i)]

$J=(-ne)\left(\frac{-eE\tau }{m}\right)$

=$\left(\frac{n{e}^2\tau }{m}\right)E$

$\Rightarrow J=\sigma$ E [from (ii)]

which is the required relation.