# (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.

a) The conductivity of a metallic wire is defined as it is the ability of the wire to allow the electric flow of electric charges as heat through it.
Numerically, the conductivity of a wire is defined as the reciprocal of resistivity.
That is, conductivity

$\sigma = \frac{1}{\rho}$
SI unit is, ohm-1 m-1 or mho m-1
b) Consider the electric field is present then the acceleration of electrons is
$a= \frac{-eE}{m}----(1)$
where e = charge of the electron
m = mass of the electron
From (1) the drift velocity of the electron is given by
Average drift velocity

$V_{d}= \frac{-eE}{m}\, \tau ---(2)$
where $\tau=$ Average force between successive collision which is known as relaxation time
Let 'n' is the number of free electrons per unit volume of a conductor the magnitude of the current is given by
$I= neA\left | V_{d}} \right |---(3)$
substitute the value of V2 from (2) get,
$I= n\frac{e^{2}A}{m}\tau \left | E \right |---(4)$
we know the current density

$J= \frac{I}{A}$
from eqn (4) we get,
$J= \frac{ne^{2}}{m}\tau \left | E \right |$
But $J= \sigma E$
so, By comparing

$\sigma = \frac{ne^{2}}{m}\tau$

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