Question

For a metallic conductor, what is the relationship between current density $\left( j \right)$, conductivity $\left( \sigma \right)$ and electric field intensity $E$?

Answer 1

The amount of electric current passing through per unit cross section area is called as current density and expressed as amperes per square meter. So we have to use the concept of current density. As current density in metals with finite resistance are directly proportional to electric field. So we are using an elementary area and passing current through it and calculating current density.

Formula used:
According to ohm’s law the potential difference between ends of a conductor is directly proportional to the current flowing in it.
i.e. $V=IR$ --------(1)
where $R$ = resistance
and $I$ = current passing through it

Complete step by step answer:
Let the current $I$ flowing in a uniform cross section area and normally to it, then:
$I=\int{j.}ds$
$=\int{j.ds.\cos (0)}$
$=j\int{ds}$
$I=jA$ ----(2) (where $\int{ds=A}$)
Resistance can be written as $R=\dfrac{\rho l}{A}$
So, equation (2) can be written as $R=\dfrac{l}{\sigma A}$ ------(3)
Where $\sigma =\dfrac{1}{\rho }$ is conductivity.
Now put the equations 2 and 3 in equation 1 we get-
$V=\left( jA \right)*\left( \dfrac{l}{\sigma A} \right)$
$V=\dfrac{jl}{\sigma }$
Or $j=\dfrac{\sigma V}{I}$ ------(4)
But the electric field intensity $E=\dfrac{V}{I}$
By putting above value of E in equation (4) we get
$j=\sigma E$

Note:
There is a difference between current and current density so use it according to the problem. As current density measures the density of electric current i.e. current passing through per unit cross section area and it has both magnitude and direction. But on the other hand, current is the flow of electrically charged particles in electron deficient atoms.