Derive the relation between drift velocity and current density.

(N/A) Consider a conductor of cross-sectional area $A$ in which an electric field $\vec{E}$ is applied. Due to this field,free electrons experience a force and drift in the opposite direction of $\vec{E}$ with a drift velocity $v_d$.
In a small time interval $\Delta t$,the distance covered by the electrons is $\Delta x = |v_d| \Delta t$.
The volume of the cylindrical element containing these electrons is $V = A \Delta x = A |v_d| \Delta t$.
If $n$ is the number density of free electrons (number of electrons per unit volume),then the total number of electrons $N$ in this volume is $N = n V = n A |v_d| \Delta t$.
The total charge $\Delta Q$ passing through the cross-section in time $\Delta t$ is $\Delta Q = N e = n A e |v_d| \Delta t$,where $e$ is the magnitude of the charge of an electron.
The current $I$ is given by $I = \frac{\Delta Q}{\Delta t} = n A e |v_d|$.
Since current density $J = \frac{I}{A}$,we have $J = \frac{n A e |v_d|}{A} = n e |v_d|$.
In vector form,since electrons drift opposite to the electric field,$\vec{J} = -n e \vec{v}_d$.