Consider a conductor XY of length L and area of cross section A (Fig ). An electric field E is applied between its ends. Let n be the number of free electrons per unit volume. The free electrons move towards the left with a constant drift velocity vd. The number of conduction electrons in the conductor = nAL The charge of an electron = e The total charge passing through the conductor q = (nAL) e
The time in which the charges pass through the conductor, \(t=L / v_{d}\) The current flowing through the conductor,
$$
\begin{aligned}
&I=\frac{q}{t}=\frac{(n A L) e}{\left(L / v_{d}\right)} \\
&I=n A e v_{d}
\end{aligned}
$$
The current flowing through a conductor is directly proportional to the drift velocity. From equation (1),
$$
\begin{aligned}
&\frac{I}{A}=n e v_{d} \\
&J=n e v_{d}
\end{aligned} \quad\left[\because J=\frac{I}{A}, \text { current density }\right]
$$
We know that \(_{2 . .}\)
$$
\begin{aligned}
&\therefore v_{d}=\frac{e E}{m} \tau=\mu E \\
&\therefore J=\left(n e^{2} / m\right) \tau E \quad \text { and } \sigma=\left(n e^{2} / m\right) \tau
\end{aligned}
$$
Hence
$$
\vec{J}=\sigma \vec{E}
$$
