\subsection{Current, Drift Velocity, and Current Density}

This subsection connects the macroscopic current $I$ in a wire to the microscopic motion of charge carriers through the current density $\vec{J}$ and the drift velocity $\vec{v}_d$.

\dfn{Current, drift velocity, and current density}{Let $\Delta q$ denote the net charge that crosses a chosen surface in a time interval $\Delta t$. The \emph{electric current} through that surface is
\[
I=\frac{\Delta q}{\Delta t}
\]
in the limit of very small time intervals.

Let $\vec{v}_d$ denote the average drift velocity of the charge carriers, let $n$ denote the number of carriers per unit volume, and let $q$ denote the charge of each carrier. The \emph{current density} $\vec{J}$ is the vector that describes current per unit area, with direction defined by the motion of positive charge. For an oriented surface element $d\vec{A}$,
\[
dI=\vec{J}\cdot d\vec{A}.
\]
Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}

\nt{Conventional current is defined to point in the direction that positive charges would move. Therefore $\vec{J}$ points in the conventional-current direction. In a metal wire, the mobile charge carriers are electrons, so $q=-e$ and the electron drift velocity $\vec{v}_d$ points opposite to $\vec{J}$. If the carriers were positive instead, then $\vec{v}_d$ and $\vec{J}$ would point in the same direction.}

\mprop{Microscopic-macroscopic current relations}{Let $n$ denote the carrier number density, let $q$ denote the charge of each carrier, let $\vec{v}_d$ denote the drift velocity, and let $S$ be a surface with oriented area element $d\vec{A}$. Then the current density is
\[
\vec{J}=nq\vec{v}_d.
\]
The total current through $S$ is
\[
I=\iint_S \vec{J}\cdot d\vec{A}.
\]
If $\vec{J}$ is uniform across a flat cross section of area $A$ and parallel to the area normal, then
\[
I=JA
\qquad \text{and} \qquad
J=\frac{I}{A}.
\]
For a straight wire with uniform carrier density, the corresponding magnitude relation is
\[
I=n|q|Av_d,
\]
where $v_d=|\vec{v}_d|$.}

\qs{Worked AP-style problem}{A long copper wire carries a steady current
\[
I=3.0\,\mathrm{A}
\]
to the right. The wire has radius
\[
r=0.80\,\mathrm{mm}=8.0\times 10^{-4}\,\mathrm{m},
\]
so its cross-sectional area is $A=\pi r^2$. Assume the conduction-electron number density is
\[
n=8.5\times 10^{28}\,\mathrm{m^{-3}},
\]
and the charge of each electron is
\[
q_e=-1.60\times 10^{-19}\,\mathrm{C}.
\]
Let $+\hat{\imath}$ point to the right.

Find:
\begin{enumerate}[label=(\alph*)]
\item the current density magnitude $J$,
\item the current density vector $\vec{J}$, and
\item the electron drift velocity vector $\vec{v}_d$.
\end{enumerate}}

\sol First compute the wire's cross-sectional area:
\[
A=\pi r^2=\pi(8.0\times 10^{-4}\,\mathrm{m})^2.
\]
Since
\[
(8.0\times 10^{-4})^2=6.4\times 10^{-7},
\]
we get
\[
A=\pi(6.4\times 10^{-7})\,\mathrm{m^2}=2.01\times 10^{-6}\,\mathrm{m^2}.
\]

For part (a), use
\[
J=\frac{I}{A}.
\]
Then
\[
J=\frac{3.0\,\mathrm{A}}{2.01\times 10^{-6}\,\mathrm{m^2}}=1.49\times 10^6\,\mathrm{A/m^2}.
\]

For part (b), the current is to the right, so conventional current and $\vec{J}$ point to the right:
\[
\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath}.
\]

For part (c), use the microscopic relation
\[
\vec{J}=nq_e\vec{v}_d.
\]
Solve for the drift velocity:
\[
\vec{v}_d=\frac{\vec{J}}{nq_e}.
\]
Because $q_e$ is negative, $\vec{v}_d$ must point opposite to $\vec{J}$, so it points left. Its magnitude is
\[
v_d=\frac{J}{n|q_e|}.
\]
Substitute the values:
\[
v_d=\frac{1.49\times 10^6}{(8.5\times 10^{28})(1.60\times 10^{-19})}\,\mathrm{m/s}.
\]
The denominator is
\[
(8.5\times 10^{28})(1.60\times 10^{-19})=1.36\times 10^{10},
\]
so
\[
v_d=\frac{1.49\times 10^6}{1.36\times 10^{10}}\,\mathrm{m/s}=1.10\times 10^{-4}\,\mathrm{m/s}.
\]
Therefore the drift velocity vector is
\[
\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
\]

Therefore,
\[
J=1.49\times 10^6\,\mathrm{A/m^2},
\qquad
\vec{J}=(1.49\times 10^6\,\mathrm{A/m^2})\hat{\imath},
\]
\[
\vec{v}_d=-(1.10\times 10^{-4}\,\mathrm{m/s})\hat{\imath}.
\]
