Current density
= Current density
{title2=$\mathbf J$}
{wiki}
= Electric current density
{synonym}
The <current density> $\mathbf J(\mathbf r,t)$ describes the amount and direction of local <electric charge> transport per <area> per <time>. Its <SI> unit is $\mathrm{A/m^2}$. Through a small <area element> with <unit normal> $\mathbf n$, the contribution to <electric current> is $dI=\mathbf J\cdot\mathbf n\,dA$: the <dot product> selects transport through the surface. A <current density> tangent to the surface contributes zero. If <current density> is uniform and perpendicular to a wire's transverse slice, its magnitude $j=|\mathbf J|$ satisfies
$$
I=jA,\qquad j=\frac IA\quad\text{with the normal chosen along conventional current}.
$$
Here $A$ is the <wire cross-sectional area>. For nonuniform <current density>, $I/A$ is only the average normal component, and the <surface integral> is needed. https://openstax.org/books/university-physics-volume-2/pages/9-2-model-of-conduction-in-metals[Ling, Sanny and Moebs, §9.2: local current density]{ref}.
Consider two separate uniform-flow cases, each carrying <electric current> $I=2\,\mathrm A$. A <wire cross-sectional area> of $1\,\mathrm{mm^2}=10^{-6}\,\mathrm{m^2}$ gives $j=2\times10^6\,\mathrm{A/m^2}$; doubling the <wire cross-sectional area> to $2\,\mathrm{mm^2}$ gives $j=10^6\,\mathrm{A/m^2}$. Equal <electric current> can therefore mean different local transport intensity. This distinction matters for <Joule heating>, whose local <power> per volume depends on <current density> and <electric field>.
\Image[/current-density-and-wire-area.png]
{title=Equal <electric current> through different <wire cross-sectional areas> gives different <current densities>}
{description=Both cases carry $2\,\mathrm A$ with uniform normal <current density>. The transverse slices use the same spatial scale; the bars compare $j=I/A$.}
{source=current-density-and-wire-area.py}
For a toy <electron> transport model, stipulate mobile <electron number density> $n=8\times10^{28}\,\mathrm{m^{-3}}$ and <drift velocity> $\mathbf v_d=-10^{-4}\,\mathrm{m/s}\,\widehat{\mathbf x}$, with the positive <unit normal> $\widehat{\mathbf x}$. Each <electron> has <electric charge> $-e$, where the <elementary charge> $e=1.602176634\times10^{-19}\,\mathrm C$ is exact in <SI>. These are defined model inputs, rather than measured parameters of a particular material. https://physics.nist.gov/cuu/Constants/Value/e.html[NIST, exact elementary charge]{ref} The <current density> and its magnitude are
$$
\mathbf J=n(-e)\mathbf v_d=j\widehat{\mathbf x},\qquad j=ne|\mathbf v_d|=1.2817413072\times10^6\,\mathrm{A/m^2}.
$$
For <wire cross-sectional area> $A=1\,\mathrm{mm^2}$ and constant transport over <time> $10\,\mathrm s$, this gives
$$
I=jA=1.2817413072\,\mathrm A,\qquad \Delta Q_{\mathrm{cross}}=I\Delta t=12.817413072\,\mathrm C.
$$
The two minus signs make <conventional current> point opposite to the <electron> <drift velocity>. A small <drift velocity> can carry an appreciable <electric current> because many <charge carriers> contribute; it does not specify the propagation speed of an electrical disturbance.