= Electric current
{title2=$I=dQ_{\mathrm{cross}}/dt$}
{wiki}
An <electric current> measures how quickly signed <electric charge> crosses a chosen fixed <oriented surface> $S$. Choose its <unit normal> $\mathbf n$; positive <electric charge> crossing along $\mathbf n$ counts positively, while crossing against it counts negatively. If $Q_{\mathrm{cross}}(t)$ is the accumulated signed <electric charge> transported through $S$, then
$$
I_S(t)=\frac{dQ_{\mathrm{cross}}}{dt}=\int_S\mathbf J(\mathbf r,t)\cdot\mathbf n\,dA,\qquad Q_{\mathrm{cross}}(t_2)-Q_{\mathrm{cross}}(t_1)=\int_{t_1}^{t_2}I_S(t)\,dt.
$$
Here $\mathbf J$ is the <current density>; the <surface integral> collects its local normal contributions. Reversing the <unit normal> reverses $I_S$. The <electric current> is a signed <scalar> associated with the whole chosen <oriented surface>, whereas <current density> is a <vector> at each point. The <SI> unit is the <ampere>, one <coulomb> per second. https://openstax.org/books/university-physics-volume-2/pages/9-1-electrical-current[Ling, Sanny and Moebs, University Physics II, §9.1: transported charge and current]{ref}; https://openstax.org/books/university-physics-volume-2/pages/9-2-model-of-conduction-in-metals[Ling, Sanny and Moebs, §9.2: current density and its surface integral]{ref}.
For a constant <electric current> of $2\,\mathrm A$ over a <time> interval of $3\,\mathrm s$, the transported <electric charge> is $6\,\mathrm C$. This need not be <electric charge> accumulating in the wire: in steady transport, <electric charge> enters and leaves. For a closed <oriented surface> with outward <unit normal>, <electric charge conservation> instead gives $I_{\mathrm{out}}=-dQ_{\mathrm{inside}}/dt$. Keeping transported and enclosed <electric charge> distinct connects <electrical circuit> bookkeeping to the local <continuity equation>.
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