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concepts/em/u11/e11-1-current-density.tex

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