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