The vector potential of a circuit
We are often interested in the magnetic fields produced by circuits of wire in which the diameter of the wire is very small compared with the dimensions of the whole system. In such cases, we can simplify the equations for the magnetic field. For a thin wire we can write our volume element as \begin{equation*} dV=S\,ds \end{equation*}  where $S$ is the cross-sectional area of the wire and $ds$ is the element of distance along the wire. In fact, since the vector $d\FLPs$ is in the same direction as $\FLPj$, as shown in Fig. 14–9 (and we can assume that $\FLPj$ is constant across any given cross section), we can write a vector equation: \begin{equation} \label{Eq:II:14:37} \FLPj\,dV=jS\,d\FLPs. \end{equation}  But $jS$ is just what we call the current $I$ in a wire, so our integral for the vector potential (14.19) becomes \begin{equation} \label{Eq:II:14:38} \FLPA(1)=\frac{1}{4\pi\epsO c^2}\int\frac{I\,d\FLPs_2}{r_{12}} \end{equation}  (see Fig. 14–10). (We assume that $I$ is the same throughout the circuit. If there are several branches with different currents, we should, of course, use the appropriate $I$ for each branch.) Again, we can find the fields from (14.38) either by integrating directly or by solving the corresponding electrostatic problems.