The vector potential of known currents
Since $\FLPB$ is determined by currents, so also is $\FLPA$. We want now to find $\FLPA$ in terms of the currents. We start with our basic equation (14.2): \begin{equation*} c^2\FLPcurl{\FLPB}=\frac{\FLPj}{\epsO}, \end{equation*}  which means, of course, that \begin{equation} \label{Eq:II:14:12} c^2\FLPcurl{(\FLPcurl{\FLPA})}=\frac{\FLPj}{\epsO}. \end{equation}  This equation is for magnetostatics what the equation \begin{equation} \label{Eq:II:14:13} \FLPdiv{\FLPgrad{\phi}}=-\frac{\rho}{\epsO} \end{equation}  was for electrostatics. Our equation (14.12) for the vector potential looks even more like that for $\phi$ if we rewrite $\FLPcurl{(\FLPcurl{\FLPA})}$ using the vector identity Eq. (2.58): \begin{equation} \label{Eq:II:14:14} \FLPcurl{(\FLPcurl{\FLPA})}=\FLPgrad{(\FLPdiv{\FLPA})}-\nabla^2\FLPA. \end{equation}  Since we have chosen to make $\FLPdiv{\FLPA}=0$ (and now you see why), Eq. (14.12) becomes \begin{equation} \label{Eq:II:14:15} \nabla^2\FLPA=-\frac{\FLPj}{\epsO c^2}. \end{equation}  This vector equation means, of course, three equations:  \begin{equation} \label{Eq:II:14:16} \nabla^2A_x=-\frac{j_x}{\epsO c^2},\quad \nabla^2A_y=-\frac{j_y}{\epsO c^2},\quad \nabla^2A_z=-\frac{j_z}{\epsO c^2}. \end{equation} 
  \begin{equation} \begin{aligned} \nabla^2A_x&=-\frac{j_x}{\epsO c^2},\\[1ex] \nabla^2A_y&=-\frac{j_y}{\epsO c^2},\\[1ex] \nabla^2A_z&=-\frac{j_z}{\epsO c^2}. \end{aligned} \label{Eq:II:14:16} \end{equation}     And each of these equations is mathematically identical to \begin{equation} \label{Eq:II:14:17} \nabla^2\phi=-\frac{\rho}{\epsO}. \end{equation}  All we have learned about solving for potentials when $\rho$ is known can be used for solving for each component of $\FLPA$ when $\FLPj$ is known! We have seen in Chapter 4 that a general solution for the electrostatic equation (14.17) is \begin{equation*} \phi(1)=\frac{1}{4\pi\epsO}\int\frac{\rho(2)\,dV_2}{r_{12}}. \end{equation*}  So we know immediately that a general solution for $A_x$ is \begin{equation} \label{Eq:II:14:18} A_x(1)=\frac{1}{4\pi\epsO c^2}\int\frac{j_x(2)\,dV_2}{r_{12}}, \end{equation}  and similarly for $A_y$ and $A_z$. (Figure 14–2 will remind you of our conventions for $r_{12}$ and $dV_2$.) We can combine the three solutions in the vector form \begin{equation} \label{Eq:II:14:19} \FLPA(1)=\frac{1}{4\pi\epsO c^2}\int\frac{\FLPj(2)\,dV_2}{r_{12}}, \end{equation}  (You can verify if you wish, by direct differentiation of components, that this integral for $\FLPA$ satisfies $\FLPdiv{\FLPA}=0$ so long as $\FLPdiv{\FLPj}=0$, which, as we saw, must happen for steady currents.) We have, then, a general method for finding the magnetic field of steady currents. The principle is: the $x$-component of vector potential arising from a current density $\FLPj$ is the same as the electric potential $\phi$ that would be produced by a charge density $\rho$ equal to $j_x/c^2$—and similarly for the $y$- and $z$-components. (This principle works only with components in fixed directions. The “radial” component of $\FLPA$ does not come in the same way from the “radial” component of $\FLPj$, for example.) So from the vector current density $\FLPj$, we can find $\FLPA$ using Eq. (14.19)—that is, we find each component of $\FLPA$ by solving three imaginary electrostatic problems for the charge distributions $\rho_1=j_x/c^2$, $\rho_2=j_y/c^2$, and $\rho_3=j_z/c^2$. Then we get $\FLPB$ by taking various derivatives of $\FLPA$ to obtain $\FLPcurl{\FLPA}$. It’s a little more complicated than electrostatics, but the same idea. We will now illustrate the theory by solving for the vector potential in a few special cases.