The law of Biot and Savart
In studying electrostatics we found that the electric field of a known charge distribution could be obtained directly with an integral, Eq. (4.16): \begin{equation*} \FLPE(1)=\frac{1}{4\pi\epsO}\int \frac{\rho(2)\FLPe_{12}\,dV_2}{r_{12}^2}. \end{equation*}  As we have seen, it is usually more work to evaluate this integral—there are really three integrals, one for each component—than to do the integral for the potential and take its gradient. There is a similar integral which relates the magnetic field to the currents. We already have an integral for $\FLPA$, Eq. (14.19); we can get an integral for $\FLPB$ by taking the curl of both sides:  \begin{equation} \label{Eq:II:14:39} \FLPB(1)=\FLPcurl{\FLPA(1)}= \FLPcurl{\biggl[\frac{1}{4\pi\epsO c^2}\!\int\! \frac{\FLPj(2)\,dV_2}{r_{12}}\biggr]}\!. \end{equation} 
  \begin{align} \FLPB(1)&=\FLPcurl{\FLPA(1)}\notag\\ \label{Eq:II:14:39} &=\FLPcurl{\biggl[\frac{1}{4\pi\epsO c^2}\!\int\! \frac{\FLPj(2)\,dV_2}{r_{12}}\biggr]}\!. \end{align}   Now we must be careful: The curl operator means taking the derivatives of $\FLPA(1)$, that is, it operates only on the coordinates $(x_1,y_1,z_1)$. We can move the $\FLPcurl{}$ operator inside the integral sign if we remember that it operates only on variables with the subscript $1$, which of course, appear only in \begin{equation} \label{Eq:II:14:40} r_{12}=[(x_1\!-\!x_2)^2\!+\!(y_1\!-\!y_2)^2\!+\!(z_1\!-\!z_2)^2]^{1/2}. \end{equation}  We have, for the $x$-component of $\FLPB$, \begin{align} B_x&=\ddp{A_z}{y_1}-\ddp{A_y}{z_1}\notag\\[1.5ex] &=\frac{1}{4\pi\epsO c^2}\!\int\biggl[ j_z\ddp{}{y_1}\biggl(\!\frac{1}{r_{12}}\!\biggr)\!-\! j_y\ddp{}{z_1}\biggl(\!\frac{1}{r_{12}}\!\biggr) \!\biggr]dV_2\notag\\[2ex] \label{Eq:II:14:41} &=-\frac{1}{4\pi\epsO c^2}\!\int\biggl[ j_z\frac{y_1-y_2}{r_{12}^3}\!-\!j_y\frac{z_1-z_2}{r_{12}^3} \biggr]dV_2. \end{align}  The quantity in brackets is just the negative of the $x$-component of \begin{equation*} \frac{\FLPj\times\FLPr_{12}}{r_{12}^3}= \frac{\FLPj\times\FLPe_{12}}{r_{12}^2} \end{equation*}  Corresponding results will be found for the other components, so we have \begin{equation} \label{Eq:II:14:42} \FLPB(1)=\frac{1}{4\pi\epsO c^2}\int \frac{\FLPj(2)\times\FLPe_{12}}{r_{12}^2}\,dV_2. \end{equation}  The integral gives $\FLPB$ directly in terms of the known currents. The geometry involved is the same as that shown in Fig. 14–2. If the currents exist only in circuits of small wires we can, as in the last section, immediately do the integral across the wire, replacing $\FLPj\,dV$ by $I\,d\FLPs$, where $d\FLPs$ is an element of length of the wire. Then, using the symbols in Fig. 14–10, \begin{equation} \label{Eq:II:14:43} \FLPB(1)=-\frac{1}{4\pi\epsO c^2}\int \frac{I\FLPe_{12}\times d\FLPs_2}{r_{12}^2}. \end{equation}  (The minus sign appears because we have reversed the order of the cross product.) This equation for $\FLPB$ is called the Biot-Savart law, after its discoverers. It gives a formula for obtaining directly the magnetic field produced by wires carrying currents. You may wonder: “What is the advantage of the vector potential if we can find $\FLPB$ directly with a vector integral? After all, $\FLPA$ also involves three integrals!” Because of the cross product, the integrals for $\FLPB$ are usually more complicated, as is evident from Eq. (14.41). Also, since the integrals for $\FLPA$ are like those of electrostatics, we may already know them. Finally, we will see that in more advanced theoretical matters (in relativity, in advanced formulations of the laws of mechanics, like the principle of least action to be discussed later, and in quantum mechanics) the vector potential plays an important role.