A straight wire
For our first example, we will again find the field of a straight wire—which we solved in the last chapter by using Eq. (14.2) and some arguments of symmetry. We take a long straight wire of radius $a$, carrying the steady current $I$. Unlike the charge on a conductor in the electrostatic case, a steady current in a wire is uniformly distributed throughout the cross section of the wire. If we choose our coordinates as shown in Fig. 14–3, the current density vector $\FLPj$ has only a $z$-component. Its magnitude is \begin{equation} \label{Eq:II:14:20} j_z=\frac{I}{\pi a^2} \end{equation}  inside the wire, and zero outside. Since $j_x$ and $j_y$ are both zero, we have immediately \begin{equation*} A_x=0,\qquad A_y=0. \end{equation*}  To get $A_z$ we can use our solution for the electrostatic potential $\phi$ of a wire with a uniform charge density $\rho=j_z/c^2$. For points outside an infinite charged cylinder, the electrostatic potential is \begin{equation*} \phi=-\frac{\lambda}{2\pi\epsO}\ln r', \end{equation*}  where $r'=\sqrt{x^2+y^2}$ and $\lambda$ is the charge per unit length, $\pi a^2\rho$. So $A_z$ must be \begin{equation*} A_z=-\frac{\pi a^2j_z}{2\pi\epsO c^2}\ln r' \end{equation*}  for points outside a long wire carrying a uniform current. Since $\pi a^2j_z=I$, we can also write \begin{equation} \label{Eq:II:14:21} A_z=-\frac{I}{2\pi\epsO c^2}\ln r' \end{equation}  Now we can find $\FLPB$ from (14.4). There are only two of the six derivatives that are not zero. We get  \begin{alignat}{6} \label{Eq:II:14:22} &B_x&&=-&&\frac{I}{2\pi\epsO c^2}\ddp{}{y}&&\ln r'= -&&\frac{I}{2\pi\epsO c^2}\,\frac{y}{r'^2}&&,\\[1ex] \label{Eq:II:14:23} &B_y&&=&&\frac{I}{2\pi\epsO c^2}\vphantom{\ddp{}{y}}\ddp{}{x}&&\ln r'= &&\frac{I}{2\pi\epsO c^2}\,\frac{x}{r'^2}&&,\\[2pt] % ebook insert: \label{Eq:II:0:0} &B_z&&=&&0\vphantom{\ddp{}{y}}.\notag \end{alignat}   We get the same result as before: $\FLPB$ circles around the wire, and has the magnitude \begin{equation} \label{Eq:II:14:24} B=\frac{1}{4\pi\epsO c^2}\,\frac{2I}{r'}. \end{equation} 