The field of a small loop; the magnetic dipole
Let’s use the vector-potential method to find the magnetic field of a small loop of current. As usual, by “small” we mean simply that we are interested in the fields only at distances large compared with the size of the loop. It will turn out that any small loop is a “magnetic dipole.” That is, it produces a magnetic field like the electric field from an electric dipole. We take first a rectangular loop, and choose our coordinates as shown in Fig. 14–6. There are no currents in the $z$-direction, so $A_z$ is zero. There are currents in the $x$-direction on the two sides of length $a$. In each leg, the current density (and current) is uniform. So the solution for $A_x$ is just like the electrostatic potential from two charged rods (see Fig. 14–7). Since the rods have opposite charges, their electric potential at large distances would be just the dipole potential (Section 6–5). At the point $P$ in Fig. 14–6, the potential would be \begin{equation} \label{Eq:II:14:28} \phi=\frac{1}{4\pi\epsO}\,\frac{\FLPp\cdot\FLPe_R}{R^2}, \end{equation}  where $\FLPp$ is the dipole moment of the charge distribution. The dipole moment, in this case, is the total charge on one rod times the separation between them: \begin{equation} \label{Eq:II:14:29} p=\lambda ab. \end{equation}  The dipole moment points in the negative $y$-direction, so the cosine of the angle between $\FLPR$ and $\FLPp$ is $-y/R$ (where $y$ is the coordinate of $P$). So we have \begin{equation*} \phi=-\frac{1}{4\pi\epsO}\,\frac{\lambda ab}{R^2}\,\frac{y}{R}. \end{equation*}  We get $A_x$ simply by replacing $\lambda$ by $I/c^2$: \begin{equation} \label{Eq:II:14:30} A_x=-\frac{Iab}{4\pi\epsO c^2}\,\frac{y}{R^3}. \end{equation}  By the same reasoning, \begin{equation} \label{Eq:II:14:31} A_y=\frac{Iab}{4\pi\epsO c^2}\,\frac{x}{R^3}. \end{equation}  Again, $A_y$ is proportional to $x$ and $A_x$ is proportional to $-y$, so the vector potential (at large distances) goes in circles around the $z$-axis, circulating in the same sense as $I$ in the loop, as shown in Fig. 14–8. The strength of $\FLPA$ is proportional to $Iab$, which is the current times the area of the loop. This product is called the magnetic dipole moment (or, often, just “magnetic moment”) of the loop. We represent it by $\mu$: \begin{equation} \label{Eq:II:14:32} \mu=Iab. \end{equation}  The vector potential of a small plane loop of any shape (circle, triangle, etc.) is also given by Eqs. (14.30) and (14.31) provided we replace $Iab$ by \begin{equation} \label{Eq:II:14:33} \mu=I\cdot(\text{area of the loop}). \end{equation}  We leave the proof of this to you. We can put our equation in vector form if we define the direction of the vector $\FLPmu$ to be the normal to the plane of the loop, with a positive sense given by the right-hand rule (Fig. 14–8). Then we can write \begin{equation} \label{Eq:II:14:34} \FLPA=\frac{1}{4\pi\epsO c^2}\,\frac{\FLPmu\times\FLPR}{R^3}= \frac{1}{4\pi\epsO c^2}\,\frac{\FLPmu\times\FLPe_R}{R^2}. \end{equation}  We have still to find $\FLPB$. Using (14.33) and (14.34), together with (14.4), we get \begin{equation} \label{Eq:II:14:35} B_x=-\ddp{}{z}\,\frac{\mu}{4\pi\epsO c^2}\,\frac{x}{R^3}= \dotsm\frac{3xz}{R^5} \end{equation}  (where by $\dotsm$ we mean $\mu/4\pi\epsO c^2$), \begin{equation} \begin{aligned} B_y&=\ddp{}{z}\biggl(-\dotsm\frac{y}{R^3}\biggr)=\dotsm\frac{3yz}{R^5},\\[1ex] B_z&=\ddp{}{x}\biggl(\dotsm\frac{x}{R^3}\biggr)- \ddp{}{y}\biggl(-\dotsm\frac{y}{R^3}\biggr)\\[1ex] &=-\dotsm\biggl(\frac{1}{R^3}-\frac{3z^2}{R^5}\biggr). \end{aligned} \label{Eq:II:14:36} \end{equation}  The components of the $\FLPB$-field behave exactly like those of the $\FLPE$-field for a dipole oriented along the $z$-axis. (See Eqs. (6.14) and (6.15); also Fig. 6–4.) That’s why we call the loop a magnetic dipole. The word “dipole” is slightly misleading when applied to a magnetic field because there are no magnetic “poles” that correspond to electric charges. The magnetic “dipole field” is not produced by two “charges,” but by an elementary current loop.  It is curious, though, that starting with completely different laws, $\FLPdiv{\FLPE}=\rho/\epsO$ and $\FLPcurl{\FLPB}=\FLPj/\epsO c^2$, we can end up with the same kind of a field. Why should that be? It is because the dipole fields appear only when we are far away from all charges or currents. So through most of the relevant space the equations for $\FLPE$ and $\FLPB$ are identical: both have zero divergence and zero curl. So they give the same solutions. However, the sources whose configuration we summarize by the dipole moments are physically quite different—in one case, it’s a circulating current; in the other, a pair of charges, one above and one below the plane of the loop for the corresponding field.