The magnetic field of a straight wire and of a solenoid; atomic currents
We can illustrate the use of Ampère’s law by finding the magnetic field near a wire. We ask: What is the field outside a long straight wire with a cylindrical cross section? We will assume something which may not be at all evident, but which is nevertheless true: that the field lines of $\FLPB$ go around the wire in closed circles. If we make this assumption, then Ampère’s law, Eq. (13.16), tells us how strong the field is. From the symmetry of the problem, $\FLPB$ has the same magnitude at all points on a circle concentric with the wire (see Fig. 13–7). We can then do the line integral of $\FLPB\cdot d\FLPs$ quite easily; it is just the magnitude of $\FLPB$ times the circumference. If $r$ is the radius of the circle, then \begin{equation*} \oint\FLPB\cdot d\FLPs=B\cdot2\pi r. \end{equation*}  The total current through the loop is merely the current $I$ in the wire, so \begin{equation} B\cdot2\pi r=\frac{I}{\epsO c^2},\notag \end{equation}  or \begin{equation} \label{Eq:II:13:17} B=\frac{1}{4\pi\epsO c^2}\,\frac{2I}{r}. \end{equation}  The strength of the magnetic field drops off inversely as $r$, the distance from the axis of the wire. We can, if we wish, write Eq. (13.17) in vector form. Remembering that $\FLPB$ is at right angles both to $\FLPI$ and to $\FLPr$, we have \begin{equation} \label{Eq:II:13:18} \FLPB=\frac{1}{4\pi\epsO c^2}\,\frac{2\FLPI\times\FLPe_r}{r}. \end{equation}  We have separated out the factor $1/4\pi\epsO c^2$, because it appears often. It is worth remembering that it is exactly $10^{-7}$ (in the mks system), since an equation like (13.17) is used to define the unit of current, the ampere. At one meter from a current of one ampere the magnetic field is $2\times10^{-7}$ webers per square meter. Since a current produces a magnetic field, it will exert a force on a nearby wire which is also carrying a current. In Chapter 1 we described a simple demonstration of the forces between two current-carrying wires. If the wires are parallel, each is at right angles to the $\FLPB$ field of the other; the wires should then be pushed either toward or away from each other. When currents are in the same direction, the wires attract; when the currents are moving in opposite directions, the wires repel. Let’s take another example that can be analyzed by Ampère’s law if we add some knowledge about the field. Suppose we have a long coil of wire wound in a tight spiral, as shown by the cross sections in Fig. 13–8. Such a coil is called a solenoid. We observe experimentally that when a solenoid is very long compared with its diameter, the field outside is very small compared with the field inside. Using just that fact, together with Ampère’s law, we can find the size of the field inside. Since the field stays inside (and has zero divergence), its lines must go along parallel to the axis, as shown in Fig. 13–8. That being the case, we can use Ampère’s law with the rectangular “curve” $\Gamma$ shown in the figure. This loop goes the distance $L$ inside the solenoid, where the field is, say, $\FLPB_0$, then goes at right angles to the field, and returns along the outside, where the field is negligible. The line integral of $\FLPB$ for this curve is just $B_0L$, and it must be $1/\epsO c^2$ times the total current through $\Gamma$, which is $NI$ if there are $N$ turns of the solenoid in the length $L$. We have \begin{equation*} B_0L=\frac{NI}{\epsO c^2}. \end{equation*}  Or, letting $n$ be the number of turns per unit length of the solenoid (that is, $n=N/L$), we get \begin{equation} \label{Eq:II:13:19} B_0=\frac{nI}{\epsO c^2}. \end{equation}  What happens to the lines of $\FLPB$ when they get to the end of the solenoid? Presumably, they spread out in some way and return to enter the solenoid at the other end, as sketched in Fig. 13–9. Such a field is just what is observed outside of a bar magnet. But what is a magnet anyway? Our equations say that $\FLPB$ comes from the presence of currents. Yet we know that ordinary bars of iron (no batteries or generators) also produce magnetic fields. You might expect that there should be some other terms on the right-hand side of (13.12) or (13.13) to represent “the density of magnetic iron” or some such quantity. But there is no such term. Our theory says that the magnetic effects of iron come from some internal currents which are already taken care of by the $\FLPj$ term. Matter is very complex when looked at from a fundamental point of view—as we saw when we tried to understand dielectrics. In order not to interrupt our present discussion, we will wait until later to deal in detail with the interior mechanisms of magnetic materials like iron: You will have to accept, for the moment, that all magnetism is produced from currents, and that in a permanent magnet there are permanent internal currents. In the case of iron, these currents come from electrons spinning around their own axes. Every electron has such a spin, which corresponds to a tiny circulating current. Of course, one electron doesn’t produce much magnetic field, but in an ordinary piece of matter there are billions and billions of electrons. Normally these spin and point every which way, so that there is no net effect. The miracle is that in a very few substances, like iron, a large fraction of the electrons spin with their axes in the same direction—for iron, two electrons of each atom take part in this cooperative motion. In a bar magnet there are large numbers of electrons all spinning in the same direction and, as we will see, their total effect is equivalent to a current circulating on the surface of the bar. (This is quite analogous to what we found for dielectrics—that a uniformly polarized dielectric is equivalent to a distribution of charges on its surface.) It is, therefore, no accident that a bar magnet is equivalent to a solenoid.