A long solenoid
Next, we consider again the infinitely long solenoid with a circumferential current on the surface of $nI$ per unit length. (We imagine there are $n$ turns of wire per unit length, carrying the current $I$, and we neglect the slight pitch of the winding.) Just as we have defined a “surface charge density” $\sigma$, we define here a “surface current density” $\FLPJ$ equal to the current per unit length on the surface of the solenoid (which is, of course, just the average $\FLPj$ times the thickness of the thin winding). The magnitude of $\FLPJ$ is, here, $nI$. This surface current (see Fig. 14–4) has the components: \begin{equation*} J_x=-J\sin\phi,\quad J_y=J\cos\phi,\quad J_z=0. \end{equation*}  Now we must find $\FLPA$ for such a current distribution. First, we wish to find $A_x$ for points outside the solenoid. The result is the same as the electrostatic potential outside a cylinder with a surface charge density \begin{equation*} \sigma=\sigma_0\sin\phi, \end{equation*}  with $\sigma_0=-J/c^2$. We have not solved such a charge distribution, but we have done something similar. This charge distribution is equivalent to two solid cylinders of charge, one positive and one negative, with a slight relative displacement of their axes in the $y$-direction. The potential of such a pair of cylinders is proportional to the derivative with respect to $y$ of the potential of a single uniformly charged cylinder. We could work out the constant of proportionality, but let’s not worry about it for the moment. The potential of a cylinder of charge is proportional to $\ln r'$; the potential of the pair is then \begin{equation} \phi\propto\ddp{\ln r'}{y}=\frac{y}{r'^2}.\notag \end{equation}  So we know that \begin{equation} \label{Eq:II:14:25} A_x=-K\,\frac{y}{r'^2}, \end{equation}  where $K$ is some constant. Following the same argument, we would find \begin{equation} \label{Eq:II:14:26} A_y=K\,\frac{x}{r'^2}. \end{equation}  Although we said before that there was no magnetic field outside a solenoid, we find now that there is an $\FLPA$-field which circulates around the $z$-axis, as in Fig. 14–4. The question is: Is its curl zero? Clearly, $B_x$ and $B_y$ are zero, and \begin{align*} B_z&=\ddp{}{x}\biggl(K\,\frac{x}{r'^2}\biggr)- \ddp{}{y}\biggl(-K\,\frac{y}{r'^2}\biggr)\\[1.5ex] &=K\biggl( \frac{1}{r'^2}-\frac{2x^2}{r'^4}+\frac{1}{r'^2}-\frac{2y^2}{r'^4} \biggr)=0. \end{align*}  So the magnetic field outside a very long solenoid is indeed zero, even though the vector potential is not. We can check our result against something else we know: The circulation of the vector potential around the solenoid should be equal to the flux of $\FLPB$ inside the coil (Eq. 14.11). The circulation is $A\cdot2\pi r'$ or, since $A=K/r'$, the circulation is $2\pi K$. Notice that it is independent of $r'$. That is just as it should be if there is no $\FLPB$ outside, because the flux is just the magnitude of $\FLPB$ inside the solenoid times $\pi a^2$. It is the same for all circles of radius $r'>a$. We have found in the last chapter that the field inside is $nI/\epsO c^2$, so we can determine the constant $K$: \begin{equation*} 2\pi K=\pi a^2\,\frac{nI}{\epsO c^2}, \end{equation*}  or \begin{equation*} K=\frac{nIa^2}{2\epsO c^2}. \end{equation*}  So the vector potential outside has the magnitude \begin{equation} \label{Eq:II:14:27} A=\frac{nIa^2}{2\epsO c^2}\,\frac{1}{r'}, \end{equation}  and is always perpendicular to the vector $\FLPr'$. We have been thinking of a solenoidal coil of wire, but we would produce the same fields if we rotated a long cylinder with an electrostatic charge on the surface. If we have a thin cylindrical shell of radius $a$ with a surface charge $\sigma$, rotating the cylinder makes a surface current $J=\sigma v$, where $v=a\omega$ is the velocity of the surface charge. There will then be a magnetic field $B=\sigma a\omega/\epsO c^2$ inside the cylinder. Now we can raise an interesting question. Suppose we put a short piece of wire $W$ perpendicular to the axis of the cylinder, extending from the axis out to the surface, and fastened to the cylinder so that it rotates with it, as in Fig. 14–5. This wire is moving in a magnetic field, so the $\FLPv\times\FLPB$ forces will cause the ends of the wire to be charged (they will charge up until the $\FLPE$-field from the charges just balances the $\FLPv\times\FLPB$ force). If the cylinder has a positive charge, the end of the wire at the axis will have a negative charge. By measuring the charge on the end of the wire, we could measure the speed of rotation of the system. We would have an “angular-velocity meter”! But are you wondering: “What if I put myself in the frame of reference of the rotating cylinder? Then there is just a charged cylinder at rest, and I know that the electrostatic equations say there will be no electric fields inside, so there will be no force pushing charges to the center. So something must be wrong.” But there is nothing wrong. There is no “relativity of rotation.” A rotating system is not an inertial frame, and the laws of physics are different. We must be sure to use equations of electromagnetism only with respect to inertial coordinate systems. It would be nice if we could measure the absolute rotation of the earth with such a charged cylinder, but unfortunately the effect is much too small to observe even with the most delicate instruments now available. 