9.4: Long Solenoid

Let us place an infinitely long solenoid of $n$ turns per unit length so that its axis coincides with the $z$-axis of coordinates, and the current $I$ flows in the sense of increasing $\phi$. In that case, we already know that the field inside the solenoid is uniform and is $\mu\, n\, I\, \hat{\textbf{z}}$ inside the solenoid and zero outside. Since the field has only a $z$ component, the vector potential $\textbf{A}$ can have only a $\phi$- component.

We'll suppose that the radius of the solenoid is $a$. Now consider a circle of radius $r$ (less than $a$) perpendicular to the axis of the solenoid (and hence to the field $\textbf{B}$). The magnetic flux through this circle (i.e. the surface integral of $\textbf{B}$ across the circle) is $\pi r^2B = \pi r^2 nI$. Now, as everybody knows, the surface integral of a vector field across a closed curve is equal to the line integral of its curl around the curve, and this is equal to $2\pi r A_\phi$. Thus, inside the solenoid the vector potential is

$$\textbf{A}=\frac{1}{2}\mu n r I \hat{\boldsymbol{\phi}}.\label{9.4.1}$$

It is left to the reader to argue that, outside the solenoid $(r > a)$, the magnetic vector potential is

$$\textbf{A}=\frac{\mu na^2 I}{2r}\hat{\boldsymbol{\phi}}.$$