# 9.4: Long Solenoid

- Page ID
- 5469

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

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