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# 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

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