Question 5.12: Find the vector potential of an infinite solenoid with n tur...

This time we cannot use Eq. 5.66, since the current itself extends to infinity. But here’s a cute method that does the job. Notice that

A=\frac{\mu _{0}}{4\pi}\int{\frac{\pmb{I}}{\eta }}d\acute{l}= \frac{\mu _{0}I}{4\pi}\int{\frac{1}{\eta }}d\acute{\pmb{I}};A=\frac{\mu _{0}}{4\pi}\int{\frac{\pmb{K}}{\eta }}d\acute{a}      (5.66)

\oint{\pmb{A}.d\pmb{I}}=\int{(∇ × A) ·da}=\int{B .da}=\phi,       (5.71)

where Φ is the flux of B through the loop in question. This is reminiscent of
Ampère’s law in integral form (Eq. 5.57),

\oint{\pmb{B} · d\pmb{I} }= μ_0 I_{enc}.     (5.57)

In fact, it’s the same equation, with B → A and μ0 Ienc → Φ. If symmetry permits, we can determine A from Φ in the same way we got B from I_{enc}, in Sect. 5.3.3. The present problem (with a uniform longitudinal magnetic field μ0nI inside the solenoid and no field outside) is analogous to the Ampère’s law problem of a fat wire carrying a uniformly distributed current. The vector potential is “circumferential” (it mimics the magnetic field in the analog); using a circular “Amperian loop” at radius s inside the solenoid, we have

so

A=\frac{\mu _{0}nI}{2}s\hat{\phi },        for       s\leq R.              (5.72)

For an Amperian loop outside the solenoid, the flux is

since the field only extends out to R. Thus

A=\frac{\mu _{0}nI}{2}\frac{R^2}{s}\hat{\phi },        for       s\geq R.              (5.73)

If you have any doubts about this answer, check it: Does ∇ × A = B? Does ∇ · A = 0? If so, we’re done.