### Induction

Let us consider the contour going along the surface of one cylinder closing at the edge, going back along the surface of the other cylinder, and closing at the other edge.
The magnetic field crossing the area of this contour is $B=2KI/r$,
so that the flux is $\Phi=l\int_a^b (2KI/r) dr=2KIl\ln(b/a)$.
The inductance, therefore, is $L=\Phi/I=2Kl\ln(b/a)$.