As mentioned in Exercise 9.15, the magnetic field inside a discharging capacitor can be calculated by summing the contributions from all elements of conduction current. This calculation is extremely tedious. We can, however, get a handle on the contribution from the plates' conduction current in a much easier way that doesn't involve a nasty integral. If we make the usual assumption that the distance s between the plates is small compared with their radius b, then any point P inside the capacitor is close enough to the plates so that they look essentially like infinite planes, with a surface current density equal to the density at the nearest point.
(a) Determine the current that crosses a circle of radius r in the capacitor plates, and then use this to find the surface current density. Hint: The charge on each plate is essentially uniformly distributed at all times.
(b) Combine the field contributions from the wires and the plates to show that the field at a point P inside the capacitor, a distance r from the axis of symmetry, equals B=μ0Ir/2πb2. (Assume s≪r, so that you can approximate the two wires as a complete infinite wire.)