We can assume that the interior magnetic field is uniform and given by equation 30.17 where n = n/l is the number of turns per unit length. b = mu_0ni = mu_0 n/l i the magnetic flux through each turn is given below, where a is the cross-sectional area of the solenoid. phi b = ba = mu_0 na/l i using this expression and equation 32.2, we find the following, which shows that l depends on the geometry of the solenoid and is proportional to the square of the number of turns. (use mu_0 for mu_0, lowercase l for l, and n and a as necessary.) l = n phi_b/i = because n = nl, we can also express the result in the following form, where v = al is the volume of the solenoid. l = mu_0(nl)^2/l a = mu_0n^2 al = mu_0n^2v what if? what would happen to the inductance if you inserted a ferromagnetic material inside the solenoid? answer the inductance would increase. for a given current, the magnetic flux in the solenoid is much greater because of the increase in the magnetic field originating from the magnetization of the ferromagnetic material. for example, if the material has a magnetic permeability of 500 mu_0, the inductance increases by a factor of 500.