Two insulated wires are wound on the same hollow cylinder, so as to form two solenoids sharing a common air-filled core.Let `I` be the length of the core. `A` the cross-sectional area of the area of the core `N_(1)` the number of times the first wire is wound around the core and `N_(2)` the number of turns the second wire is wound around the cora.Find the mutual inductance of the two solenoids neglecting the end effects.

Explanation

If a current `I_(1)` flows around the first wire then a uniform aaxial magnetic field of strength `B_(1)=(mu_(0)N_(1)I_(1))/(l)` is generated in the core. The magnetic field in the region outsides the core is of neglifible magnitude. The flux linking a single turn of the second wire is `B_(1)A`. Thus, the flux linking all `N_(2) ` turns of the second wire is
`phi_(2)=N_(2)B_(1)A= (mu_(0)N_(1)N_(2)I_(1))/(l)=MI_(1) " " :. M=(mu_(0)N_(1)N_(2)A)/(l)`
As described previously, M iks a geometric quantity dependng on the dimensions of the core and the manner in which the two wires are wound around the core, but not on the actual currents flowing through the wires.