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

Explanation

If a current `l_(1)` flows around the first wire then a unifrom axial magnetic field of strength `B_(1)=(mu_(0)N_(1)l_(1))/l` is generated in the core.The magnetic field in the region outside the core is of negligible 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)AI_(1))/l=MI_(1)`
`therefore M=(mu_(0)N_(1)N_(2)A)/l`
As described prevoiusly, `M` is a geometric quantity depending 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