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Self Inductance of a Long Solenoid

Consider a long air-cored solenoid of length l, total number of turns N, and area of cross-section A. The number of turns per unit length is N/l. Let a current I flow through it. Then the magnetic field inside the solenoid

            B = μ0 (N/l) l
. : Flux through each turn = ΙΈ B = BA = μ0 N/A / l

Total flux through N turns
Φ = N Φ B = N x μ0 NIA / l = μ0 N2IA / l

. : self-inductance of the solenoid} = L = Φ /I = μ0 N2/A / l

If the solenoid is wound on a core of constant of constantly permeability μ, then
        L = μ0 N2 A / l
Where        μ = μ0 μr

In general, if there is a core consisting of a number of media of relative permeability, μr1, μr2,… etc., and areas of cross-section, A1, A2, A3….. etc., then,
        L = μ0 N2 / l [μr1A1 + μr2A2 + μr3 A3 + …]

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