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 = ɸ 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, μ_r1, μ_r2,… etc., and areas of cross-section, A₁, A₂, A₃….. etc., then,
        L = μ₀ N₂ / l [μ_r1A₁ + μ_r2A₂ + μ_r3 A₃ + …]


For more help in Secondary Cells click the button below to submit your homework assignment