Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.


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Magnetic field inside the solenoid

B =μ0nI
n → number of turns per unit length, I → current flowing,
ϕ' = Magnetic field through each turn = BA = μ0nIA
Total magnetic flux linked with solenoid
ϕ = Nϕ' = nl x μ0nIA =μ0n2AlI LI =>L =μ0n2Al 
L → self inductance of solenoid and l → length of solenoid.