Question

Derive expression for self-inductance of a long air-cored solenoid of length l, cross-sectional area A and having number of turns N.

Answer 1

Self-Inductance of a long air-cored solenoid:

Consider a long air solenoid having ‘n’ number of turns per unit length. If current in solenoid is I, then magnetic field within the solenoid, B = µ₀ nl ...(i) 

Where µ₀ = 4π x 10^–7 henry/metre is the permeability of free space.

If A is cross-sectional area of solenoid, then effective flux linked with solenoid of length l′ Φ = NBA where N = nl is the number of turns in length ‘l’ of solenoid. 

∴ Φ = (nl BA) 

Substituting the value of B from (i)

∴ Φ = (µ0 nI) A = µ0 n² Al...(ii) 

∴ Self-inductance of air solenoid

L = \(\frac{Φ}{I}\)= µ₀ n² Al  ....(ii)

If N is total number of turns in length I then

n = \(\frac{N}{I}\)

∴ Self - inductance L = µ₀ \(\Big(\frac{N}{I}\Big)^2\)  AI ....(iv)

Remark : If solenoid contains a core of ferromagnetic substance of relative Permeability µ_r then self inductance L = \(\frac{\mu_r\mu_0N^2A}{l}\)