Question

Find expression for coefficient of self induction for solenoid

Answer 1

this is the above expressions

Answer 2

Considering 'R' as the radius and 'L' length of the solenoid , such that R<< L having 'n' number of turns per unit length.

Hence n = (N/L) -----(1) ; N = total numbers of turns.

When current 'I' flows through the coil, than the magnetic field will be given by

B = μ₀ N I ; μ₀ = permeability

Thus magnetic flux in each turn will be given by

Φ =B A = μ₀ n I A

and the total magnetic flux in the solenoid will be

NΦ=( μ₀ n I A) N ------(A)

Also we know NΦ = L I -------(B); here L=  coefficient of self induction

Reducing (A) in (B) we get,

( μ₀ n I A) N = L I

L = ( μ₀ n A) N

L = ( μ₀ I A) N²/L  (From 1)