Question

define self-inductance and give its SI unit. Derive expression for self inductance of a long air-cored solenoid of length l,radius r and having N turns.

Answer 1

The phenomenon in which changing the current in a coil an opposing emf is indused is called self induction 

S.I. unit is Henry

L=NΦ/i

Φ for a solenoid is μniA where n is num of turns per unit length n=N/L
L=N×N/L×μiA/i=N²μA/L

Answer 2

Hey !!

FIRST PART OF YOUR QUESTION !!

When the current in a coil is changed, an induced emf is produced in the same coil. This phenomenon is called seld-induction.

It's SI unit is ''HENRY''.

SECOND PART OF YOUR QUESTION !!

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 = μ₀nl   --------> (1)

where μ₀ = 4π × 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 of length 'l' of solenoid

∴      Ф = (nl BA)

 Substituting the value of B from (i)

 ∴ Ф = (μ₀ nl) A = μ₀n²Al ------> (ii)

Self inductance of air solenoid

               L = Φ / I = μ₀n²Al --------> (iii)

If N is total number of turns in length l then

        n = N / l

∴ Self - inductance L = μ₀ (N/l)² Al

       = μ₀N²A / l --------------> (iv)

NOTE :- If solenoid contains a core of ferromagnetic substance of relative permeability μ, then

     Self conductance, L = μrμ₀N²A / l

Good luck !!