Consider a solenoid of length \(l\) and area of cross-section A with fixed number of turns. The self-inductance of the s…

The self inductance L of a solenoid of length I and area of cross-section A, with a fixed number of turns N increases as \(\underline{\mathbf{l}}\) decreases and A increases.

Explanation:
The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.

\(\mathrm{L}=\mu_{\mathrm{r}} \mu_0 \mathrm{~N}^2 \mathrm{Al}\)

Where \(\mathrm{n}=\mathrm{N} / \mathrm{l}\) (no. of turns per unit length)
1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross-section: Larger the area of cross-section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases \(\mu \mathrm{r}\) times if it is wound over an iron core of relative permeability \(\mu_r\).

The long solenoid of cross-sectional area A and length 1 , having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is \(L=\mu_r \mu_0 N^2 A / l\)

So, the self inductance \(L\)of a solenoid increases as 1 decreases and \(A\) increases because \(L\) is directly proportional to the area and inversely proportional to length.

Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.