Derive expression for the self-inductance of a solenoid. What factors affect it?

Hint:
- You should know Self-inductance.
- You should be aware of vector calculus.
- You should know the basic equations of flux, self-inductance…

Complete step by step solution:
Self-inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In other words inductance of the coil is defined as the property of the coil due to which it opposes the change of current flowing through it In the case of self-inductance; the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit.
Self-Inductance of a Current Carrying Solenoid:
Consider a solenoid of radius r metre, length l metre, having total number of turns N and carrying a current I ampere. The magnetic field produced on its axis inside the solenoid is
\[B = \dfrac{{{\mu _0}NI}}{l}N{A^{ - 1}}{m^{ - 1}}\]
If it is assumed that the solenoid is very long, the intensity of magnetic field B can be assumed to be uniform inside it at each point, then the total magnetic flux linked with the solenoid is
\[B = \dfrac{{{\mu _0}m}}{{4\pi }}\left[ {\dfrac{{{d^2} + {l^2} + 2dl - \left( {{d^2} + {l^2} + 2dl} \right)}}{{{{\left\{ {\left( {d - l} \right)\left( {d + l} \right)} \right\}}^2}}}} \right]\]
\[\phi = B \times \]total effective area of solenoid
\[\phi = B \times \left( {NA} \right)\]
where \[A = \pi {r^2}\]is the area of the cross section of solenoid.
\[\phi = \dfrac{{{\mu _0}NI}}{l} \times NA\]
\[\therefore \phi = \dfrac{{{\mu _0}{N^2}AI}}{l}weber\]
But by definition \[\phi = LI\]where L is the self-inductance of solenoid,
Therefore
Self-inductance, L
\[L = \dfrac{\phi }{I}\]
\[L = \dfrac{{{\mu _0}{N^2}A}}{l}henry\]
If inside the solenoid instead of air or vacuum there is a substance of relative permeability
\[L = \dfrac{{{\mu _0}{\mu _r}{N^2}A}}{l}henry\]
Thus, the self-inductance of a solenoid depends on the following factors:

1. On the number of turns in the solenoid: The self-inductance increases on increasing the number of turns.

2. On the area of the cross section of solenoid \[\left( {i.e.,{\text{ }}on{\text{ }}the{\text{ }}radius{\text{ }}of{\text{ }}solenoid} \right)\]: The self-inductance increases on increasing the area of cross section.

3. On the length of solenoid: On increasing the length of solenoid, its self-inductance decreases.

4. On the relative permeability of the substance placed inside the solenoid: If a soft iron rod is placed inside the solenoid, its self-inductance increases.

Remember that if the number of turns per unit length in the solenoid is n, then \[n = N/l\;\]and the self-inductance of solenoid is
\[L = {\mu _0}{n^2}lA\] henry

Note:
- Memorize basic units of magnetism.
- All units should be in SI units.
- Should have a good conceptual clarity in self-inductance.