Step 1:Self-inductance is defined as the property of a coil (or an electrical circuit) by which a changing current in the coil induces an electromotive force (emf) in the same coil itself. This induced emf opposes the change in current according to Lenz's law. The self-inductance \( L \) quantifies the ratio of the induced emf to the rate of change of current:
\[
\mathcal{E} = -L \frac{dI}{dt}.
\]
Step 2: Consider a solenoid of length \( l \), cross-sectional area \( A \), and \( N \) turns:
The magnetic field inside a long solenoid carrying current \( I \) is uniform and given by:
\[
B = \mu_0 \frac{N}{l} I,
\]
where \( \mu_0 \) is the permeability of free space.
The magnetic flux \( \phi \) through one turn of the solenoid is:
\[
\phi = B \times A = \mu_0 \frac{N}{l} I \times A.
\]
Since the solenoid has \( N \) turns, the total magnetic flux linked with the coil is:
\[
\Phi = N \phi = N \times \left( \mu_0 \frac{N}{l} I A \right) = \mu_0 \frac{N^2 A}{l} I.
\]
Step 3: By definition, the self-inductance \( L \) is the ratio of the total magnetic flux linked to the current:
\[
L = \frac{\Phi}{I} = \mu_0 \frac{N^2 A}{l}.
\]
This expression shows that the self-inductance of a solenoid depends on the number of turns squared, the cross-sectional area, and inversely on its length.
