Write down the formula for the self-inductance of a very long solenoid.

(N/A) For a very long solenoid of length $l$,cross-sectional area $A$,and number of turns $N$,the magnetic field $B$ inside the solenoid is given by $B = \mu_0 n I$,where $n = N/l$ is the number of turns per unit length and $I$ is the current.
The magnetic flux $\phi$ through each turn is $\phi = B \cdot A = \mu_0 (N/l) I A$.
The total magnetic flux linkage is $N\phi = N \cdot \mu_0 (N/l) I A = (\mu_0 N^2 A / l) I$.
Since the total flux linkage is equal to $LI$,where $L$ is the self-inductance,we have $L = \frac{\mu_0 N^2 A}{l}$.