A tightly-wound solenoid of radius a and length l has n turns per...

The magnetic field at the centre of a finite solenoid of length $l$ and radius $a$ having $n$ turns per unit length and carrying current $i$ is given by

$B = \dfrac{\mu_0 n i}{2} (\sin \theta_1 + \sin \theta_2)$

where $\theta_1$ and $\theta_2$ are the semi-vertical angles subtended by the ends at the centre.

At the centre of the solenoid, $\theta_1 = \theta_2 = \theta$. From the geometry of the solenoid,

$\sin \theta = \dfrac{\dfrac{l}{2}}{\sqrt{a^2 + \left(\dfrac{l}{2}\right)^2}} = \dfrac{l}{\sqrt{4a^2 + l^2}}$

Substituting this into the expression for $B$,

$B = \dfrac{\mu_0 n i}{2} \left( \dfrac{2l}{\sqrt{4a^2 + l^2}} \right) = \dfrac{\mu_0 n i l}{\sqrt{4a^2 + l^2}}$

In the limit $l \gg a$, the term $4a^2$ can be neglected compared to $l^2$.

$B \rightarrow \dfrac{\mu_0 n i l}{\sqrt{l^2}} = \mu_0 n i$

In the limit $a \gg l$, the term $l^2$ can be neglected compared to $4a^2$.

$B \rightarrow \dfrac{\mu_0 n i l}{\sqrt{4a^2}} = \dfrac{\mu_0 n i l}{2a}$

The magnetic field tends to $\mu_0 n i$ and $\dfrac{\mu_0 n i l}{2a}$ respectively.