a long solenoid with closely wound turn has n turns per unit of its length.a steady current/flow through this solenoid.use ampere circuital law to obtain an expression for the magnetic field at point on its axis and closed to its midpoint.

Let us suppose a rectangular amperian loop PQRS in the middle of the solenoid, PQ = L.
Let magnetic field along PQ is B and Along RS is zero. Path along QR and SP is perpendicular to axis of solenoid, therefore magnetic field component along these path is zero.
Therefore Path QR and SP will not contribute to the line integral of magnetic fied B.
L is length of the solenoid and n be the no. of turns per unit length of the solenoid.

The line integral of the magnetic field B over the closed path PQRS is:


$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dl} = {\int }_P^Q\overrightarrow{B}.\overrightarrow{dl} + {\int }_Q^R\overrightarrow{B}.\overrightarrow{dl} + {\int }_R^S\overrightarrow{B}.\overrightarrow{dl} + {\int }_S^P\overrightarrow{B}.\overrightarrow{dl} \\ Now \\ {\int }_P^Q\overrightarrow{B}.\overrightarrow{dl } {\int }_P^{Q}Bdl \cos 0 = BL \\ {\int }_Q^R\overrightarrow{B}.\overrightarrow{dl} = {\int }_Q^{R}Bdl \cos {90}^o = 0 \\ similarly {\int }_S^P\overrightarrow{B}.\overrightarrow{dl} = 0 \\ {\int }_S^R\overrightarrow{B}.\overrightarrow{dl} = 0 because no field outside the solenoid. \\ therefore \\ \oint \overrightarrow{B}.\overrightarrow{dl} = BL \\ From Ampere\text{'}s law: \\ \oint \overrightarrow{B}.\overrightarrow{dl} = {\mu }_o\times total current through the path PQRS. \\ \oint \overrightarrow{B}.\overrightarrow{dl} = {\mu }_o\times total no of turns \times current \\ BL = {\mu }_{0}nLI \\ B = {\mu }_{0}nI \end{gathered}$$

This is the magnetic field inside the solenoid near the mid point.