A single -layer coil (solenoid) has length l and cross-section radius R. Number of turns per unit length is equal to n. Find the magnetic induction at the centre of the coil when a current I flows through it.                           

We consider a circular coil of radius R and thickness dx, the number of turns in the considered element is $\text{dn}=\text{n}.\text{d}x$

The magnetic field at point P due to considered element is

$\text{dB}=\frac{\text{dn}{\mu }_0{\text{IR}}^{\text{2}}}{2\left({\text{R}}^2+x^2\right)}=\frac{{\mu }_{0}I{\text{R}}^{\text{2}}\text{nd}x}{2{\left({\text{R}}^{\text{2}}+x^2\right)}^{\text{3/2}}}$

Its direction is along the axis of solenoid

$\cot \theta =\frac{x}{\text{R}}\Rightarrow x=\text{R}\cot \theta$

$\Rightarrow \text{d}x=-{\text{Rcosec}}^2\theta .\text{d}\theta$

$\text{dB}=-\frac{1}{2}{\mu }_0\text{n}I\sin \theta .\text{d}\theta$

The magnetic field at point P due to entire solenoid

$\text{B}=\frac{1}{2}{\mu }_0\text{n}I{\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}\left(-\sin \theta \right)\text{d}\theta }$

$=\frac{1}{2}{\mu }_0\text{n}I\left(\cos {\theta }_2-\cos {\theta }_1\right)$

It point P is at the centre of the coil

$\cos {\theta }_1=\frac{l}{2\sqrt{{\text{R}}^{\text{2}}+\frac{l^2}{4}}}=\frac{l}{\sqrt{4{\text{R}}^{\text{2}}+l^2}}$

Similarly$\cos \left(180°-{\theta }_2\right)=\frac{l}{\sqrt{4{\text{R}}^2-l^2}}$

$\cos {\theta }_2=\frac{-l}{\sqrt{4{\text{R}}^{\text{2}}+l^2}}$

$\text{B}=\frac{{\mu }_0\text{n}I}{\sqrt{1+{\left(\frac{I\text{R}}{l}\right)}^2}}$