A thin insulated wire forms a plane spiral of $N=100$ tight turns carrying a current $I=8 mA.$ The radii inside and outside turns (Fig.) are equal to $a=50mm$ and $b=100mm$. Find the magnetic induction at the Centre of the spiral.

We consider a circular element of radius $\mathrm{r}$ and thickness $\mathrm{dr}$.

The number of turns in the considered element is

$\mathrm{dN}=\frac{\mathrm{N}}{\mathrm{b}-\mathrm{a}}\mathrm{dr}$

The magnetic field as its Centre due to considered element is 

$\mathrm{dB}=\frac{{\mathrm{dN\mu }}_0\mathrm{I}}{2\mathrm{r}}$

$\mathrm{B}=\frac{{\mathrm{N\mu }}_0\mathrm{I}}{2(\mathrm{b}-\mathrm{a})}{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{\mathrm{dr}}{\mathrm{r}}=\frac{{\mathrm{N\mu }}_0}{2(\mathrm{b}-\mathrm{a})}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}=7\mathrm{\mu T}$