A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I pass through the coil, the magnetic field at the centre is

Number of turns per unit width $\mathrm{h}=\frac{N}{b-a}$

Consider an elemental ring of radius x and with thickness dx Number of turns in the ring $=dN=\frac{Ndx}{b-a}$

Magnetic field at the centre due to the ring element

$dB=\frac{{\mu }_0\left(dN\right)i}{2x}=\frac{{\mu }_{0}i}{2}\cdot \frac{Ndx}{(b-a)}\cdot \frac{1}{x}$

$\therefore$ Field at the centre

$=\int dB=\frac{{\mu }_{0}Ni}{2(b-a)}{\int }_a^b \frac{dx}{x}$

$=\frac{{\mu }_{0}Ni}{2(b-a)}\ln \frac{b}{a}$