$9.2.10.$ A current $I$ flows through a ring of radius $R$. Determine the magnetic field induction at the center of the ring and on its axis at a distance $h$ from the center of the ring.
The magnetic field $B$ will be directed perpendicular to the plane of the ring. Let's consider a section of the ring of length $Rd\theta$ for it the formula from 9.2.8 will be valid then $$ dB = \frac{\mu_0 I R}{4\pi (R^2 + h^2)}\cdot d\theta $$ but since the field will be directed along the axis, its projection onto this axis is needed there $$ dB\cdot \cos\alpha= \frac{\mu_0 I Rd\theta}{4\pi (R^2 + h^2)} \cdot \frac{R}{\sqrt{R^2 + h^2}} $$ $$ dB_z = \frac{\mu_0 I R^2d\theta}{4\pi (R^2 + h^2)^{3/2}} $$ When integrating $dB_z$ from $0$ to $2\pi$ we get: $$ B(h)=\int_0^{2\pi}dB_z=\int_0^{2\pi}\frac{\mu_0 I R^2}{4\pi (R^2 + h^2)^{3/2}}~d\theta $$ $$ B(h) = \frac{\mu_0 I R^2}{2 (R^2 + h^2)^{3/2}} $$ substituting $h = 0$ we get the formula you know for the inductance in the center of the ring: $$ B(0) = \frac{\mu_0 I}{2R} $$
$$ B=\frac{\mu_0 qv}{4\pi r^2} \sin\alpha; \quad B = \frac{\mu_0 Il}{4\pi r^2}\sin\alpha$$