The magnetic field at the centre of a circular coil of radius r is $\pi$ times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases; in all cases the circular part has radius r and straight ones are infinitely long. For same current, the B field at the
centre P in cases 1, 2, 3 have the ratio

Case 1:

$\begin{array}{l}{\mathrm{B}}_{\mathrm{A}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{\mathrm{i}}{\mathrm{r}}\otimes \\ {\mathrm{B}}_{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{\mathrm{\pi i}}{\mathrm{r}}\odot \\ {\mathrm{B}}_{\mathrm{C}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{\mathrm{i}}{\mathrm{r}}\odot \end{array}$

So, net magnetic field at the centre of case 1,

${\mathrm{B}}_1={\mathrm{B}}_{\mathrm{B}}-{\mathrm{B}}_{\mathrm{C}}-{\mathrm{B}}_{\mathrm{A}}\Rightarrow {\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{\mathrm{\pi i}}{\mathrm{r}}\odot ...\left(\mathrm{i}\right)$

Case 2: Magnetic field at the centre O in this case

${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{\mathrm{\pi i}}{\mathrm{r}}\otimes ...\left(\mathrm{ii}\right)$

Case 3: B_A = 0

$\begin{array}{l}{\text{B}}_{\mathrm{B}}=\frac{{\text{μ}}_0}{4\text{π}}.\frac{\left(2\mathrm{\pi }-\mathrm{\pi }/2\right)\mathrm{i}}{\text{r}}\otimes =\frac{{\text{μ}}_0}{4\text{π}}.\frac{3\mathrm{\pi i}}{2\mathrm{r}}\otimes \\ {\text{B}}_{\mathrm{C}}=\frac{{\text{μ}}_0}{4\text{π}}.\frac{\mathrm{i}}{\mathrm{r}}\odot \end{array}$

So, net magnetic field at the centre of case 3,

${\text{B}}_3=\frac{{\text{μ}}_0}{4\text{π}}.\frac{\mathrm{i}}{\mathrm{r}}\left(\frac{3\mathrm{\pi }}{2}-1\right)\otimes .....\left(\mathrm{iii}\right)$

From equations (i), (ii) and (iii),

$\begin{array}{l}{\mathrm{B}}_1:{\mathrm{B}}_2:{\mathrm{B}}_3=\mathrm{\pi }\odot :\mathrm{\pi }\otimes :\left(\frac{3\mathrm{\pi }}{2}-1\right)\otimes \\ =-\frac{\mathrm{\pi }}{2}:\frac{\mathrm{\pi }}{2}:\left(\frac{3\mathrm{\pi }}{4}-\frac{1}{2}\right)\end{array}$