An annular wire loop ABCD carries a current I as shown in Fig. O is the common centre of the curved parts AD and BC. If OA = r and OB  = 2r, The magnitude of the magnetic field at O is:

 

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2 \mathrm{R}}\times \frac{\mathrm{\theta }}{2\mathrm{\pi }} \left[\because \mathrm{\theta } = 60° = \frac{\mathrm{\pi }}{3}\right] \\ \Rightarrow \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{12\mathrm{R}} \end{gathered}$$

The magnetic field at O due to straight segments AB and CD is zero.

Magnetic field at O due to current I in AD is:

${\mathrm{B}}_{\mathrm{AD}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{12 \mathrm{r}}$  directed out of the page.

Magnetic field at O due to current I in BC is:

${\mathrm{B}}_{\mathrm{BC}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{12\left(2\mathrm{r}\right)}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{24\mathrm{r}}$  directed into the page.

Therefore, Magnetic field at O due to ABCD is:

$$\begin{gathered} \mathrm{B} = {\mathrm{B}}_{\mathrm{AD}}-{\mathrm{B}}_{\mathrm{BC}} \\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{12\mathrm{r}}-\frac{{\mathrm{\mu }}_0\mathrm{I}}{24\mathrm{r}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{24\mathrm{r}} \end{gathered}$$ 

directed out of the page.