Two mutually perpendicular concentric current carrying coils have the same radius $2\pi \mathrm{cm}\text{. }$ If currents 3 A and 4 A flow in each coil respectively, then calculate the magnitude of magnetic induction (in Wb/m²) at the common centre of the two coils.

$\left({\mu }_0=4\pi \times {10}^{-7}{\mathrm{WbAm}}^{-1}\right)$

The magnetic field at the centre of a circular of a circular loop is given by ;

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{r}}$

Thus,

${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1}{2\mathrm{r}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{i}}_1}{2\left(2\mathrm{\pi }\times {10}^{-2}\right)}$$=\frac{{\mathrm{\mu }}_{\mathrm{o}}\times 3\times {10}^2}{4\mathrm{\pi }}$

 ${\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_2}{2\mathrm{r}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}{\mathrm{i}}_2}{2\left(2\mathrm{\pi }\times {10}^{-2}\right)}$ $=\frac{{\mathrm{\mu }}_{\mathrm{o}}\times 4\times {10}^2}{4\mathrm{\pi }}$

Since B₁ and B₂ are perpendicular to each other, resultant magnetic field is :

${\mathrm{B}}_{\mathrm{net}}=\sqrt{{\mathrm{B}}_1^2+{\mathrm{B}}_2^2}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}5\times {10}^2$

${\mathrm{B}}_{\text{net }}={10}^{-7}\times 5\times {10}^2$

${\mathrm{B}}_{\mathrm{net}}=5\times {10}^{-5}\mathrm{Wb}/{\mathrm{m}}^2$