(a) State Biot – Savart law and express this law in the vector form.
(b) Two identical circular coils, P and Q each of radius R, carrying currents 1 A and √3 A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.

(a) Biot-Savart’s law states that the magnetic field at point P is proportional to the length of the element dl, the current flowing through it I and inversely proportional to the square of the distance r. The direction of the field is perpendicular to the plane containing both dl and r. 

The magnetic field along the direction of dl is always zero.
We can derive the formula for the Biot-Savart’s law as –
The above form gives the equation of Biot-Savart’s law in vector form. The magnitude of the magnetic field is given by,
The proportionality constant is the permeability of free space. So, the required vector form of the law is – And the direction of the magnetic field is perpendicular to both dl and r.
$\begin{array}{l}\left|\mathrm{dB}\right|-\underset{\_}{{\mathrm{\mu }}_0}\underset{\_}{\mathrm{Idllsin}\mathrm{\theta }}\\ \mathrm{d}\mathbf{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}{\frac{\mathrm{Id}\mathbf{l}\times \mathbf{r}}{{\mathrm{r}}^3}}_{4\mathrm{\pi }}\frac{{ }^7\mathrm{Tm}}{{\mathrm{r}}^3}/\mathrm{A}\\ \end{array}$
$\begin{array}{l}\overrightarrow{{\mathrm{B}}_{\mathrm{P}}}=\frac{{\mathrm{\mu }}_0\sqrt{3}\mathrm{I}}{2\mathrm{R}}\\ \therefore \overrightarrow{{\mathrm{B}}_{\text{net }}}=\overrightarrow{{\mathrm{B}}_{\mathrm{P}}}+\overrightarrow{{\mathrm{B}}_{\mathrm{Q}}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}+\frac{{\mathrm{\mu }}_0\sqrt{3}\mathrm{I}}{2\mathrm{R}}\\ \overrightarrow{{\mathrm{B}}_{\mathrm{P}}}=\overrightarrow{{\mathrm{B}}_{\mathrm{net}}}=\sqrt{{\left(\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}\right)}^2+{\left(\frac{{\mathrm{\mu }}_0\sqrt{3}\mathrm{I}}{2\mathrm{R}}\right)}^2}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}\times 2=\frac{{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{R}}\\ \therefore \tan \mathrm{\theta }=\frac{\left|\overrightarrow{{\mathrm{B}}_{\mathrm{P}}}\right|}{\left|\overrightarrow{{\mathrm{B}}_{\mathrm{Q}}}\right|}=\left(\frac{\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}}{\frac{{\mathrm{\mu }}_0\sqrt{3}\mathrm{I}}{2\mathrm{R}}}\right)=\frac{1}{\sqrt{3}}\\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^{\circ }\\ \end{array}$
(b) Given that two identical coils are lying in perpendicular planes and having common centre. P and Q carry current I and √3 I respectively. Now, magnetic field at the centre of P due to its current I is:
And, magnetic field at the centre of Q due to its current √3I is