Biot-Savart's law is used to determine the strength of the magnetic field at any point due to a current carrying conductor.
Consider a very small element AB of length dl of a conductor carrying current I. The strength of magnetic field dB due to this small current element $(I\overrightarrow{dl})$ at point P, distant r from the element is found to depend upon the quantities as under:
$$(i) dB \propto dl$$
$$(ii) dB \propto I$$
$$(iii) dB \propto sin\theta$$
where $\theta$ is the angle between $\overrightarrow{dl}$ and $\overrightarrow{r}$.
$$(iv) dB \propto \frac{1}{r^{2}}$$
Combining (i) to (iv) , we get
$$dB \propto \frac{Idlsin\theta}{r^{2}}$$
$$dB = k \frac{Idl sin\theta}{r^{2}}$$
Where k is a constant of proportionality.
In S.I units, $$k = \frac{\mu_{0}}{4\pi}$$
where , $\mu_{0}$ is called absoulte permeability of free space i.e . vacuum.
Hence , equ (i) becomes
$$dB = \frac{\mu_{0}}{4\pi}. \frac{Idlsin\theta}{r^{2}}$$
Value of $\mu_{0}$ in S.I units = $4\pi \times 10^{-7} Tm A^{-1}$ or $Wb m^{-1} A^{-1}$
$$\frac{\mu_{0}}{4\pi} = 10^{-7} TmA^{-1}$$
Biot-Savart's Law in Vector form :
If we consider length of element as $\overrightarrow{dl}$ , distance of point P as displacement vector $\overrightarrow{r}$ and unit vector along CP as $\hat{r}$ , then
$$\overrightarrow{dB} = \frac{\mu_{0}}{4\pi}. \frac{I\overrightarrow{dl}\times \hat{r}}{r^{2}}$$
now
$$\overrightarrow{r} = \left| \overrightarrow{r} \right|\hat{r}$$
$$\hat{r} = \frac{\overrightarrow{r}}{\left| \overrightarrow{r} \right|} = \frac{\overrightarrow{r}}{r}$$
$$\overrightarrow{dB} = \frac{\mu_{0}}{4\pi}. \frac{I\overrightarrow{dl}\times \overrightarrow{r}}{r^{3}}$$
Magnitude of the Strength of the magnetic field is given by :
$$\left| \overrightarrow{dB} \right| = \frac{\mu_{0}}{4\pi}. \frac{\left|I\overrightarrow{dl}\times \overrightarrow{r} \right|}{r^{3}}$$
Direction of $\overrightarrow{dB}$ is same as that of direction of $\overrightarrow{dl}\times \overrightarrow{r}$ which can be determinded by using Rigth Handed Screw Rule for the vector - product.
Therefore , at point P , direction of $\overrightarrow{dB}$ is perpendicular to the plane containing $\overrightarrow{dl}$ and $\overrightarrow{r}$ and is directed into plane of paper and represented by $\otimes$.
The resultant magnetic field at point P due to the whole conductor can be obtained by intergrating equ (iv) over the whole conductor.