State and explain Biot-Savart law for the magnetic field produced by a current element. Give the direction of magnetic field and define the unit of it

Biot-Savart Law : Magnetic field of position vector of point relative to element "I $d \vec{l}$ which is current carrying element at $\vec{r}$ position vector is given by $d \overrightarrow{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\mathrm{I} d \vec{l} \times \vec{r}}{r^{2}}$.

According to Biot-Savart law, the magnitude of the field is $d \overrightarrow{\mathrm{B}}$.

$(1)$Directly proportional to the current I through the conductor,

$\therefore d \mathrm{~B} \propto \mathrm{I}$

$(2)$ Directly proportional to the length $|d \vec{l}|$ of the current element,

$\therefore d \mathrm{~B} \propto d \mathrm{I}$

$(3)$ Directly proportional to $\sin \theta$,

$\therefore d \mathrm{~B} \propto \sin \theta$

$(4)$ Inversely proportional to the square of the distance $r$ of the point P from the current element,

$\therefore d \mathrm{~B} \propto \frac{1}{r^{2}}$

Combining all these four factors, we get

$d \mathrm{~B} \propto \frac{\mathrm{I} d l \sin \theta}{r^{2}}$

$d \mathrm{~B} \propto \frac{\mathrm{I} d l r \sin \theta}{r^{3}}$

$d \overrightarrow{\mathrm{B}} \propto \frac{\mathrm{I} d \vec{l} \times \vec{r}}{r^{3}}$