Derive an Expression For Biot Savart's Law in Vector form - Param Himalaya

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.

$$\overrightarrow{B}= \int\overrightarrow{dB} = \frac{\mu_{0}}{4\pi}\int_{}^{} \frac{I (\overrightarrow{dl} \times \overrightarrow{r})}{r^{3}}$$