Question
Explain Biot-Savart Law.

Answer

Law : "The intensity of magnetic field due to an electric current element I $\overrightarrow{d l}$ at a point having position vector $\vec{r}$ with respect to the electric current element is given by the formula, $d \overrightarrow{ B }=\frac{\mu_0}{4 \pi}=\frac{ I \overrightarrow{d l} \times \hat{ r }}{r^2}, \because$
→As shown in the figure, a finite conductor XY carries current I.


Image

→A point P is located at some distance from the conductor. We want to find magnetic field at point P . For this, consider the conductor as divided in to an small elements and choose one small current element $I d \vec{l}$ from conductor.
→$\vec{r}$ is the position vector of point P from this current element $I d \vec{l}$. The angle between $I d \vec{l}$ and the position vector $\vec{r}$ is $\theta$.
→According to Biot-Savart's law, the magnitude of magnetic field $d \overrightarrow{ B }$ is proportional to the current I , the element length $\mid \vec{d} \vec{l}$ and inversely proportional to the square of the distance $r$. It's direction is perpendicular to the plane of $d \vec{l}$ and $\vec{r}$.
→The direction of the magnetic field can be found using the right hand screw rule.
→Vector form,
$\begin{array}{l}
\text { Vector form, } \\
\qquad d \overrightarrow{ B } \propto \frac{ I d \vec{l} \times \vec{r}}{r^3} \\
\therefore \quad d\overrightarrow{ B }=\frac{\mu_0}{4 \pi} \cdot \frac{ I d \vec{l} \times \vec{r}}{r^3}
\end{array}$
Where, $\frac{\mu_0}{4 \pi}$ is a constant of proportionality and its SI unit has the exact value,
$\frac{\mu_0}{4 \pi}=10^{-7} \frac{ Tm }{ A }$
$\mu_0=$ the permeability of free space.
Equation (1) holds when the medium is only vacuum.
→The magnitude of this field is :
$\begin{aligned}
d B & =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l r \sin \theta}{r^3} \\
\therefore \quad d B & =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l \sin \theta}{r^2}
\end{aligned}$
→To obtain the total magnetic field at point $P$ due to entire wire, equation (1) has to be integrated.
$\begin{aligned}
\overrightarrow{ B } & =\int d \overrightarrow{ B } \\
\overrightarrow{ B } & =\frac{\mu_0}{4 \pi} \int \frac{ I d \vec{l} \times \hat{r}}{r^2}
\end{aligned}$

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