Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMOVING CHARGES AND MAGNETISM2 Marks
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.
→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. $\begin{aligned} \overrightarrow{ B } & =\int d \overrightarrow{ B } \\ \overrightarrow{ B } & =\frac{\mu_0}{4 \pi} \int \frac{ I d \vec{l} \times \hat{r}}{r^2} \end{aligned}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.