Question
Explain Biot$-$Savart Law.
✓
Answer
$\rightarrow$ 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, $d \overrightarrow{ B }=\frac{\mu_0}{4 \pi}=\frac{ I \overrightarrow{d l} \times \hat{ r }}{r^2},$
$\rightarrow$ As shown in the figure, a finite conductor $XY$ carries current $I$.
$\rightarrow A$ 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.
$\rightarrow$ $\vec{r}$ 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 $\theta$.
$\rightarrow$ 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 $\vec{r}$.
$\rightarrow$ The direction of the magnetic field can be found using the right hand screw rule.
$\rightarrow$ Vector form,
Vector form,
$\therefore d \overrightarrow{ B } \propto \frac{ I d \vec{l} \times \vec{r}}{r^3}$
$\therefore d\overrightarrow{ B }=\frac{\mu_0}{4 \pi} \cdot \frac{ I d \vec{l} \times \vec{r}}{r^3}$
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.
$\rightarrow$ The magnitude of this field is :
$d B =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l r \sin \theta}{r^3}$
$\therefore d B =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l \sin \theta}{r^2}$
$\rightarrow$ To obtain the total magnetic field at point $P$ due to entire wire, equation $(1)$ has to be integrated.
$\overrightarrow{ B } =\int d \overrightarrow{ B }$
$\overrightarrow{ B } =\frac{\mu_0}{4 \pi} \int \frac{ I d \vec{l} \times \hat{r}}{r^2}$
$\rightarrow$ As shown in the figure, a finite conductor $XY$ carries current $I$.
$\rightarrow A$ 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.
$\rightarrow$ $\vec{r}$ 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 $\theta$.
$\rightarrow$ 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 $\vec{r}$.
$\rightarrow$ The direction of the magnetic field can be found using the right hand screw rule.
$\rightarrow$ Vector form,
Vector form,
$\therefore d \overrightarrow{ B } \propto \frac{ I d \vec{l} \times \vec{r}}{r^3}$
$\therefore d\overrightarrow{ B }=\frac{\mu_0}{4 \pi} \cdot \frac{ I d \vec{l} \times \vec{r}}{r^3}$
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.
$\rightarrow$ The magnitude of this field is :
$d B =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l r \sin \theta}{r^3}$
$\therefore d B =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l \sin \theta}{r^2}$
$\rightarrow$ To obtain the total magnetic field at point $P$ due to entire wire, equation $(1)$ has to be integrated.
$\overrightarrow{ B } =\int d \overrightarrow{ B }$
$\overrightarrow{ B } =\frac{\mu_0}{4 \pi} \int \frac{ I d \vec{l} \times \hat{r}}{r^2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Answer the following:
The discharging current in the atmosphere due to the small conductivity of air is known to be 1800A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
The discharging current in the atmosphere due to the small conductivity of air is known to be 1800A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
Define wavefront of a travelling wave. Using Huygens principle, obtain the law of refraction at a plane interface when light passes from a denser to rarer medium.
Current in a circuit falls from $5.0A$ to $0.0A$ in $0.1s.$ If an average emf of $200V$ induced, give an estimate of the self$-$inductance of the circuit.
→(a) The peak voltage of an ac supply is 300 V. What is its rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
What will be the path of a charged particle moving in a region of crossed (or transverse) uniform electrostatic and magnetic fields with initial velocity zero?
In the given figure, point A is at zero potential. Use Kirchhoff's rules to find out the potential at B.
In a photoelectric effect, the yellow light is just able to emit electrons, will green light emit photoelectrons? What about red light?
A nearsighted person cannot clearly see beyond 200cm. Find the power of the lens needed to see objects at large distances.
Can we physically combine a strip of P-type semiconductor with an N-type semiconductor to form a P-N junction?
An electric dipole of dipole moment $20 \times 10^{-6}\ C$ is enclosed by closed surface. What is the net electric flux coming out of this surface?
→