Question
State and explain the Biot – Savart Law.
✓
Answer
Biot and Savart experimentally observed that the magnitude of magnetic field $d \vec{B}$ at a point $P$ at a distance $r$ from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element $\overrightarrow{d l}$
(iii) directly as the sine of the angle (say, $\theta$ ) between $\overrightarrow{d l}$ and $\hat{r}$.
(iv) inversely as the square of the distance between the point $P$ and length element $\overrightarrow{d l}$. This is expressed as
$dB \propto \frac{I d l}{r^2} \sin \theta$
$dB = k \frac{I d l}{r^2} \sin \theta$
Where $K =\frac{\mu_0}{4 \pi}$ in SI units and $k =1$ in CGS units.
In vector notation,
$d \vec{B}=\frac{\mu_0}{4 \pi} \frac{ I d \vec{l} \times \hat{r}}{r^2}$
Here vector $d \vec{B}$ is perpendicular to both $I \overrightarrow{d l}$ (pointing current carrying conductor the direction of current flow) and the unit vector and $\hat{r}$ directed from $\overrightarrow{d l}$ toward point P The equation 1 is used to compute the magnetic field only due to a small elemental length $\overrightarrow{d l}$ of the conductor. The net magnetic field at $P$ due to the conductor is obtained from principle of superposition by considering the contribution from all current elements $I \overrightarrow{d l}$. Hence integrating equation (1), we get
$\vec{B}=\int d \vec{B}=\frac{\mu_0 I}{4 \pi} \int \frac{\overrightarrow{d l} \times \quad \hat{r}}{r^2}$
where the integral is taken over the entire current distribution.
Case:
1. If the pont $P$ lies on the ckonductor, then $\theta=0^{\circ}$. Therefore, $d \vec{B}$ is zero.
2. If the point lies perpendicular to the conductor, then $\theta=90^{\circ}$. Therefore, $d \vec{B}$ is maximum and is given by $d \vec{B}=\frac{I d l}{r^2} \hat{n}$. where $\hat{n}$ is the unit vector perpendicular to both $\mid \overrightarrow{d l}$ and
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element $\overrightarrow{d l}$
(iii) directly as the sine of the angle (say, $\theta$ ) between $\overrightarrow{d l}$ and $\hat{r}$.
(iv) inversely as the square of the distance between the point $P$ and length element $\overrightarrow{d l}$. This is expressed as
$dB \propto \frac{I d l}{r^2} \sin \theta$
$dB = k \frac{I d l}{r^2} \sin \theta$
Where $K =\frac{\mu_0}{4 \pi}$ in SI units and $k =1$ in CGS units.
In vector notation,
$d \vec{B}=\frac{\mu_0}{4 \pi} \frac{ I d \vec{l} \times \hat{r}}{r^2}$
Here vector $d \vec{B}$ is perpendicular to both $I \overrightarrow{d l}$ (pointing current carrying conductor the direction of current flow) and the unit vector and $\hat{r}$ directed from $\overrightarrow{d l}$ toward point P The equation 1 is used to compute the magnetic field only due to a small elemental length $\overrightarrow{d l}$ of the conductor. The net magnetic field at $P$ due to the conductor is obtained from principle of superposition by considering the contribution from all current elements $I \overrightarrow{d l}$. Hence integrating equation (1), we get
$\vec{B}=\int d \vec{B}=\frac{\mu_0 I}{4 \pi} \int \frac{\overrightarrow{d l} \times \quad \hat{r}}{r^2}$
where the integral is taken over the entire current distribution.
Case:
1. If the pont $P$ lies on the ckonductor, then $\theta=0^{\circ}$. Therefore, $d \vec{B}$ is zero.
2. If the point lies perpendicular to the conductor, then $\theta=90^{\circ}$. Therefore, $d \vec{B}$ is maximum and is given by $d \vec{B}=\frac{I d l}{r^2} \hat{n}$. where $\hat{n}$ is the unit vector perpendicular to both $\mid \overrightarrow{d l}$ and
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