Question

A long curved conductor carrier a current $\overrightarrow{I}$($\overrightarrow{I}$ is a vector). A small current element of length $d\overrightarrow{l}$, on the wire, induces a magnetic field at a point, away from the current element. If the position vector between the current element and the point is $\overrightarrow{r}$, making an angle $\theta$ with current element then, the induced magnetic field density, $d\overrightarrow{B}$ at the point is $({\mu }_0=$permeability of free space):

• A. $\frac{{\mu }_{o}I(d\overrightarrow{l}\times \overrightarrow{r})}{4\pi r^3}$ perpendicular to the current element $d\overrightarrow{l}$
• B. $\frac{{\mu }_{o}I\times \overrightarrow{r}.d\overrightarrow{l}}{4\pi r^2}$ perpendicular to the current element $d\overrightarrow{l}$
• C. $\frac{{\mu }_{o}I\times d\overrightarrow{l}}{r}$ perpendicular to the plane containing the current element and position vector $\overrightarrow{r}$
• D. $\frac{{\mu }_{o}I\times d\overrightarrow{l}}{4\pi r^2}$ perpendicular to the plane containing current element and position vector $\overrightarrow{r}$

Answer 1

The correct option is A $\frac{{\mu }_{o}I(d\overrightarrow{l}\times \overrightarrow{r})}{4\pi r^3}$ perpendicular to the current element $d\overrightarrow{l}$

Biot Savart law provides the general expression for magnetic field induced by a small current carrying element at a distance from the element in a given direction. It states,

$d\overrightarrow{B}=\frac{{\mu }_{0}I(d\overrightarrow{l}\times \overrightarrow{r})}{4\pi r^3}$.

Since the direction is that of ($d\overrightarrow{l}\times \overrightarrow{r}$), it is perpendicular to the plane containing current element and the position vector $\overrightarrow{r}$.