Question

A long curved conductor carries a current $I$ (I is a vector). A small current element of length $d l$, 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 $r$, making an angle with current element then, the induced magnetic field density; $d B$ (vector) at the point is $\left(\mu_{0}=\right.$ permeability of free space $)$

• A. $\frac{\mu_{0} \text { Id } l \times r }{4 \pi r}$ (perpendicular to the current element $d l$ )
• B. $\frac{\mu_{0} I \times r \times d l}{4 \pi r^{2}}$ (perpendicular to the current element $d l$ )
• C. $\frac{\mu_{0} I \times d l}{r}$ (perpendicular to the plane containing the current element and position vector $r$ )
• D. $\frac{\mu_{0} I \times d l}{4 \pi r^{2}}$ (perpendicular to the plane containing current element and position vector $r$ )

Answer 1

Answer: (B)

$\frac{\mu_{0} I \times r \times d l}{4 \pi r^{2}}$ (perpendicular to the current element $d l$ )

Answer 2

The magnetic field
$d B=\frac{\mu_{0} I \times r \times d l}{4 \pi r^{2}}$