The magnetic field dB→ due to a small current element dl→ at a distance r→ and element carrying current i is

dB=μ04π. idl sin θr2 ⇒ dB→ = μ04π. i dl→ × r→ r3

The magnetic field dB→ due to a small current element dl→ at a distance r→ and element carrying current i is

dB=μ04π. idl sin θr2 ⇒ dB→ = μ04π. i dl→ × r→ r3

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The magnetic field $d \vec{B}$ due to a small current element $d \vec{l}$ at a distance $\vec{r}$ and element carrying current i is

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$\vec{d B} = \frac{μ_{0}}{4 π} i (\frac{d \vec{l} \times \vec{r}}{r})$

$\vec{d B} = \frac{μ_{0}}{4 π} i^{2} (\frac{d \vec{l} \times \vec{r}}{r})$

$\vec{d B} = \frac{μ_{0}}{4 π} i^{2} (\frac{d \vec{l} \times \vec{r}}{r^{2}})$

$\vec{d B} = \frac{μ_{0}}{4 π} i (\frac{d \vec{l} \times \vec{r}}{r^{3}})$

answer is D.

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$dB = \frac{μ_{0}}{4 π} . \frac{idl \sin θ}{r^{2}} \Rightarrow d \vec{B} = \frac{μ_{0}}{4 π} . \frac{i (d \vec{l} \times \vec{r})}{r^{3}}$

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The magnetic field $d \vec{B}$ due to a small current element $d \vec{l}$ at a distance $\vec{r}$ and element carrying current i is