An element $\Delta l=\Delta \mathrm{xi}$ is placed at the origin and carries a large current $\mathrm{I}=10 \mathrm{~A}$. The magnetic field on the $y$-axis at a distance of $0.5 \mathrm{~m}$ from the elements $\Delta \mathrm{x}$ of $1 \mathrm{~cm}$ length is:
JEE MAIN 2024, Diffcult
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$\overrightarrow{\mathrm{dB}}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{(\overrightarrow{\mathrm{d} l} \times \overrightarrow{\mathrm{r}})}{\mathrm{r}^3} \text { (Tesla) }$
$=\frac{10^{-7} \times 10 \times\left(\frac{1}{2} \times \frac{1}{100}\right)(+\hat{\mathrm{k}})}{\left(\frac{1}{2}\right)^3}=4 \times 10^{-8} \mathrm{~T}(+\hat{\mathrm{k}})$
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