Two parallel long current carrying wire separated by a distance $2 \mathrm{r}$ are shown in the figure. The ratio of magnetic field at $\mathrm{A}$ to the magnetic field produced at $C$ is $\frac{x}{7}$. The value of $x$ is $\qquad$
JEE MAIN 2024, Diffcult
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$\mathrm{B}_{\mathrm{A}}=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}+\frac{\mu_0(2 \mathrm{i})}{2 \pi(3 \mathrm{r})}=\frac{5 \mu_0 \mathrm{i}}{6 \pi \mathrm{r}}$

$\mathrm{B}_C=\frac{\mu_0(2 \mathrm{i})}{2 \pi \mathrm{r}}+\frac{\mu_0 \mathrm{i}}{2 \pi(3 \mathrm{r})}=\frac{7 \mu_0 \mathrm{i}}{6 \pi \mathrm{r}}$

$\therefore \frac{\mathrm{B}_A}{\mathrm{~B}_C}=\frac{5}{7}$

$\therefore \mathrm{x}=5$

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