A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is$......$
NEET 2019, Medium
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$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\ell_{1}}{\ell_{2}}= \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\theta_{1}}{\theta_{2}}$
$ \Rightarrow \theta_{1} \mathrm{I}_{1}=\theta_{2} \mathrm{I}_{2}$
$\left.\begin{array}{l}{\mathrm{B}_{1}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}} \\ {\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}}\end{array}\right\} $
$\Rightarrow \mathrm{B}_{1}=\mathrm{B}_{2}$
$ \Rightarrow \theta_{1} \mathrm{I}_{1}=\theta_{2} \mathrm{I}_{2}$
$\left.\begin{array}{l}{\mathrm{B}_{1}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}} \\ {\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{1} \theta_{1}}{\mathrm{r}}}\end{array}\right\} $
$\Rightarrow \mathrm{B}_{1}=\mathrm{B}_{2}$
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