Question 23 of 276

Figure shows a current carrying square loop ABCD of edge length is ' $a$ ' lying in a plane. If the resistance of the $A B C$ part is $r$ and that of $A D C$ part is 2 r , then the magnitude of the resultant magnetic field at centre of the square loop is

A$\frac{3 \pi \mu_0 \mathrm{I}}{\sqrt{2} \mathrm{a}}$B$\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}}$C$\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}$D$\frac{2 \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}$

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Answer: (C) $\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}$

$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{BC}}+\overrightarrow{\mathrm{B}}_{\mathrm{CD}}+\overrightarrow{\mathrm{B}}_{\mathrm{DA}}$
$\begin{aligned} \overrightarrow{\mathrm{B}}= & {\left[\frac{-\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}-\frac{\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right.} \\ & \left.+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right] \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}= & {\left[\frac{-2 \sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}+\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\right] \hat{\mathrm{k}} } \\ \overrightarrow{\mathrm{B}}= & \frac{-\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}} \hat{\mathrm{k}}\end{aligned}$