Question 88 of 276

Two identical wires $A$ and $B$, each of length $l$, carry the same current $I$ . Wire $A$ is bent into a circle of radius $R$ and wire $B$ is bent to form a square of side $a$. If ${B}_{A}$ and ${B}_{B}$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $\dfrac{{B}_{A}}{{B}_{B}}$ is

A$\dfrac{{\pi }^{2}}{16}$B$\dfrac{{\pi }^{2}}{8\sqrt{2}}$C$\dfrac{{\pi }^{2}}{8}$D$\dfrac{{\pi }^{2}}{16\sqrt{2}}$

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Answer: (B) $\dfrac{{\pi }^{2}}{8\sqrt{2}}$

As length of both the wires is same, $2\pi R=4a$

$\Rightarrow \dfrac{a}{R}=\dfrac{2\pi }{4}$
$\Rightarrow \dfrac{a}{R}=\dfrac{\pi }{2}$

${B}_{A}=\dfrac{{\mu }_{0}i}{2R}$ and

${B}_{B}=\dfrac{{\mu }_{0}i}{\pi a}\left(2\sqrt{2}\right)$

Therefore, ratio will be $\dfrac{{B}_{A}}{{B}_{B}}=\dfrac{{\mu }_{0}i}{2R}\times \dfrac{\pi a}{2\sqrt{2}{\mu }_{0}i}$

$=\dfrac{\pi a}{4\sqrt{2}R}=\dfrac{\pi }{4\sqrt{2}}\left(\dfrac{\pi }{2}\right)=\dfrac{{\pi }^{2}}{8\sqrt{2}}$