Question

Two identical wires $A$ and $B$ have the same length $l$ and 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 $B_1$ and $B_2$ are the values of magnetic induction at the center of the circle and the center of the square, respectively, then the ratio $B_1/B_2$ is :

• A. $\left({\pi }^2/8\right)$
• B. $\left({\pi }^2/8\sqrt{2}\right)$
• C. $\left({\pi }^2/16\right)$
• D. $\left({\pi }^2/16\sqrt{2}\right)$

Answer 1

The correct option is B $\left({\pi }^2/8\sqrt{2}\right)$

$B=\frac{{\mu }_0}{4\pi }\times \frac{2\pi I}{R}=\frac{{\mu }_0}{4\pi }\times \frac{2\pi l\times 2\pi }{L}.........\left(1\right)$
$(\because L=2\pi R,forcircularloop)$
$B_2=\frac{{\mu }_0}{4\pi }\times \frac{1}{\left(a/2\right)}[\sin {45}^o+\sin {45}^o]\times 4$
Where $a=\left(L/4\right)$
$\therefore B_2=\frac{{\mu }_0}{4\pi L}\times 8\times 4\times [\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}]=\frac{{\mu }_0}{4\pi L}\times \frac{64}{\sqrt{2}}$
$\therefore \frac{B_1}{B_2}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{4{\pi }^{2}I}{L}╱\frac{{\mu }_0}{4\pi L}\times \frac{64}{\sqrt{2}}=\frac{{\pi }^2}{8\sqrt{2}}$