Question

Two identical wires $A$ and $B$, each of length $l$, carry the same current $l$. 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 $\frac{B_A}{B_B}$ is :

• A. $\frac{{\pi }^2}{16\sqrt{2}}$
• B. $\frac{{\pi }^2}{16}$
• C. $\frac{{\pi }^2}{8\sqrt{2}}$
• D. $\frac{{\pi }^2}{8}$

Answer 1

The correct option is C $\frac{{\pi }^2}{8\sqrt{2}}$

Magnetic field in case of circle of radius $R$, we have,

$B_A=\frac{{\mu }_{0}I}{2R}$

As, $2\pi R=l$ ($l$ is length of a wire)

$R=\frac{l}{2\pi }$

$B_A=\frac{{\mu }_{0}I}{2\times \frac{l}{2\pi }}$ .............(1)

Magnetic field in case of square of side $a$, we get,

$B_B=4\times \frac{{\mu }_0}{4\pi }\times \frac{I}{\left(a/2\right)}\times (\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$

$⟹B_B=\frac{4I{\mu }_0}{\pi a\sqrt{2}}=\frac{{\mu }_{0}2\sqrt{2I}}{a\pi }$

As $4a=l,a=\frac{l}{4}⟹B_B=\frac{8\sqrt{2}{\mu }_{0}i}{\pi l}$ .............(2)

Dividing equation (1) and equation (2), we get

$\frac{B_A}{B_B}=\frac{{\pi }^2}{8\sqrt{2}}$