A square frame of side $l$ carries a current $I$ and produces a magnetic field $B$ at its center, The same current is passed through a circular coil having same perimeter as that of square. The field at the center of the circular coil is $B_1$. Then the ratio $\frac{B_1}{B}$ is-

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


When current $I$ passes through a square loop, the net magnetic induction at its center is

$B=4\times B_{oneside}$



$B_{oneside}=\left[\frac{{\mu }_{0}I}{4\pi r}(\sin \alpha +\sin \beta )\right]$

Here, $r=\frac{l}{2};\alpha ={45}^{\circ };\beta ={45}^{\circ }$

$B=4\left[\frac{{\mu }_{0}I}{4\pi \left(\frac{l}{2}\right)}(\sin {45}^{\circ }+\sin {45}^{\circ })\right]$

$B=\frac{{\mu }_{0}I}{4\pi l}\left(8\sqrt{2}\right)......\left(1\right)$

For a circular having perimeter $4l$ carries a current $I$, then magnetic induction at its center is,

$2\pi r=4l\Rightarrow r=\frac{2l}{\pi }$

$B_1=\frac{{\mu }_{0}I}{2r}$

$=\frac{{\mu }_{0}I}{2\left[\frac{2l}{\pi }\right]}=\frac{{\mu }_{0}I\pi }{4l}.......\left(2\right)$


Now, from $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{B_1}{B}=\frac{{\mu }_{0}I\pi }{4l}\times \frac{4\pi l}{8\sqrt{2}{\mu }_{0}I}=\frac{{\pi }^2}{8\sqrt{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.