Question

A square frame of side l carries a current produces a field $B$ at its centre. The same current is passed through a circular loop having same perimeter as the square. The field at its centre is $B^{\prime }$ , the ratio of $\frac{B}{B^{\prime }}$ is :

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

Answer 1

Answer: (A)

$\frac{8\sqrt{2}}{{\pi }^2}$

B due to square $=4\times \frac{{\mu }_{0}i}{4\pi (\frac{l}{2})}(\cos {45}^o+\cos {45}^o)$
$=\frac{2\sqrt{2}{\mu }_{0}i}{\mu l}$
Now, $4l=2\pi r$
$r=\frac{2l}{\pi }$
$B^1=\frac{{\mu }_{0}i}{2\times \frac{2l}{\pi }}=\frac{\pi {\mu }_{0}i}{4l}$