Question

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 $\frac{B_A}{B_B}$ is :

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

Answer 1

Answer: (D)

$\frac{ForA}{2\pi R=\ell }$
$R=\frac{\ell }{2\pi }$
$B_A=\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}I}{2\times \frac{\ell }{2\pi }}$
$B_A=\frac{{\mu }_0\pi I}{\ell }$
$\frac{ForB}{4a=\ell }$
$a=\frac{\ell }{4}$
$$\begin{gathered} B_B=4\times \frac{{\mu }_{0}I\frac{\ell }{4}}{2\pi \frac{\ell }{8}\sqrt{{\left(\frac{\ell }{4}\right)}^2+4{\left(\frac{\ell }{8}\right)}^2}}=\frac{4{\mu }_{0}I}{\pi .\sqrt{2}\frac{l \\ ell}{4}} \end{gathered}$$
$B_B=\frac{16{\mu }_{0}I}{\sqrt{2}\pi \ell }$
$\frac{B_A}{B_B}=\frac{\ell }{\frac{16{\mu }_{0}I}{\sqrt{2}\pi \ell }}=\frac{\sqrt{2}{\pi }^2}{16}=\frac{{\pi }^2}{8\sqrt{2}}$