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

The Correct Option is D

$\frac{For A}{2 \pi R = \,\ell}$
$ 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{For B}{4a =\ell} $
$a = \frac{\ell}{4} $
$B_{B} = 4 \times \frac{\mu_{0}I \frac{\ell}{4}}{2 \pi \frac{\ell}{8} \sqrt{\left(\frac{\ell}{4}\right)^{2}+4 \left(\frac{\ell}{8}\right)^{2}}} = \frac{4\mu_{0}I}{\pi.\sqrt{2} \frac{l\\ell}{4}}$
$ 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}} $