question_answer

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:      [JEE MAIN - I 3-4-2016]

A) \[\frac{{{\pi }^{2}}}{8\sqrt{2}}\]

B) \[\frac{{{\pi }^{2}}}{8}\]

C) \[\frac{{{\pi }^{2}}}{16\sqrt{2}}\]

D) \[\frac{{{\pi }^{2}}}{16}\]

Correct Answer: A

Solution :

[a]

\[{{B}_{1}}=\frac{{{\mu }_{0}}i}{2r}\]            \[{{B}_{2}}{{=}^{4}}\times \frac{{{\mu }_{0}}}{4\pi }\times \frac{i}{\left( \frac{a}{2} \right)}\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)\]

\[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\pi a}{4\sqrt{2}r}\]                        \[\ell =2\pi r=4a\]

\[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\pi }{4\sqrt{2}}\frac{\pi }{2}\]                    \[\frac{a}{r}=\frac{2\pi }{4}=\frac{\pi }{2}\]

\[=\frac{{{\pi }^{2}}}{8\sqrt{2}}\]