A) \[\frac{{{\pi }^{2}}}{8}\]
B) \[\frac{{{\pi }^{2}}}{8\sqrt{2}}\]
C) \[\frac{{{\pi }^{2}}}{16}\]
D) \[\frac{{{\pi }^{2}}}{16\sqrt{2}}\]
Correct Answer: B
Solution :
At the centre of circle \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I}{R}\] or \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I}{L/2\pi }\] (\[\because \,L=2\,\pi \,R\]for circular loop) or \[{{B}_{1}}=\frac{{{\mu }_{0}}\pi I}{L}\] ... (i) At the centre of square \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{I}{a/2}[\sin \,{{45}^{o}}+\sin {{45}^{o}}]\times 4\] where \[a=L/4\] \[\therefore \] \[{{B}_{2}}=\frac{{{\mu }_{0}}I}{4\pi L}\times 8\times 4\times \left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right]\] or \[{{B}_{2}}=\frac{{{\mu }_{0}}I}{4\pi L}\times \frac{64}{\sqrt{2}}\] ?. (ii) Dividing Eq. (i) by Eq. (ii), we obtain \[\frac{{{B}_{1}}}{{{B}_{2}}}={\frac{{{\mu }_{0}}\pi I}{L}}/{\frac{{{\mu }_{0}}I}{4\pi L}\times \frac{64}{\sqrt{2}}}\;\] \[=\frac{{{\pi }^{2}}}{8\sqrt{2}}\]
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