A) \[({{\pi }^{2}}/8)\]
B) \[({{\pi }^{2}}/8\sqrt{2})\]
C) \[({{\pi }^{2}}/16)\]
D) \[({{\pi }^{2}}/16\sqrt{2})\]
Correct Answer: B
Solution :
\[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi i}{R}\] \[=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi i\times 2\pi }{L}\] \[(\because L=2\pi R,\]for circular loop) \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{i}{(a/2)}\] \[[\sin {{45}^{o}}+\sin {{45}^{o}}]\times 4\] where \[a=L/4\] \[\therefore \]\[{{B}_{2}}=\frac{{{\mu }_{0}}i}{4\pi L}\times 8\times 4\times \left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right]\] \[=\frac{{{\mu }_{0}}i}{4\pi L}\times \frac{64}{\sqrt{2}}\] ...(ii) Hence, \[\frac{{{B}_{1}}}{{{B}_{2}}}=\left( \frac{{{\mu }_{0}}}{4\pi } \right)\times {\frac{4{{\pi }^{2}}i}{L}}/{\frac{{{\mu }_{0}}i}{4\pi L}}\;\times \frac{64}{\sqrt{2}}\] Or \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\pi }^{2}}}{8\sqrt{2}}\]
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