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 :
\[{{B}_{1}}=\frac{{{\mu }_{0}}i}{2r}\] \[{{B}_{2}}{{=}^{4}}\times \frac{{{\mu }_{0}}}{4\pi }\times \frac{i}{\left( \frac{a}{2} \right)}\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\pi a}{4\sqrt{2}r}\] \[\ell =2\pi r=4a\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\pi }{4\sqrt{2}}\frac{\pi }{2}\] \[\frac{a}{r}=\frac{2\pi }{4}=\frac{\pi }{2}\] \[=\frac{{{\pi }^{2}}}{8\sqrt{2}}\]You need to login to perform this action.
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