A) \[\frac{{{\mu }_{0}}i}{24}\]
B) \[\frac{{{\mu }_{0}}i}{16}\]
C) \[\frac{{{\mu }_{0}}i}{12}\]
D) \[\frac{{{\mu }_{0}}i}{8}\]
Correct Answer: A
Solution :
[a] \[B=\frac{{{\mu }_{0}}i}{2\pi \sqrt{{{x}^{2}}+{{a}^{2}}}}\] \[\int{\overrightarrow{B}}.\overrightarrow{d\ell }=\int\limits_{a/\sqrt{5}}^{a}{\frac{{{\mu }_{0}}idx\cos \,(90-\theta )}{2\pi \,\sqrt{{{x}^{2}}+{{a}^{2}}}}}\] \[=\frac{{{\mu }_{0}}ia}{2\pi }\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right)_{\frac{a}{\sqrt{3}}}^{a}=\frac{{{\mu }_{0}}i}{2\pi }\left[ \frac{\pi }{4}-\frac{\pi }{6} \right]=\frac{{{\mu }_{0}}i}{24}\]You need to login to perform this action.
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