A) circle only
B) square only
C) equilateral triangle only
D) all shapes
Correct Answer: C
Solution :
For circle, \[\ell =2\pi r\] \[{{B}_{1}}=\frac{{{\mu }_{0}}i}{2r}=\frac{{{\mu }_{0}}i\times 2\pi }{2\ell }=\frac{{{\mu }_{0}}\pi i}{\ell }\] For equilateral triangle, \[\ell =3a\] \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{\ell }{6\sqrt{3}} \right)}(\sin {{60}^{o}}+\sin {{60}^{o}})\times 3\] \[=\frac{{{\mu }_{0}}i}{\pi \ell }\left( \frac{27}{2} \right)\] For square, \[\ell =4a\] \[{{B}_{3}}=\frac{{{\mu }_{0}}i}{4\pi d}\sqrt{2}\times 4=\frac{{{\mu }_{0}}i\,4\sqrt{2}}{4\pi \frac{\ell }{8}}=\frac{{{\mu }_{0}}i8\sqrt{2}}{\pi \ell }\]You need to login to perform this action.
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