question_answer

A straight wire of length f. carrying a current i, is bent into following shapes as shown in figure, keeping length and current constant.   The magnetic field at point 0 is maximum for

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, \[\ell =4a\] \[{{B}_{3}}=\frac{{{\mu }_{0}}i}{4\pi d}\sqrt{2}\times 4=\frac{{{\mu }_{0}}i\,4\sqrt{2}}{4\pi \frac{\ell }{8}}=\frac{{{\mu }_{0}}i8\sqrt{2}}{\pi \ell }\]