question_answer

A current I flows in the anticlockwise direction through a square loop of side a lying in the XOY plane with its center at the origin. The magnetic induction at the center of the square loop is

A) \[\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi a}{{\hat{e}}_{x}}\] 

B) \[\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi a}{{\hat{e}}_{z}}\]

C) \[\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi {{a}^{2}}}{{\hat{e}}_{z}}\]          

D) \[\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi {{a}^{2}}}{{\hat{e}}_{x}}\]

Correct Answer: B

Solution :

[b] Field due to one side of loop at \[O=\frac{{{\mu }_{0}}I}{4\pi \left( \frac{a}{2} \right)}(2\,\sin \,{{45}^{o}})\] Field at O due to all four sides is along unit vector \[{{\hat{e}}_{z}}\] \[\therefore \]Tool field \[=4.\frac{{{\mu }_{0}}I}{4\pi \left( \frac{a}{2} \right)}(2\,\sin \,{{45}^{o}})=\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi a}\]