question_answer

    The magnetic field at the point of intersection of diagonals of a square loop of side L carrying a current\[I\]is

A)  \[\frac{{{\mu }_{0}}I}{\pi L}\]                                    

B)  \[\frac{2{{\mu }_{0}}I}{\pi L}\]

C)  \[\frac{\sqrt{2}{{\mu }_{0}}I}{\pi L}\]                    

D)  \[\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi L}\]

Correct Answer: D

Solution :

                From Biot-Savarfs law, magnetic field due to current carrying conductor is where, \[{{\phi }_{1}}=45{}^\circ ,{{\phi }_{2}}=45{}^\circ \]                 \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)=\frac{{{\mu }_{0}}i\sqrt{2}}{4\pi r}\] Since, square consists of four sides, total magnetic field is                 \[B=4B=\frac{4}{4\pi r}.{{\mu }_{0}}i\sqrt{2}=\frac{{{\mu }_{0}}i\sqrt{2}}{\pi r}\] From figure, \[r=\sqrt{{{\left( \frac{L\sqrt{2}}{2} \right)}^{2}}-{{\left( \frac{L}{2} \right)}^{2}}}\]                 \[=\sqrt{\frac{2{{L}^{2}}}{4}-\frac{{{L}^{2}}}{4}}\]                 \[=\sqrt{\frac{{{L}^{2}}}{4}}=\frac{L}{2}\] \[\therefore \]  \[B=\frac{{{\mu }_{0}}I\sqrt{2}}{\pi \frac{L}{2}}=\frac{{{\mu }_{0}}I.2\sqrt{2}}{\pi L}\]