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, \[r=\sqrt{{{\left( \frac{L\sqrt{2}}{2} \right)}^{2}}-{{\left( \frac{L}{2} \right)}^{2}}}\] \[=\sqrt{\frac{2{{L}^{2}}}{4}-\frac{{{L}^{2}}}{4}}\] \[=\sqrt{\frac{{{L}^{2}}}{4}}=\frac{L}{2}\] \[\therefore \] \[B=\frac{{{\mu }_{0}}I\sqrt{2}}{\pi \frac{L}{2}}=\frac{{{\mu }_{0}}I.2\sqrt{2}}{\pi L}\]You need to login to perform this action.
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