question_answer

The magnetic field at the point of intersection of diagonals of a square wire 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 :

                 We can have the figure as follows The net magnetic field at the point of intersection of the diagonal would be given by \[{{B}_{net}}=({{B}_{L/2}}sin{{45}^{o}}+{{B}_{L/2}}sin{{45}^{o}})\]                 \[+({{B}_{L/2}}sin{{45}^{o}}+{{B}_{L/2}}sin{{45}^{o}})\] \[+\,({{B}_{L/2}}sin{{45}^{o}}+{{B}_{L/2}}sin{{45}^{o}})\] \[+\,({{B}_{L/2}}sin{{45}^{o}}+{{B}_{L/2}}sin{{45}^{o}})\] \[=\,4\,({{B}_{L/2}}sin{{45}^{o}}+{{B}_{L/2}}sin{{45}^{o}})\] \[=8{{B}_{L/2}}\sin {{45}^{o}}=8\frac{{{\mu }_{0}}}{4\pi }\frac{I}{\left( \frac{L}{2} \right)}\frac{1}{\sqrt{2}}=\frac{{{\mu }_{0}}}{\pi }\frac{2\sqrt{2I}}{L}\] Alter \[{{B}_{net}}=4\times \frac{{{\mu }_{0}}}{4\pi }.\frac{I}{\frac{L}{2}}(sin{{45}^{o}}+\sin {{45}^{o}})\] \[=4\times \frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{a}\times \frac{2}{\sqrt{2}}=\frac{2{{\sqrt{2\mu }}_{0}}I}{\pi L}\] +71     0.     VZ         Til.