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.You need to login to perform this action.
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