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 \[=4.\frac{{{\mu }_{0}}I}{4\pi \left( \frac{a}{2} \right)}(2\,\sin \,{{45}^{o}})=\frac{2\sqrt{2}{{\mu }_{0}}I}{\pi a}\]You need to login to perform this action.
You will be redirected in
3 sec