A) \[\frac{\sqrt{2}{{\mu }_{0}}nI}{\pi l}\]
B) \[\frac{\sqrt{2}{{\mu }_{0}}nI}{2\pi l}\]
C) \[\frac{\sqrt{2}{{\mu }_{0}}nI}{4\pi l}\]
D) \[\frac{2{{\mu }_{0}}nI}{\pi l}\]
Correct Answer: A
Solution :
Magnetic field due to one side of the square at centre O \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i\sin {{45}^{o}}}{a/2}\]Þ \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\sqrt{2}\,i}{a}\] Hence magnetic field at centre due to all side \[B=4{{B}_{1}}=\frac{{{\mu }_{0}}(2\sqrt{2}i)}{\pi a}\] Magnetic field due to n turns \[{{B}_{net}}=nB=\frac{{{\mu }_{0}}2\sqrt{2}ni}{\pi a}=\frac{{{\mu }_{0}}2\sqrt{2}ni}{\pi (2l)}\]\[=\frac{\sqrt{2}{{\mu }_{0}}ni}{\pi l}\] \[(\because \,a=2l)\]
You need to login to perform this action.
You will be redirected in
3 sec
