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: B
Solution :
Magnetic field due to one side of the square at centre\[O\] \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i\sin 45{}^\circ }{a/2}\] \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\sqrt{2}i}{a}\] Hence, magnetic field at centre due to all side \[B=4{{B}_{1}}=\frac{{{\mu }_{0}}(2\sqrt{2}i)}{\pi a}\] Magnetic field at centre due to n turns \[B=nB=\frac{{{\mu }_{0}}(2\sqrt{2}i)}{\pi a}\] \[=\frac{{{\mu }_{0}}2\sqrt{2}ni}{\pi (2l)}=\frac{\sqrt{2}{{\mu }_{0}}ni}{\pi l}\] \[(\because a=2l)\]
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