A) \[\frac{{{\mu }_{0}}IN}{4r}\]
B) \[\frac{{{\mu }_{0}}IN}{8r}\]
C) \[\frac{{{\mu }_{0}}IN}{16\pi r}\]
D) \[\frac{{{\mu }_{0}}IN}{32r}\]
Correct Answer: C
Solution :
The magnetic induction at any point distance \[x\]from its centre \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi NI{{r}^{2}}}{{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}\] ?(i) where r is the radius of the circular coil. Then, \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi NI{{r}^{2}}}{{{[{{r}^{2}}+{{(\sqrt{3}r)}^{2}}]}^{3/2}}}\] or \[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{NI{{r}^{2}}}{{{({{r}^{2}}+3{{r}^{2}})}^{3/2}}}\] [Given, \[x=\sqrt{3r}\]] or \[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{NI{{r}^{2}}}{{{(4{{r}^{2}})}^{3/2}}}\] or \[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{NI{{r}^{2}}}{{{(2r)}^{3}}}\] or \[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{NI{{r}^{2}}}{8{{r}^{3}}}\] or \[B=\frac{{{\mu }_{0}}NI}{16\pi r}\]
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