A) 1.0 m
B) 0.1 m
C) 0.15 m
D) 0.2 m
Correct Answer: B
Solution :
\[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi Ni{{R}^{2}}}{{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\Rightarrow B\propto \frac{1}{{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}\] \[\Rightarrow \frac{8}{1}=\frac{{{({{R}^{2}}+x_{2}^{2})}^{3/2}}}{{{({{R}^{2}}+x_{1}^{2})}^{3/2}}}\Rightarrow {{\left( \frac{8}{1} \right)}^{2/3}}=\frac{{{R}^{2}}+0.04}{{{R}^{2}}+0.0025}\] \[\Rightarrow \frac{4}{1}=\frac{{{R}^{2}}+0.04}{{{R}^{2}}+0.0025}\]. On solving \[R=0.1m\]
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