A) 1.0 m
B) 0.1 m
C) 0.15 m
D) 0.2 m
Correct Answer: B
Solution :
Magnetic field at a point on the axis of a current carrying circular coil of radius r is \[B=\frac{{{\mu }_{0}}n\,l{{a}^{2}}}{2{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}{{{({{r}^{2}}+x_{1}^{2})}^{3/2}}}\] \[\frac{8}{1}=\left[ \frac{[{{r}^{2}}+{{(0.2)}^{3/2}}]}{[{{r}^{2}}+{{(0.05)}^{3/2}}]} \right][\because {{(8)}^{2/3}}={{2}^{2}}=4]\] \[4=\frac{{{r}^{2}}+0.04}{{{r}^{2}}+0.0025}\] Thus, r = 0.1 m
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