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 \[B=\frac{{{\mu }_{0}}nl{{a}^{2}}}{2{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{({{r}^{2}}+{{x}^{2}})}^{3/2}}}{{{({{r}^{2}}+x_{1}^{2})}^{3/2}}}\] \[\frac{8}{1}=\left[ \frac{[{{r}^{2}}+{{(0.2)}^{3/2}}]}{{{r}^{2}}+{{(0.05)}^{3/2}}} \right]\] \[[\because \,{{(8)}^{2/3}}={{2}^{2}}=4]\] \[4=\frac{{{r}^{2}}+0.04}{{{r}^{2}}+0.0025}\] thus, \[r=0.1\,m\]
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