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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from the center is given by $B = \frac{\mu_0 N i R^2}{2(R^2 + x^

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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from the center is given by $B = \frac{\mu_0 N i R^2}{2(R^2 + x^2)^{3/2}}$.Since $B \propto (R^2 + x^2)^{-3/2}$,we have the ratio $\frac{B_1}{B_2} = \left( \frac{R^2 + x_2^2}{R^2 + x_1^2} \right)^{3/2}$.Given $\frac{B_1}{B_2} = 8$,$x_1 = 0.05\,m$,and $x_2 = 0.2\,m$,we substitute these values:$8 = \left( \frac{R^2 + (0.2)^2}{R^2 + (0.05)^2} \right)^{3/2}$.Taking the power of $2/3$ on both sides:$8^{2/3} = \frac{R^2 + 0.04}{R^2 + 0.0025} \Rightarrow 4 = \frac{R^2 + 0.04}{R^2 + 0.0025}$.$4(R^2 + 0.0025) = R^2 + 0.04 \Rightarrow 4R^2 + 0.01 = R^2 + 0.04$.$3R^2 = 0.03 \Rightarrow R^2 = 0.01 \Rightarrow R = 0.1\,m$.

Magnetic fields at two points on the axis of a circular coil at a distance of $0.05\,m$ and $0.2\,m$ from the centre are in the ratio $8 : 1$. The radius of the coil is.....$m$

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