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(B) The magnetic field at the center of a circular coil is given by $B_{center} = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ on the axis

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$R\sqrt{3}$

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(B) The magnetic field at the center of a circular coil is given by $B_{center} = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ on the axis of the coil is given by $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Taking the ratio,we get $\frac{B_{center}}{B_{axis}} = \frac{(R^2 + x^2)^{3/2}}{R^3} = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.Given that $B_{axis} = \frac{1}{8} B_{center}$,so $\frac{B_{center}}{B_{axis}} = 8$.Substituting this into the equation: $8 = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.Taking the power of $2/3$ on both sides: $8^{2/3} = 1 + \frac{x^2}{R^2}$.$4 = 1 + \frac{x^2}{R^2} \implies 3 = \frac{x^2}{R^2}$.Therefore,$x^2 = 3R^2$,which gives $x = \sqrt{3}R$.

$A$ current flows through a circular coil of radius $R$. Find the distance from the center on the axis where the magnetic field is $\frac{1}{8}$th of the magnetic field at the center.

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