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

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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 $B = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ along the axis of the coil is given by $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.We are given that $B_x = \frac{B}{8}$. Substituting the expression for $B$:$\frac{B}{8} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$\frac{1}{8} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$\frac{1}{8R} = \frac{R^2}{(R^2 + x^2)^{3/2}}$$\frac{1}{8R^3} = \frac{1}{(R^2 + x^2)^{3/2}}$Taking the cube root of both sides:$\frac{1}{2R} = \frac{1}{(R^2 + x^2)^{1/2}}$Squaring both sides:$\frac{1}{4R^2} = \frac{1}{R^2 + x^2}$$R^2 + x^2 = 4R^2$$x^2 = 3R^2$$x = R\sqrt{3}$.

$A$ circular coil carrying current $I$ has a radius $R$ and magnetic field at the centre is $B$. The distance from the centre along the axis of the same coil where the magnetic field will be $\frac{B}{8}$ is

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