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

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$R\sqrt{3}$

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(B) The magnetic field at the centre of a circular coil is given by $B_{centre} = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ from the centre 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 $\frac{B_{centre}}{B_{axis}} = \frac{(R^2 + x^2)^{3/2}}{R^3} = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.Given that $B_{axis} = \frac{1}{8} B_{centre}$,we have $\frac{B_{centre}}{B_{axis}} = 8$.Therefore,$8 = \left(1 + \frac{x^2}{R^2}\right)^{3/2}$.Taking the cube root of both sides,$2 = \left(1 + \frac{x^2}{R^2}\right)^{1/2}$.Squaring both sides,$4 = 1 + \frac{x^2}{R^2}$.This gives $\frac{x^2}{R^2} = 3$,so $x^2 = 3R^2$,which implies $x = R\sqrt{3}$.

$A$ circular current-carrying coil has a radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{8}$th of its value at the centre of the coil is:

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