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(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_{centre} = \frac{\mu_0 I}{2R}$.The magnetic

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$2 \sqrt{2} R$

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(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ 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 $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.According to the problem,$B_{axis} = \frac{1}{27} B_{centre}$.Substituting the expressions: $\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{27} 	imes \frac{\mu_0 I}{2R}$.Simplifying,we get: $\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{27R}$.This simplifies to $\frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{27}$.Taking the cube root of both sides: $\frac{R}{(R^2 + x^2)^{1/2}} = \frac{1}{3}$.Squaring both sides: $\frac{R^2}{R^2 + x^2} = \frac{1}{9}$.$9R^2 = R^2 + x^2$,which gives $x^2 = 8R^2$.Therefore,$x = \sqrt{8}R = 2\sqrt{2}R$.

$A$ circular coil carrying current has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{27}$th of its value at the centre of the coil is:

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