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(B) 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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$R \sqrt{3}$

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(B) 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,$B_{axis} = \frac{1}{8} B_{centre}$.Substituting the expressions,we get: $\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} 	imes \frac{\mu_0 I}{2R}$.Simplifying the equation: $\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R}$.This gives $(R^2 + x^2)^{3/2} = 8R^3$.Taking the cube root of both sides: $(R^2 + x^2)^{1/2} = 2R$.Squaring both sides: $R^2 + x^2 = 4R^2$.Therefore,$x^2 = 3R^2$,which implies $x = R \sqrt{3}$.

$A$ circular current-carrying coil has radius $R$. At what distance from the centre of the coil on the axis,the magnetic induction will become $\frac{1}{8}$ th of its value at the centre of the coil?

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