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(A) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_2 = \frac{\mu_0 I}{2R}$.The magnetic field a

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$R \sqrt{3}$

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(A) The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by $B_2 = \frac{\mu_0 I}{2R}$.The magnetic field at an axial point $P$ at a distance $x$ from the centre is given by $B_1 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given $B_1 = \frac{B_2}{8}$,we substitute the expressions:$\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} 	imes \frac{\mu_0 I}{2R}$.Simplifying the equation:$\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R}$.$8R^3 = (R^2 + x^2)^{3/2}$.Taking the power of $2/3$ on both sides:$(8R^3)^{2/3} = R^2 + x^2$.$4R^2 = R^2 + x^2$.$x^2 = 3R^2$.$x = R \sqrt{3}$.

$A$ current carrying circular coil of radius $R$ produces magnetic fields $B_1$ and $B_2$ at an axial point $P$ at a distance $x$ from its centre and at a point $Q$ placed at its centre respectively. If $B_1 = \frac{B_2}{8}$,the value of $x$ is

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