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(B) The magnetic field $B$ on the axis of a circular coil of radius $R$ carrying current $I$ at a distance $x$ from its center is given by: $B = \frac

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$\sqrt{3} R$

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(B) The magnetic field $B$ on the axis of a circular coil of radius $R$ carrying current $I$ at a distance $x$ from its center is given by: $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.At the center of the coil $(x = 0)$,the magnetic field $B_0$ is: $B_0 = \frac{\mu_0 I}{2R}$.According to the problem,the magnetic field at point $P$ is $\frac{1}{8}$ of the field at the center: $B_P = \frac{1}{8} B_0$.Substituting the expressions: $\frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 I}{2R} \right)$.Simplifying the equation: $\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \Rightarrow \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8}$.Taking the cube root of both sides: $\frac{R}{(R^2 + x^2)^{1/2}} = \frac{1}{2}$.Squaring both sides: $\frac{R^2}{R^2 + x^2} = \frac{1}{4}$.Cross-multiplying: $4R^2 = R^2 + x^2 \Rightarrow x^2 = 3R^2 \Rightarrow x = \sqrt{3}R$.

$A$ current carrying circular coil of radius $R$ has a point $P$ situated on its axis at a distance $x$ from its centre $O$ of the coil. The magnetic induction at point $P$ is $\left(\frac{1}{8}\right)^{\text{th}}$ of the magnetic field at its centre $O$. The value of $x$ is

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