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A circular current carrying coil has a radius $R$. The distance from the centre of the coil, on the axis, where $B$ will be $\frac{1}{8}$ of its value at the centre of the coil is
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The correct answer is:
$\sqrt{3} R$
$B=\frac{\mu_0 N i R^2}{2\left(R^2+x^2\right)^{3 / 2}}$
$\begin{aligned} & =\frac{\mu_0 N i}{2 R} \cdot \frac{1}{\left[1+\frac{x^2}{R^2}\right]^{3 / 2}} \\ & =B_C \cdot \frac{1}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\text { Let } B_C=\frac{\mu_0 N i}{2 R}\right]\end{aligned}$
$\Rightarrow \quad \frac{B_C}{8}=\frac{B_C}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\because B=\frac{B_C}{8}\right]$
$\begin{array}{rlrl}\Rightarrow & \left(1+\frac{x^2}{R^2}\right)^{3 / 2} & =8 \\ \Rightarrow & 1+\frac{x^2}{R^2} & =4 \\ \Rightarrow & \frac{x^2}{R^2} & =3 \\ & \therefore & x & =R \sqrt{3}\end{array}$
$\begin{aligned} & =\frac{\mu_0 N i}{2 R} \cdot \frac{1}{\left[1+\frac{x^2}{R^2}\right]^{3 / 2}} \\ & =B_C \cdot \frac{1}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\text { Let } B_C=\frac{\mu_0 N i}{2 R}\right]\end{aligned}$
$\Rightarrow \quad \frac{B_C}{8}=\frac{B_C}{\left(1+\frac{x^2}{R^2}\right)^{3 / 2}} \quad\left[\because B=\frac{B_C}{8}\right]$
$\begin{array}{rlrl}\Rightarrow & \left(1+\frac{x^2}{R^2}\right)^{3 / 2} & =8 \\ \Rightarrow & 1+\frac{x^2}{R^2} & =4 \\ \Rightarrow & \frac{x^2}{R^2} & =3 \\ & \therefore & x & =R \sqrt{3}\end{array}$
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