Q102 / 150 · MHT CET 2022 (11 Aug Shift 1) · Physics

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

  • A
    $\frac{R}{2 \sqrt{3}}$
  • Correct answer
    B
    $\sqrt{3} R$
  • C
    $\frac{R}{\sqrt{3}}$
  • D
    $\frac{2}{\sqrt{3}} R$

Solution

Magnetic field on the axis of a circular coil, at a location $x$ away from the center of the coil is given by, $B=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}$ For, $x=0$ the induction at the centre of the coil is $B_0=\left(\frac{\mu_0 I}{2 R}\right)$ $\begin{aligned} & \therefore \frac{1}{8} \times\left(\frac{\mu_0 I}{2 R}\right)=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}} \\ & \Rightarrow(2 R)^3=\left(R^2+x^2\right)^{\frac{3}{2}} \\ & \Rightarrow(2 R)=\left(R^2+x^2\right)^{\frac{1}{2}} \\ & \Rightarrow(2 R)^2=\left(R^2+x^2\right) \\ & \Rightarrow x=\sqrt{3} R \end{aligned}$

Hence, the correct answer is option: 'B'.

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