Q140 / 180 · KCET 2024 · Physics

The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is 64 times the magnetic field at a distance $x$ on its axis from the centre of the coil. Then, the value of $x$ is

  • A
    $\frac{R}{4} \sqrt{15}$
  • B
    $R \sqrt{3}$
  • C
    $\frac{R}{4}$
  • Correct answer
    D
    $R \sqrt{15}$

Solution

Given, $B_{\text {centre }}=64 \times B_{\text {axis }}$ $\Rightarrow \quad \frac{\mu_0 I}{2 R}=64 \cdot \frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}$ $\Rightarrow \quad\left(R^2+x^2\right)^{\frac{3}{2}}=64 R^3=(4 R)^3 \Rightarrow\left(R^2+x^2\right)^{\frac{1}{2}}=4 R$ $\Rightarrow \quad R^2+x^2=16 R^2 \Rightarrow 15 R^2=x^2$ $\Rightarrow \quad x=\sqrt{15} R=R \sqrt{15}$

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

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