Q132 / 155 · WBJEE 2026 · Physics
A circular coil, carrying current, has radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\dfrac{1}{27}$ th of its value at the centre of the coil is
- Correct answerA$2\sqrt{2}R$
- B$3\sqrt{2}R$
- C$3R$
- D$2\sqrt{3}R$
Solution
The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is given by
$B_c = \dfrac{\mu_0 I}{2R}$
The magnetic field at a distance $x$ on the axis of the coil is given by
$B_a = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$
Given that $B_a = \dfrac{1}{27} B_c$, we have
$\dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \dfrac{1}{27} \left( \dfrac{\mu_0 I}{2R} \right)$
$\dfrac{R^2}{(R^2 + x^2)^{3/2}} = \dfrac{1}{27R}$
$\dfrac{R^3}{(R^2 + x^2)^{3/2}} = \dfrac{1}{27}$
Taking the cube root on both sides, we get
$\dfrac{R}{\sqrt{R^2 + x^2}} = \dfrac{1}{3}$
Squaring both sides, we get
$\dfrac{R^2}{R^2 + x^2} = \dfrac{1}{9}$
$9R^2 = R^2 + x^2$
$x^2 = 8R^2$
$x = 2\sqrt{2}R$
Answer: $2\sqrt{2}R$
$B_c = \dfrac{\mu_0 I}{2R}$
The magnetic field at a distance $x$ on the axis of the coil is given by
$B_a = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$
Given that $B_a = \dfrac{1}{27} B_c$, we have
$\dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \dfrac{1}{27} \left( \dfrac{\mu_0 I}{2R} \right)$
$\dfrac{R^2}{(R^2 + x^2)^{3/2}} = \dfrac{1}{27R}$
$\dfrac{R^3}{(R^2 + x^2)^{3/2}} = \dfrac{1}{27}$
Taking the cube root on both sides, we get
$\dfrac{R}{\sqrt{R^2 + x^2}} = \dfrac{1}{3}$
Squaring both sides, we get
$\dfrac{R^2}{R^2 + x^2} = \dfrac{1}{9}$
$9R^2 = R^2 + x^2$
$x^2 = 8R^2$
$x = 2\sqrt{2}R$
Answer: $2\sqrt{2}R$
Hence, the correct answer is option: 'A'.