Q177 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
For a circular coil of radius $a$, at what distance $x$ along its axis the magnetic field is exactly $\dfrac{1}{27}$ of its value at the center?
- A$\sqrt{2}a$
- Correct answerB$2\sqrt{2}a$
- C$2a$
- D$\dfrac{1}{\sqrt{2}}a$
Solution
The magnetic field at the center of a circular coil of radius $a$ is given by $B_c = \dfrac{\mu_0 I}{2a}$.
The magnetic field at a distance $x$ along the axis is given by $B_x = \dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.
Given that $B_x = \dfrac{1}{27} B_c$, we have:
$\dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} = \dfrac{1}{27} \left( \dfrac{\mu_0 I}{2a} \right)$
$\dfrac{a^3}{(a^2 + x^2)^{3/2}} = \dfrac{1}{27}$
Taking the cube root on both sides:
$\dfrac{a}{\sqrt{a^2 + x^2}} = \dfrac{1}{3}$
Squaring both sides:
$\dfrac{a^2}{a^2 + x^2} = \dfrac{1}{9}$
$9a^2 = a^2 + x^2$
$x^2 = 8a^2$
$x = 2\sqrt{2}a$
Answer: $2\sqrt{2}a$
The magnetic field at a distance $x$ along the axis is given by $B_x = \dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.
Given that $B_x = \dfrac{1}{27} B_c$, we have:
$\dfrac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}} = \dfrac{1}{27} \left( \dfrac{\mu_0 I}{2a} \right)$
$\dfrac{a^3}{(a^2 + x^2)^{3/2}} = \dfrac{1}{27}$
Taking the cube root on both sides:
$\dfrac{a}{\sqrt{a^2 + x^2}} = \dfrac{1}{3}$
Squaring both sides:
$\dfrac{a^2}{a^2 + x^2} = \dfrac{1}{9}$
$9a^2 = a^2 + x^2$
$x^2 = 8a^2$
$x = 2\sqrt{2}a$
Answer: $2\sqrt{2}a$
Hence, the correct answer is option: 'B'.