Q 12-04-243JEE MainJEE Main 2025 (2 Apr, Shift 1)Easy
Let $B_1$ be the magnitude of the magnetic field at the center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of the magnetic field at an axial distance $x$ from the center. For $x : R = 3 : 4$, $\dfrac{B_2}{B_1}$ is:
Answer: (C) $64 : 125$
$$B_1 = \frac{\mu_0 I}{2R},\qquad B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
$$\frac{B_2}{B_1} = \left(\frac{R}{\sqrt{R^2 + x^2}}\right)^3$$
With $x = 3k$, $R = 4k$: $\sqrt{R^2 + x^2} = 5k$, so
$$\frac{B_2}{B_1} = \left(\frac45\right)^3 = \frac{64}{125}$$
Solution by Sreeraj P, M.Sc Physics