Let B₁ be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B₂ be the magnitude of magnetic field at an axial distance ' x ' from the center. For x: R = 3: 4, (B₂)/(B₁) is: (a) 64: 125 (b) 25: 16 (c) 4: 5 (d) 16: 25

Answer: (a). (B₂)/(B₁) = R³(R²+x²)3/2 with x = 3/4R. R²+x² = 25/16R² ⇒ (R²+x²)3/2 = 125/64R³ (B₂)/(B₁) = 64/125 So the answer is (a) 64: 125.

Let B₁ be the magnitude of magnetic field at center of a circular… | JEE Main 2025 (2 Apr, Shift 1)

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Biot-Savart Law & Circular Coils

Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current I. Let $B_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x : R = 3 : 4,\ \frac{B_2}{B_1}$ is:

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Answer: (a)

$\dfrac{B_2}{B_1} = \dfrac{R^3}{(R^2+x^2)^{3/2}}$ with $x = \dfrac{3}{4}R$ .
$R^2+x^2 = \dfrac{25}{16}R^2 \Rightarrow (R^2+x^2)^{3/2} = \dfrac{125}{64}R^3$
$\dfrac{B_2}{B_1} = \dfrac{64}{125}$
So the answer is (a) $64 : 125$ .

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