A current carrying circular coil of radius 'R' produces magnetic field B₁ at an axial point P at a distance 'x' from its centre and B₂ at point Q placed at its centre respectively. If B₂ = 8B₁, the value of x is (a) 3R (b) √(3) R (c) (R)/(2√(3)) (d) (2R)/(√(3))

Answer: (b). On the axis B = μ₀ I R²2(R²+x²)3/2 and at the centre Bc = (μ₀ I)/(2R), so (B₂)/(B₁) = (R²+x²)3/2R³. 8 = (R²+x²)3/2R³ ⇒ R²+x² = 4R² x² = 3R² ⇒ x = √(3)R So the answer is (b) √(3) R.

A current carrying circular coil of radius 'R' produces magnetic… | MHT CET 2026 (19 Apr, Shift 2)

Q15MHT CET 2026 · 19 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A current carrying circular coil of radius 'R' produces magnetic field $B_1$ at an axial point P at a distance 'x' from its centre and $B_2$ at point Q placed at its centre respectively. If $B_2 = 8B_1$ , the value of x is

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Answer: (b)

On the axis $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ and at the centre $B_c = \dfrac{\mu_0 I}{2R}$ , so $\dfrac{B_2}{B_1} = \dfrac{(R^2+x^2)^{3/2}}{R^3}$ .
$8 = \dfrac{(R^2+x^2)^{3/2}}{R^3} \Rightarrow R^2+x^2 = 4R^2$
$x^2 = 3R^2 \Rightarrow x = \sqrt{3}R$
So the answer is (b) $\sqrt{3}\,R$ .

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Biot-Savart Law & Circular Coils question from MHT CET 2026 (19 Apr, Shift 2)

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