Two conducting circular loops of radii r₁ and r₂ are placed in the same plane with their centres coinciding. If r₁ r₂, then the mutual inductance M between them is proportional to [ I₁ = current flowing through coil 1 and I₂ = current flowing through coil 2] (a) (r₁)/(r₂) (b) √((r₁²)/(r₂)) (c) (r₂²)/(r₁) (d) (r₂)/(r₁)

Answer: (c). Field at centre of big loop: B = (μ₀ I₁)/(2r₁), nearly uniform over the small loop since r₁ r₂ Flux through small loop: Φ = Bπ r₂² = (μ₀π r₂²)/(2r₁)I₁ M = (Φ)/(I₁) = (μ₀π r₂²)/(2r₁) ∝ (r₂²)/(r₁) So the answer is (c) (r₂²)/(r₁).

Two conducting circular loops of radii r₁ and r₂ are placed in the… | MHT CET 2026 (19 May, Shift 2)

Q10MHT CET 2026 · 19 MAY · SHIFT 2Easy

Mutual Inductance

Two conducting circular loops of radii $r_1$ and $r_2$ are placed in the same plane with their centres coinciding. If $r_1 \gg r_2$ , then the mutual inductance $M$ between them is proportional to [ $I_1 =$ current flowing through coil 1 and $I_2 =$ current flowing through coil 2]

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Answer: (c)

Field at centre of big loop: $B = \dfrac{\mu_0 I_1}{2r_1}$ , nearly uniform over the small loop since $r_1 \gg r_2$
Flux through small loop: $\Phi = B\pi r_2^2 = \dfrac{\mu_0\pi r_2^2}{2r_1}I_1$
$M = \dfrac{\Phi}{I_1} = \dfrac{\mu_0\pi r_2^2}{2r_1} \propto \dfrac{r_2^2}{r_1}$
So the answer is (c) $\dfrac{r_2^2}{r_1}$ .

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Mutual Inductance question from MHT CET 2026 (19 May, Shift 2)

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