Two conducting circular loops of radii ' R₁ ' and ' R₂ ' are placed in the same plane with their centres coinciding. If R₁ > R₂, the mutual inductance M between them will be directly proportional to (a) (R₁)/(R₂) (b) (R₂)/(R₁) (c) (R₁²)/(R₂) (d) (R₂²)/(R₁)

Answer: (d). Field of loop R₁ at centre: B = (μ₀ I)/(2R₁), taken uniform over loop R₂ Φ = Bπ R₂² ⇒ M = (μ₀π R₂²)/(2R₁) So M ∝ (R₂²)/(R₁); the answer is (d).

Two conducting circular loops of radii ' R₁ ' and ' R₂ ' are placed… | MHT CET 2023 (10 May, Shift 1)

Q41MHT CET 2023 · 10 MAY · SHIFT 1Easy

Mutual Inductance

Two conducting circular loops of radii ' $R_1$ ' and ' $R_2$ ' are placed in the same plane with their centres coinciding. If $R_1 > R_2$ , the mutual inductance $M$ between them will be directly proportional to

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Answer: (d)

Field of loop $R_1$ at centre: $B = \dfrac{\mu_0 I}{2R_1}$ , taken uniform over loop $R_2$
$\Phi = B\pi R_2^2 \Rightarrow M = \dfrac{\mu_0\pi R_2^2}{2R_1}$
So $M \propto \dfrac{R_2^2}{R_1}$ ; the answer is (d).

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Mutual Inductance question from MHT CET 2023 (10 May, Shift 1)

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