1. R₁/R₂ 2. R₂/R₁ 3. R₁²/R₂ 4. R₂²/R₁

Answer: 4. R₂²/R₁ Key idea: the large loop R₁ produces a nearly uniform field at its centre, which passes through the small loop. B=μ₀I/2R₁ , flux through small loop φ=Bπ R₂²=μ₀Iπ R₂²/2R₁ . M=dfracφ I=μ₀π R₂²/2R₁∝R₂²/R₁ . Option 4.

Topic: Self and mutual inductance· Electromagnetic Induction and Alternating Currents

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 :

1. \(\frac{R_1}{R_2}\)

2. \(\frac{R_2}{R_1}\)

3. \(\frac{R_1^2}{R_2}\)

4. \(\frac{R_2^2}{R_1}\)

Doubtly Asked 30 Oct 2026 NEET UG 2021 Practice test neet Save WhatsApp Report

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Answer: 4. \(\frac{R_2^2}{R_1}\)

Key idea: the large loop \(R_1\) produces a nearly uniform field at its centre, which passes through the small loop.

\(B=\dfrac{\mu_0I}{2R_1}\), flux through small loop \(\phi=B\pi R_2^2=\dfrac{\mu_0I\pi R_2^2}{2R_1}\).

\(M=\dfrac\phi I=\dfrac{\mu_0\pi R_2^2}{2R_1}\propto\dfrac{R_2^2}{R_1}\). Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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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