A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The center of the small loop is on the axis of the bigger loop. The distance between their centers is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is (a) 9.1 × 10⁻¹¹ weber (b) 6 × 10⁻¹¹ weber (c) 3.3 × 10⁻¹¹ weber (d) 6.6 × 10⁻⁹ weber

Answer: (a). Use M₁₂ = M₂₁: find flux through the small loop when current I flows in the big loop. On the axis: B = μ₀ I R²2(R²+x²)3/2, with R = 0.2, x = 0.15: (R²+x²)3/2 = 0.06253/2 = 0.015625 M = μ₀ R² π r²2(R²+x²)3/2 = 4π×10⁻⁷×0.04×π×9×10⁻⁶0.03125 ≈ 4.55×10⁻¹¹ H Flux in big loop = M I = 4.55×10⁻¹¹×2 ≈ 9.1×10⁻¹¹ Wb So the answer is (a).

A circular loop of radius 0.3 cm lies parallel to a much bigger… | JEE Main 2013

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Mutual Inductance

A circular loop of radius $0.3\text{ cm}$ lies parallel to a much bigger circular loop of radius $20\text{ cm}$ . The center of the small loop is on the axis of the bigger loop. The distance between their centers is $15\text{ cm.}$ If a current of $2.0\text{ A}$ flows through the smaller loop, then the flux linked with bigger loop is

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Answer: (a)

Use $M_{12} = M_{21}$ : find flux through the small loop when current $I$ flows in the big loop.
On the axis: $B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ , with $R = 0.2$ , $x = 0.15$ : $(R^2+x^2)^{3/2} = 0.0625^{3/2} = 0.015625$
$M = \frac{\mu_0 R^2 \pi r^2}{2(R^2+x^2)^{3/2}} = \frac{4\pi\times10^{-7}\times0.04\times\pi\times9\times10^{-6}}{0.03125} \approx 4.55\times10^{-11}\text{ H}$
Flux in big loop $= M I = 4.55\times10^{-11}\times2 \approx 9.1\times10^{-11}\text{ Wb}$
So the answer is (a).

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