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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:

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15/09/2021 1:40 pm

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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:

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

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

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

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

Jaiyush19

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neet neet 2021

1 Answer

15/09/2021 1:47 pm

Correct answer: (4) \(\frac{R_2^2}{R_1}\)

Explanation:

figure.png

M = \(\frac{ϕ_{12}}{I_1}\) = \(\frac{B_1A_2}{I_1}\)

= \(\frac{\frac{(\mu_0I_1}{2R_1})(\pi R_2^2)}{I_1}\)

= \(\frac{\mu_0 \pi R_2^2}{2R_1}\)

M ∝ \(\frac{R_2^2}{R_1}\)

nimit

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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:

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