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 be - Physics MCQ NEET 2021 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Let current I_{1} flow in the outer loop.Magnetic field at center B_{1} = \frac{μ _{0} I _{1}}{2 R _{1}} .Flux through inner loop ϕ_{2} = B_{1} A_{2} = \frac{μ _{0} I _{1}}{2 R _{1}} \times π R_{2}^{2} .Mutual Inductance M = \frac{ϕ _{2}}{I _{1}} = \frac{μ _{0} π R _{2}^{2}}{2 R _{1}} .So, M \propto \frac{R _{2}^{2}}{R _{1}} .

Try it yourself before checking the explanation above.

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:

\frac{R _{1}}{R _{2}}

\frac{R _{2}^{2}}{R _{1}}

\frac{R _{1}}{R _{2}}

\frac{R _{2}}{R _{1}}

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