Inductance Puzzle involving coplanar conducting rings

[Hak]

Homework Statement

We have a small ring made of thin wire having radius $R$ and its inductance is $L$. Find the inductance of a ring having $n$-times the dimensions as this ring. If in the plane of the ring, we place another superconducting ring of half the geometric dimensions so that the planes of the rings and their centers coincide, then the inductance of the ring with radius $R$ comes out to be $L_1$. What will the inductance $L_2$ of the ring with radius $R$ be when it is placed inside a superconducting ring with twice the geometric dimensions? The planes and centers of the rings also coincide in this case.

Relevant Equations

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I solved the first point as $L_n = \frac{\mu_0 n R}{2}$. How to proceed for the other point?

 

[berkeman]

Homework Statement: We have a small ring made of thin wire having radius $R$ and its inductance is $L$. Find the inductance of a ring having $n$-times the dimensions as this ring. If in the plane of the ring, we place another superconducting ring of half the geometric dimensions so that the planes of the rings and their centers coincide, then the inductance of the ring with radius $R$ comes out to be $L_1$. What will the inductance $L_2$ of the ring with radius $R$ be when it is placed inside a superconducting ring with twice the geometric dimensions? The planes and centers of the rings also coincide in this case.
Relevant Equations: /

I solved the first point as $L_n = \frac{\mu_0 n R}{2}$. How to proceed for the other point?

Welcome to PF.

Please show us your work on the first question, so we can understand your approach. Thanks.

 

[Hak]

Let us recall the formula for the inductance of a ring of radius $R$ and wire cross-sectional area $A$, as given by $L = \frac{\mu_0 N^2A}{l}$, where $\mu_0$ is the permeability of free space, $N$ is the number of turns of the wire, and $l$ is the length of the wire. For a single-turn ring, we have $N = 1$ and $l = 2 \pi R$, so the formula simplifies to $L = \frac{\mu_0 A}{2 \pi R}$. Since $A = A_{circle} = \pi R^2$, we have: $L = \frac{\mu_0 R}{2}$.

Now, if we have another ring that has $n$-times the dimensions of the original ring, that means its radius is $R_n = nR$ and its cross-sectional area is $A_n = \pi n^2 R^2 = n^2 A$. Plugging these values into the formula, we get: $$L_n = \frac{\mu_0 A_n}{2 \pi n R} = \frac{\mu_0 \pi n^2 R^2}{2 \pi n R} \Rightarrow L_n = \frac{\mu_0 n R}{2} = nL$$

 

[Hak]

Is it correct?

 

[Hak]

Is it correct?

Could someone please tell me whether this result is correct? In that case, I would post my approach to calculate $L_1$, although I firmly believe it is wrong, as it does not lead to $L_2$.

 

[Hak]

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