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Two concentric coplanar circular conducting loops have radii \(R\) and \(r(R \gg r)\). Their mutual inductance is proportional to
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Verified Answer
The correct answer is:
\(\frac{r^2}{R}\)
Let us consider a coil having \(N_1\) turns and radius \(r\) is surrounded by another coil having \(N_2\) turns and of radius \(R\) as shown in the figure below,
Magnetic field due to coil 2 at the centre,
\(B_2=\frac{\mu_0 N_2 I}{2 R}\)
Magnetic flux through the coil,
\(\phi_{B_1}=N_1 B_2 A=\frac{N_1 \mu_0 N_2 I \cdot \pi r^2}{2 R}\)
But, \(\phi_{B_1}=M I\)
From Eqs. (i) and (ii), we get
\(M I=\frac{N_1 \mu_0 N_2 I. \pi r^2}{2 R} \Rightarrow M=\frac{N_1 \mu_0 N_2 \cdot \pi r^2}{2 R} \Rightarrow M \propto \frac{r^2}{R}\)
Magnetic field due to coil 2 at the centre,
\(B_2=\frac{\mu_0 N_2 I}{2 R}\)
Magnetic flux through the coil,
\(\phi_{B_1}=N_1 B_2 A=\frac{N_1 \mu_0 N_2 I \cdot \pi r^2}{2 R}\)
But, \(\phi_{B_1}=M I\)
From Eqs. (i) and (ii), we get
\(M I=\frac{N_1 \mu_0 N_2 I. \pi r^2}{2 R} \Rightarrow M=\frac{N_1 \mu_0 N_2 \cdot \pi r^2}{2 R} \Rightarrow M \propto \frac{r^2}{R}\)
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