Step 1: Understanding the Concept:
Mutual inductance ($M$) relates the magnetic flux linked with one coil due to a current flowing in a neighboring coil.
Because $r_1 \gg r_2$ typically applies for such textbook approximations (though not explicitly stated, it's the standard derivation path), we assume the magnetic field produced by the large coil is uniform across the small area of the inner coil.
Step 2: Key Formula or Approach:
1. The magnetic field $B_1$ at the center of a circular loop of radius $r_1$ carrying current $I$ is: $B_1 = \frac{\mu_0 I}{2r_1}$.
2. The magnetic flux $\Phi_2$ passing through the smaller inner loop (radius $r_2$, Area $A_2 = \pi r_2^2$) is approximately: $\Phi_2 \approx B_1 \times A_2$.
3. Mutual inductance $M$ is defined by: $\Phi_2 = M \times I$.
Step 3: Detailed Explanation:
Calculate the magnetic field created by the larger coil at its center:
\[ B_1 = \frac{\mu_0 I}{2r_1} \]
Since $r_2$ is usually considered much smaller than $r_1$, we assume this field $B_1$ is uniform over the entire area of the smaller inner coil.
The area of the smaller coil is:
\[ A_2 = \pi r_2^2 \]
Calculate the magnetic flux linked with the smaller coil:
\[ \Phi_2 = B_1 \times A_2 \]
\[ \Phi_2 = \left( \frac{\mu_0 I}{2r_1} \right) \times (\pi r_2^2) \]
\[ \Phi_2 = \frac{\mu_0 \pi r_2^2}{2r_1} I \]
By the definition of mutual inductance ($\Phi_2 = M I$), we equate the terms:
\[ M I = \frac{\mu_0 \pi r_2^2}{2r_1} I \]
Canceling the current $I$:
\[ M = \frac{\mu_0 \pi r_2^2}{2r_1} \]
Step 4: Final Answer:
The mutual inductance between the coils is $\frac{\mu_0 \pi \text{r}_2^2}{2\text{r}_1}$.