Solution:
The magnetic field produced by the larger loop 1 at the center is \(B_1 = \frac{\mu_0 I_1}{2 R_1}\). The magnetic flux through the smaller loop 2 is \(\phi_2 = B_1 A_2 = \frac{\mu_0 I_1}{2 R_1} \pi R_2^2\). Therefore, \(M = \frac{\phi_2}{I_1} = \frac{\mu_0 \pi R_2^2}{2 R_1}\), which means \(M \propto \frac{R_2^2}{R_1}\).
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