Mutual inductance \( M \) quantifies the magnetic flux linkage between two coils. It is defined as the ratio of the induced electromotive force (emf) in one coil to the rate of change of current in the other. For two coaxial solenoids, the mutual inductance \( M \) is derived as follows: Let \( N_1 \) and \( N_2 \) be the number of turns of the first and second solenoids, respectively. Let \( L \) be their common length, \( R_1 \) and \( R_2 \) be their radii, and \( \mu \) be the permeability of the interior material. The magnetic field \( B_1 \) within the first solenoid is given by \( B_1 = \frac{\mu N_1 I_1}{L} \), where \( I_1 \) is the current in the first solenoid. The flux linkage \( \Phi_2 \) through the second solenoid is \( \Phi_2 = B_1 \cdot A_2 \), where \( A_2 = \pi R_2^2 \) is the cross-sectional area of the second solenoid. Therefore, \( \Phi_2 = \frac{\mu N_1 I_1 \pi R_2^2}{L} \). The induced emf \( \mathcal{E}_2 \) in the second solenoid is \( \mathcal{E}_2 = -N_2 \frac{d\Phi_2}{dt} \). Since \( M = \frac{\mathcal{E}_2}{dI_1/dt} \), substituting for \( \mathcal{E}_2 \) yields \( M = \frac{\mu N_1 N_2 \pi R_2^2}{L} \). This equation represents the mutual inductance between two coaxial solenoids.