Mutual Inductance of Two Coaxial Solenoids

The geometry. Consider two long solenoids of the SAME length $l$, wound coaxially -- one directly inside the other, sharing a common central axis, as shown in the figure for this section. Let the inner solenoid $S_1$ have radius $r_1$ and $n_1$ turns per unit length, and the outer solenoid $S_2$ have a larger radius $r_2>r_1$ and $n_2$ turns per unit length.

Why the SMALLER radius decides the answer. Suppose a current $I_2$ flows in the outer solenoid $S_2$. Being a long solenoid in its own right, $S_2$ produces a uniform field $B_2=\mu_0 n_2 I_2$ everywhere WITHIN its own interior -- and crucially, the inner solenoid $S_1$ sits entirely within this interior, so every part of $S_1$ experiences this same field $B_2$. The flux linked with $S_1$, however, is determined by $S_1$'s OWN cross-sectional area, $\pi r_1^2$ -- NOT by the larger area $\pi r_2^2$ of $S_2$ itself -- because flux is only counted where there is actually a turn of $S_1$'s own wire for it to link. This is the essential qualitative point WBCHSE's syllabus asks for: however large the outer solenoid is made, it is always the SMALLER solenoid's cross-sectional area that limits how much flux can actually be linked between the pair.

The qualitative result and reciprocity. Working through the flux-linkage calculation (in exactly the same style as the self-inductance derivation of Section 6.7.1, but now crediting the flux produced by ONE solenoid's current to the number of turns on the OTHER solenoid) gives a mutual inductance of the form

using the inner radius $r_1$ regardless of which solenoid's current is treated as the source -- consistent with the reciprocity theorem of Section 6.7 ($M_{12}=M_{21}=M$), since swapping the roles of $S_1$ and $S_2$ in the reasoning above (current now in the INNER solenoid, field confined within it, linking only the AREA $\pi r_1^2$ that both solenoids' windings actually surround) leads to exactly the same expression. …