Q.Two long solenoids of the same length $l$ are wound coaxially, one inside the other, with the inner solenoid of radius $r_1$ having $n_1$ turns per unit length and the outer solenoid of radius $r_2>r_1$ having $n_2$ turns per unit length. Explain qualitatively why their mutual inductance depends on $n_1$, $n_2$, $l$ and the SMALLER radius $r_1$, and not on the larger radius $r_2$.

If current flows in the outer solenoid $S_2$. Being a long solenoid, $S_2$ on its own produces a uniform field $B_2=\mu_0n_2I_2$ everywhere throughout ITS OWN interior -- which includes the space occupied by the inner solenoid $S_1$, since $S_1$ sits entirely inside $S_2$. So every part of $S_1$'s winding does experience this field $B_2$. But the flux that actually LINKS $S_1$'s turns is worked out using $S_1$'s OWN cross-sectional area, $\pi r_1^2$ -- not $S_2$'s larger area $\pi r_2^2$ -- because flux linkage only counts where a turn of $S_1$'s wire genuinely encircles the flux; $S_1$'s wire never reaches out as far as radius $r_2$, so the field present between $r_1$ and $r_2$ (also produced by $S_2$) contributes nothing to $S_1$'s flux linkage.

If current flows in the inner solenoid $S_1$ instead. Now $S_1$'s field $B_1=\mu_0n_1I_1$ exists only WITHIN $S_1$'s own smaller radius $r_1$ (a long solenoid's field is essentially confined to its own interior); outside radius $r_1$ (in the gap out to $S_2$'s radius $r_2$) there is no field from $S_1$ at all. So the flux linking $S_2$'s (much larger) turns is again set by the SMALLER area $\pi r_1^2$ -- the only region where $S_1$'s field actually exists. …

Consider each solenoid carrying the current in turn, and identify in each case which …

• Assuming the mutual inductance should involve BOTH radii symmetrically, like an average or a sum, rather than only the smaller one. …