Question 7 of 8: Self-Inductance of Two Combined Solenoids

Question 7: Self-Inductance of Two Combined Solenoids (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Single solenoid: area $A$, length $d$, turns $N$, $L_0=\mu_0N^2A/d$ (valid for $d\gg\sqrt A$, $N\gg d/\sqrt A$). Two identical solenoids of this kind are joined end-to-end into one physical solenoid of length $2d$, total turns $2N$, same turn density $n=N/d$ throughout. Case (i): both halves wound in the SAME sense (fields aid). Case (ii): the second half is wound in the OPPOSITE sense (fields oppose locally).

Find. $L_{(i)}$ and $L_{(ii)}$, the self-inductances of the combined $2d$-long, $2N$-turn solenoid in each case.

Approach. Use the magnetic energy method, $L=2U_m/I^2$ with $U_m=\int B^2/(2\mu_0)\,dV$, exploiting that in the long-thin-solenoid limit ($d\gg\sqrt A$) the interior field magnitude in each half is set by the LOCAL turn density and current only — the fringing field from the far half decays within a few radii and is negligible over a length $d$.

1. Case (i): direct formula. A single continuous winding of length $2d$ and $2N$ turns (same turn density $n=N/d$ throughout) is exactly the situation the given formula describes, just with $d\to2d$, $N\to2N$: $$L_{(i)}=\mu_0\frac{(2N)^2A}{2d}=2\,\frac{\mu_0N^2A}{d}=\boxed{2L_0}.$$
2. Case (ii): field magnitude in each half. Deep inside a long thin solenoid the field is uniform, $B\approx\mu_0nI$, set by the LOCAL current sheet; reversing the second half's winding sense flips the LOCAL field direction there to $-\mu_0nI\hat z$ but does not change its magnitude, because (for $d\gg\sqrt A$) the far half's fringing influence has decayed to a negligible correction within a few radii of the junction — far less than the length $d$ of either half.
3. Energy in each case. Since $|B|=\mu_0nI$ throughout BOTH halves in both cases (only the sign flips in case ii), the volume integral $\int B^2\,dV$ over the whole $2d$-long solenoid is identical in the two cases: $$U_m=2\times\frac{(\mu_0nI)^2}{2\mu_0}\,(Ad)=\mu_0n^2I^2Ad\qquad\text{(same expression either way).}$$
4. Case (ii) inductance. $$L_{(ii)}=\frac{2U_m}{I^2}=2\mu_0n^2Ad=2\mu_0\left(\frac{N}{d}\right)^2Ad=\frac{2\mu_0N^2A}{d}=\boxed{2L_0}.$$ So, to the leading order set by the problem's own approximation, $L_{(i)}=L_{(ii)}=2\mu_0N^2A/d$ — reversing the winding sense of one half changes the LOCAL field direction but, because mutual coupling between two remote long-thin-solenoid sections is a higher-order (fringing) effect, it does not change the self-inductance at this order. (Confirmed independently via flux linkage: each turn, referenced to its OWN winding's right-hand sense, links the same $\mu_0nIA$ regardless of which half it is in, so $\Lambda=2N\mu_0nIA=2L_0I$ in both cases.)