Question 7 of 8: Self-inductance of two stacked solenoids for like and opposed senses of circulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 16-Elec-A7 Electromagnetics. Three hours, closed book (one Casio or Sharp approved calculator). Eight questions of equal value; the rubric states that any five questions constitute a complete paper and that only the first five answered will be marked. All eight are solved here, because the set is a study resource rather than a sitting. Aids printed on the paper: $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stub networks, waveguides); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, polarisation, Biot–Savart, inductance); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (magnetostatics, guided waves); F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (wave polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields).
Constants used throughout. The aid sheet's $\varepsilon_0$ and $\mu_0$ give $c = 2.9986 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 376.82\ \Omega$, while every round figure the paper quotes (10 km at $3 \times 10^{8}\ \text{m/s}$, 300 and 600 MHz, 10 GHz giving a 3 cm wavelength) is built on $c = 3.00 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 376.99\ \Omega$. Those are the values used below; the two sets differ by less than 0.05 % and no answer or conclusion changes.
Question 7: Self-inductance of two stacked solenoids for like and opposed senses of circulation
Given. Two identical long air-core solenoids butted coaxially end to end, connected in series.
Given data
Quantity
Symbol
Value
Turns per solenoid
$N$
as given
Length of each solenoid
$d$
as given
Cross-sectional area
$A$
circular, as given
Single-solenoid inductance
$L_1$
$\mu_0 N^{2}A/d$
Combined length
—
$2d$
Combined turns
—
$2N$
Stated premises
—
$d \gg A^{1/2}$ and $N \gg d/A^{1/2}$
Find. The self-inductance of the series combination when the two currents circulate in the same sense (case i) and in opposite senses (case ii).
Figure 7.1 — the two solenoids stacked coaxially. Red arrows show the interior H: aligned in case (i), opposed in case (ii).
Approach. Apply the supplied formula to the like-sense stack read as a single solenoid, then show that the formula's own scaling forces the mutual inductance to vanish at this order — which is what makes the two cases come out equal.
Record the single-unit value. Each solenoid on its own has $L_1 = \mu_0 N^{2}A/d$. Both premises are statements that the winding is a long, densely wound current sheet: $d \gg A^{1/2}$ makes it long compared with its bore, and $N \gg d/A^{1/2}$ makes the turn pitch small compared with the bore.
Case (i): read the stack as one solenoid. With the currents circulating the same way the two windings form a single uniform current sheet of $2N$ turns over a length $2d$, with the same turn density $N/d$ and hence the same interior field $H = NI/d$ throughout. The formula applies directly:$$L_{(i)} = \frac{\mu_0 (2N)^{2}A}{2d} = \boxed{\frac{2\mu_0 N^{2}A}{d} = 2L_1}$$
Read off the mutual inductance. The series combination also obeys $L = L_1 + L_2 + 2M = 2L_1 + 2M$. Comparing with the previous step gives $M = 0$ to the accuracy of the supplied formula. That is not an accident: because $L \propto N^{2}/d$, doubling both $N$ and $d$ exactly doubles $L$, and pure additivity is precisely the statement that the halves are uncoupled.
Confirm it physically. Outside a long solenoid the field falls away over a distance of order the bore radius $a = (A/\pi)^{1/2}$, so the flux one half sends through the other is smaller than its own flux by a factor of order $a/d$. Hence $M/L_1 = \mathcal{O}(a/d)$, which the premise $d \gg A^{1/2}$ makes negligible. Each half therefore links essentially only its own flux.
Case (ii): reverse one winding. Now the two interior fields oppose, so the stack is no longer a uniform solenoid. But each half still produces $H = NI/d$ inside itself, and each half's turns still link that flux with the sense of their own current, so each contributes $N \times \mu_0 (NI/d)A$ to the flux linkage exactly as before:$$L_{(ii)} = L_1 + L_2 - 2M = 2L_1 - 0 = \boxed{\frac{2\mu_0 N^{2}A}{d} = 2L_1}$$Reversing the sense flips the sign of the mutual term, but that term is zero, so the inductance is unchanged. The two cases are equal.
Put numbers on it. Taking an illustrative $N = 500$, $A = 4\ \text{cm}^{2}$ and $d = 20\ \text{cm}$ (bore radius $a = 1.13\ \text{cm}$, so $d/a = 17.7$ and $d/A^{1/2} = 10 \ll N$, both premises comfortably met): $L_1 = 0.628\ \text{mH}$, and both cases give $1.257\ \text{mH}$. The neglected mutual term is of order $(a/d)L_1 = 0.056 \times 0.628 = 0.035\ \text{mH}$, i.e. roughly $\pm 3\ \%$ on the combined value — comfortably inside the "approximately" of the formula the paper supplies.
Check: the equality of the two cases holds under the paper's own stated premises, which make the halves magnetically independent. It is not a universal result. For short, fat coils the mutual inductance is a real fraction of $L_1$ and the two cases differ by $4M$; and if the two coils shared a closed magnetic circuit (a core ring rather than air), the coupling would approach unity and the opposed case would collapse toward zero while the like case approached $4L_1$.
