The inductance of a long solenoid with \$N_o\$ turns and a length \$l_o\$ is $$L_o=\pi r^2 \mu_0\frac{N_o^2}{l_o}$$

If I now make a new solenoid, \$L_{new}\$, with double the length of the original (\$l_{new}=2l_o\$) and double the number of turns as well (in order to keep the turns per unit length constant), I clearly have a new inductor with double the inductance of the original solenoid so that \$L_{new}=2L_o\$. But doubling the length with constant \$\frac{N}{l}\$ is the same as putting two long solenoids of inductance \$L_o\$ together in series (with perfect coupling). Yet I am told that when two inductors (\$L_1\$ and \$L_2\$) are placed in a series aiding configuration, the combined inductance is (Fundamentals of electric circuits, by Alexander): $$L_{Tot}=L_1+L_2+2M$$ where \$M\$ is the mutual inductance between the two inductors with \$M=k\sqrt{L_1\cdot L_2}\$ (\$k\$ being the coupling coefficient). Suppose these inductors both have inductance \$L_o\$. Then when placed in a series aiding fashion, we have $$L_{Tot}=2L_o+2\cdot k\cdot L_o$$ The further these inductors are apart (still being in series though), presumably the smaller the coupling coefficient \$k\$ so that in the limit that they are infinitely far apart, we have \$k=0\$ and \$L_{Tot}=2L_o\$. But conversely, the closer they are the larger \$k\$ gets, so that when the end of the first just touches the beginning of the second, we surely have \$0 <k\leq 1\$. Thus, in this case we must have that \$L_{Tot}>2L_o\$. But this case is identical to the first case where we simply doubled the length of the original inductor and doubled the number of turns to get \$L_{new}=2L_o\$ so we have a contradiction. How can this be? How can we get two different values for the inductance of the same solenoid?


3 Answers 3


Regarding the air solenoid described by your first formula: -

  1. An air solenoid will not have 100% coupling between successive turns.
  2. It might be about 95% for two fairly tightly wound turns
  3. The coupling between turns spaced further apart drops significantly
  4. Turns at one end and turns at the other end have a coupling that might be less than k = 0.01

So, given what I've said above, extending the length of the solenoid by adding more turns is quite literally adding two series inductors with minimal influence of the coupling factor.

On the other hand, an inductor that is wound around a ferrite core will have a fairly decent coupling between all the turns that are used. This is because a ferrite core of high permeability will concentrate the flux. But, it's still an open-ended solenoid and will still return flux (between north to south) through the air.

However, when you refer to this formula: -


It's a totally different game. That formula can describe quite closely coupled inductors where the flux produced by one inductor is very-well coupled to the flux of the other. That is not what can be achieved with an air solenoid.

In short, you are comparing apples with pears.

  • \$\begingroup\$ Thanks for the response. You've pretty much solved my issue for air core inductors. My understanding is that the formula \$L=\pi r^2 \mu_0\frac{N^2}{l}\$ should still apply to an iron core inductor, However, we need to replace \$\mu_0\$ with \$\mu_{iron}\$. If we do this then we should still end up with the same inconsistency. The formula \$L=\pi r^2 \mu_{iron}\frac{N^2}{l}\$ will yield only \$2L_0\$ when the two inductors meet but the formula \$L_{tot}=L_1+L_2+2M\$ will yield greater than \$2L_0\$ because now \$k\$ is no longer negligible given that we have an iron core? \$\endgroup\$ Commented Sep 12, 2022 at 12:14
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    \$\begingroup\$ I don't think so. As mentioned above "But, it's still an open-ended solenoid and will still return flux (between north to south) through the air" <-- this means there will still be local flux paths that don't couple end-to-end. It's not as profound as an air solenoid but it's still not as good as a closed core with decent permeability @SalahTheGoat \$\endgroup\$
    – Andy aka
    Commented Sep 12, 2022 at 12:18

"Perfect coupling" doesn't mean quite what you think it means :)

See this figure from Radiotron Designer's Handbook 4th ed.:

Solenoid coupling (RDH4)

For two solenoids butted end-to-end, S = 0 (top curve) and the coupling factor is given by the L/D ratio. k approaches 1 only for \$L \approx 0\$, which is to say, a something like a two-turn loop (one turn for each solenoid).

The coupling factor is only 1 if the wires occupy literally the same space. The next best you can get, for simple wire helices, is to thread them together (either coaxially as two layers, one on top of the other; or interleaved as a double helix).

In which case, the leakage inductance can be approximated as the inductance of the corresponding length of twin-lead wire. Notice that, in either case: since turns, wire dia., and helix radius (nearly), are all equal, we can unwind the pair of wires together as a twin-lead transmission line. Which therefore has some characteristic impedance and velocity factor, and corresponding capacitance and inductance per length.

Leakage can then be converted back into coupling factor as: \$k = \sqrt{1-\frac{L_L}{L_M}}\$, for leakage \$L_L\$ and magnetizing inductance (the total inductance of one solenoid) \$L_M\$.

  • \$\begingroup\$ Thanks for the response. According to your figure, when \$S=\infty \$, \$k=0\$ and so by my equation \$L_{Tot}=2L_o+2\cdot k\cdot L_o=2L_o\$. Furthermore, lets assume \$L_o\$ is non zero. Now according to your figure again, if \$S=0\$ whilst \$L_o\$ is non zero, \$k\$ is not quite equal to 1 but still greater than 0 . Hence we still get that the inductance of two identical, finite, coaxial inductors placed end-to-end is larger than \$2L_o\$. Yet by the first equation in my OP, it should be equal to exactly \$2L_o\$. How can we get different inductances for the same inductor? \$\endgroup\$ Commented Sep 12, 2022 at 11:00
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    \$\begingroup\$ My mistake -- I shouldn't say anything about L, actually. A "long solenoid" means L >> D. This is the assumption which underlies the formula you used, so it is misapplication to consider the coupling between ends of such solenoids -- consider two infinitely long solenoids, how could you wrangle them into position anyway, you can never reach the ends. :) \$\endgroup\$ Commented Sep 12, 2022 at 11:02
  • \$\begingroup\$ Okay so the first formula simply fails in this situation then? This link is a visual representation of my issue ( imgur.com/V74wqZt ). Why would the formula fail though? I am still examining only solenoids whose lengths are substantially longer than their diameters. So the first formula should still apply? \$\endgroup\$ Commented Sep 12, 2022 at 11:15
  • \$\begingroup\$ In the diagram labelled (C) in my image above, the formula for \$L_o\$ must definitely apply because we have two inductors, both with length>>Diameter and no mutual inductance. Similarly, in the diagram labelled (D), the Formula for \$L\$ must again definitely apply because we now have one inductor with an even greater length/diameter ratio. Clearly the formula fails for situations in between these two extremes. But if we imagine the separation between the two inductors to be infinitesimally small, then the formula should still approximately apply and we should get \$2L_o\$, but the other ... \$\endgroup\$ Commented Sep 12, 2022 at 11:32
  • \$\begingroup\$ ... formula still gives us an inductance larger than \$2L_o\$ because \$k \neq 0\$ \$\endgroup\$ Commented Sep 12, 2022 at 11:32

You're cheating yourself in your simplifications. :-) You correctly write:

$$\begin{align} L_{\text{tot}}&=L_1+L_2+2M \\ &\stackrel{L_1=L_2=L_0}{\Rightarrow} \\ L_{\text{tot}}&=2L_0+2kL_0 \tag{1} \end{align}$$

But now you consider that, when \$k\rightarrow 1\$ you get \$L_{\text{tot}}\rightarrow 2L_0\$, but you're forgetting the \$2\$ from the other term:

$$\begin{align} L_{\text{tot}}&=2L_0+2kL_0 \\ &=2(L_0+kL_0) \\ &\begin{cases} k\rightarrow 0\quad\Rightarrow\quad L_{\text{tot}}=2(L_0+0\cdot L_0)=2L_0 \\ k\rightarrow 1\quad\Rightarrow\quad L_{\text{tot}}=2(L_0+1\cdot L_0)=4L_0 \end{cases} \tag{2} \end{align}$$

So it checks out. Of course, as the others have pointed out, you will never have perfect coupling, so this is the ideal case (which is seemingly what you're after).


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