Mutual inductance between an infinite solenoid and a surrounding loop—A paradox resolved

Misconceptions or errors in physics understanding often lead to paradoxes. This is particularly true in electromagnetism where errors appear to be of two types, which I will call “incomplete system” and “unjustified simplification.” Examples of the former can be found in the paradox of a force on a stationary charge due to a current-carrying wire,¹ where the paradox arises from neglect of the part of the system composed of the source and sink of current, and in the paradox of non-conservation of angular momentum,² which arises from neglect of the angular momentum of electromagnetic fields. An example of the second type is the paradox of differing readings on different voltmeters connected between the same points of a circuit.³ The simplification of equal values for the relevant line integrals, which leads to equal readings, is unjustified in a multiply connected region. In this note, I discuss another example of the latter type of error, the paradox here being that, rather than giving the wrong answer, the unjustified simplification gives the correct one.

In the undergraduate electromagnetism curriculum, as an application of Faraday's law, students are often asked to find the mutual inductance M₁₂ of the axially symmetric system shown in Fig. 1, where a varying current, $I 2$⁠, in the outer circular loop (the primary) induces an emf in the inner, co-axial and closely wound circular solenoid (the secondary) composed of n circular loops per unit length. The center of the outer loop defines the origin, and we assume that the radius R₂ of the loop is greater than the radius R₁ of the solenoid.

In the author's experience, students attempting a solution fall into two categories: those who use Method A—solving the problem as it stands but by incorrectly assuming that the field due to the loop takes its on-axis value everywhere inside the solenoid, and those who use Method B—solving the reciprocal problem with the solenoid as the primary and the loop as the secondary.

Method B considers the reciprocal system where a current $I 1$ in the solenoid produces a (constant) field B₁ =  $μ 0 n I 1$ inside the solenoid (here the capital B denotes the field due to an infinite stack of turns), producing a flux Φ₂₁ =  $μ 0 n I 1 π R 1 2$ through the loop, and leading to the mutual inductance M₂₁ =  $μ 0 nπ R 1 2$⁠. Using the principle of reciprocity⁴ one obtains M₁₂ = M₂₁ =  $μ 0 nπ R 1 2$⁠.

There is no doubt that Method B is correct, based as it is on sound fundamental principles. However, Method A's result is the same as Method B's result. Imagine a Method A solver's surprise, dismay, and confusion when told that their answer is correct but that their method is not justified. They might reasonably ask, “Why am I wrong?” (Awarding marks here presents an interesting challenge for the instructor.)

As far as the author is aware, this apparent paradox has never been noted in the treatment of mutual inductance in the literature.

For z = 0, the radial minimum in b_2z occurs on the axis, with the value at the minimum decreasing with z as expected. However, this position of the radial minimum only persists out to an axial distance |z| ≈ 0.8R₁. Beyond this distance, the on-axis value of b_2z becomes a maximum. This observation should reduce the student's surprise and eliminate the paradox, since using the axial field underestimates the flux for |z|< 0.8a but overestimates it for larger values of |z|, making it possible for the axial field to correctly predict the flux. The careful calculation of the flux requires a quadruple integral⁵ and yields M₁₂ =  $μ 0 nπ R 1 2$ in agreement with Eqs. (2) and (3) (see the supplementary material).⁶ 

While the agreement between Methods A and B is explained by this integration, one cannot avoid the feeling that there must be a deeper reason. An explanation is provided here, which applies even if the cross sections are non-circular, as long as the solenoid has translational symmetry and the loop lies in a plane perpendicular to the axis of translational symmetry (taken to lie within the solenoid). Under these conditions is is straightforward to show using Ampere's law that the magnetic field inside the solenoid is uniform and axial.^7,8

Here is the key insight. Even though they represent two different situations, the integral in Eq. (6) is mathematically identical to the interior integral in Eq. (5). Since we know from Ampère's law that the integral in Eq. (6) yields B₂ =  $μ 0 n I 2$⁠, independent of the position $(ρ, ϕ)$ inside the loop, it follows that the interior integral in Eq. (5) is also equal to $μ 0 n I 2$⁠, independent of $(ρ, ϕ)$⁠. That makes the full integral in Eq. (5) trivial to compute: the flux $Φ 12 = μ 0 n I 2 S 1$⁠.

Because the integrated effect of the single-loop field in Eq. (6) has a form independent of ρ and ϕ, any combination of the coordinates lying within the cross section of the solenoid could be used to evaluate the integral in Eq. (5). The simplest choice is ρ = 0 (i.e., using the on-axis field of the original loop, replacing $b 2 z (ρ, ϕ, z)$ in Eq. (5) by $b 2 z (z)$ as given by Eq. (1)), leading to a trivial integration over ρ and ϕ. This was the implicit choice of students using Method A, and it works because of the translational symmetry of the original solenoid. Generally, given the field distribution as a function of z along any axis parallel to the axis of translational symmetry, Eq. (5) can be evaluated to give the correct flux.

This result is applied to the example of a square loop in supplement 2 of the supplementary material. The proviso that the solenoid lies within the loop for the success of Method A translates into R₁ < R₂. If it were otherwise, then as pointed out in supplement 3 of the supplementary material for the example of the analogous scenario of Fig. 1, it would not be clear whether the integrated on-axis loop field should be multiplied by πR₁² or πR₂² to obtain $Φ 12 .$

What do we conclude from all this? Including the first-principles calculation, we now have three methods at our disposal, all giving the same correct answer. Which of these should be adopted? Although the first-principles calculation is sound and yields the correct answer, we immediately rule it out because it involves a quadruple integration. In deciding between Method A and the reciprocity method (B), the latter wins “hands down” because, notwithstanding its correct result, Method A contains an “unjustified simplification” type of misconception. The result of this method can only be justified and trusted with the hindsight of the reciprocity result.

However, there is a more important principle at stake if Method A is accepted. In the example of the configuration dual to Fig. 1, where the loop lies within the solenoid, supplement 3 of the supplementary material shows that this method is ambiguous with regard to which cross-sectional area should be used to obtain $Φ 12$⁠. Allowing the use of Method A to go unchallenged might lead its adherents to the erroneous belief that their method is always applicable, but, as we have seen, this cannot always be guaranteed.

The author acknowledges several anonymous students whose assignment solutions inspired this article. He also acknowledges the valuable comments and suggestions of the anonymous reviewers.

The author has no conflicts of interest to disclose.