10
$\begingroup$

I would like to get some elementary intuition into the problem a solenoid fed with a time-dependent current, and the resulting current that such the solenoid field would induce in a loop completely outside the solenoid.

The classic situation is to imagine a perfect solenoid of $n_1$ turns per meter, having radius $a$ and fed by a current $I(t)=I_0e^{-t/\tau}$. One places $N_2$ coils (all having radius $R>a$) around the solenoid as illustrated by this picture from Haliday&Resnick 10th Edition.

enter image description here

Using the usual Ampère's law argument shows the $\vec B$ field of the solenoid would be completely contained inside the solenoid, and homogenous across the cross section of the solenoid.

If there is a slowly varying current in the solenoid, will there be an induced current?

Presumably the answer is yes (at least according the HR solution manual): if we take the $\vec B$ field of the solenoid to be changing with time as a result of the current changing in time, but $\vec B$ still homogenous across its cross-section and $0$ outside, there is a change in the flux of this field through the surface bounded by the coils, and even though there is no magnetic field outside the solenoid where the coils are located. The resulting change in flux induces an EMF independent of $R$, the radius of the coils, or for that matter the shape of the coil, provided it completely contains the solenoid.

If you are suspicious about the use of Ampère's law for slowly varying currents, the same solenoid field is obtained explicitly by Das Gupta in "Magnetic field due to a solenoid." American Journal of Physics 52 (1984): 258-258, starting from Biot-Savart, which certainly holds for quasi-static currents.

But how can one intuitively grasp that no field outside the solenoid induces a current in a stack of coils located completely outside this solenoid, in the region where $\vec B=0$?

A "classic" explanation might that the field might be $0$ but the vector potential is $\ne 0$, yet it seems this is invoking a lot of heavy machinery for a 1st year physics problem. Moreover, Aharonov-Bohm-like explanations are really quantum in their nature and show that in quantum mechanics the potentials are the essential quantities.

Nota: a possible path of solution would be to invoke hidden momentum, along the argument in this file by K.T. McDonald. Is there a simpler explanation?

$\endgroup$
5
  • 2
    $\begingroup$ The induced E-field extends outside the solenoid according to Faraday's Law. $\endgroup$
    – K_inverse
    Mar 18, 2019 at 2:54
  • $\begingroup$ I have always thought that one can never explain this problem satisfactorily without using the vector potential to be the underlying cause of induction and introduce it as a real field not just an auxiliary mathematical tool to solve a differential equation. If you do so then there is no problem because $\textbf{A}$ extends all over the space. The handwaving starts when you try to explain (away) why only its line integral is significant... $\endgroup$
    – hyportnex
    Mar 18, 2019 at 12:11
  • $\begingroup$ there is no heavy machinery needed to introduce $\textbf{A}$ becasue it is related to $\textbf{J}$ in a completely analogous way $\phi$ is related to $\rho$ $\endgroup$
    – hyportnex
    Mar 18, 2019 at 12:16
  • $\begingroup$ @hyportnex The point here is that the potential vector is not a usual concept presented in 1st year physics, so I'm looking for an explanation that would avoid this, if there is such an explanation. $\endgroup$ Mar 18, 2019 at 12:36
  • $\begingroup$ If you stick to that point then I think you will be stuck with action at distance explanations and this I find more difficult to swallow than the potentials $\phi$ and $\textbf{A}$. After all, $\phi$ is usually introduced immediately after charge, Coulomb law, etc. Why not do the same for current and its vector potential, they will look very similar. $\endgroup$
    – hyportnex
    Mar 18, 2019 at 12:47

2 Answers 2

1
$\begingroup$

There is a magnetic field outside the solenoid and as the current changes there is a change in the magnetic field strength both inside and outside the solenoid.

This animation from MIT shows the effect of increasing the speed of positive charges (current) in the coils and here are three stills from it.

enter image description here

I think the point is that the magnetic field which is "generated" as the current increases in "compressed" within the solenoid but spreads out outside the solenoid so this is where the idea of no magnetic field outside the solenoid comes from?
In this way anything outside the solenoid, eg your outer coil, will know what is going on inside the solenoid because of the flow of electromagnetic energy out of the coil.


I am tempted to think of magnetic field lines in the form of loops passing through the inside and then outside of the solenoid being created and the part formed outside the solenoid flowing out from the solenoid?

$\endgroup$
8
  • $\begingroup$ This is very nice but this is not an ideal soleinoid since the winding is not tight; in an ideal solenoid there would be no field outside. Would you know what happens when we increase the number and density of rings? Basically the ideal solenoid doesn't have any space between the rings... still this is an interesting animation that would certainly supply an answer for the non-ideal case. $\endgroup$ Mar 18, 2019 at 15:36
  • $\begingroup$ @ZeroTheHero Would not the magnetic field lines being loops mean that there would still be a (very small) magnetic field outside the solenoid? $\endgroup$
    – Farcher
    Mar 18, 2019 at 15:40
  • $\begingroup$ For an ideal solenoid no it would be exactly $0$. Maybe the Ampère's law argument is somewhat overly qualitative but the more mathematical derivation (see the Dasgupta paper for instance) is airtight. Agree that if one could show some leakage in the case of time-varying current that would be a great step forward. $\endgroup$ Mar 18, 2019 at 17:13
  • $\begingroup$ @ZeroTheHero I think that the Dasgupta paper deals with an infinite solenoid? $\endgroup$
    – Farcher
    Mar 18, 2019 at 17:26
  • 1
    $\begingroup$ Even moderatly dense and moderately long solenoids have very little external field. Something I used to make student measure directly (Hall probes cheap enough to use in undergraduate lab are so cool). And of course the theory claims that you get induced current even with the ideal device (checked out from the physics stock room right next to the massless, rigid, semi-infinite rods, of course). $\endgroup$ Mar 19, 2019 at 0:30
1
$\begingroup$

There’s an azimuthal electric field outside of the solenoid. This electric field is what causes current to flow through the outer coils (e.g. as given by Ohm’s law $\bf{J} = \sigma \bf{E}$).

There's no single physical explanation for the origin of this external electric field; "why" questions in physics rarely have a single answer if there are multiple mathematical derivations of the same result, and it's often a matter of personal opinion which derivation is the most illuminating. Here are four possible arguments/explanations for this external electric field:

  1. The integral form of Faraday's law. (This is the most mathematically straightforward derivation - it's essentially the same one that you laid out in your question, with the additional identification of the emf with $\oint {\bf E} \cdot d{\bf l}$ - but as you say, it doesn't provide much physical intuition because it isn't manifestly local.)
  2. The differential forms of Faraday's law and Gauss's law for electricity require that outside of the solenoid, any electric field must be both curl-free and divergence-free. Zero electric field would trivially satisfy these requirements, but it wouldn't satisfy the boundary conditions at the solenoid: there's an azimuthal electric field inside of the solenoid induced locally by the changing internal magnetic field via Faraday's law, and this internal electric field is nonzero at the solenoid. And there's no net charge at the solenoid, so the electric field must be continuous across it, so it must be nonzero and azimuthal just outside of the solenoid and decay with distance away from it.
  3. The magnetic vector potential changes over time outside of the solenoid, and so ${\bf E} = -\partial {\bf A}/ \partial t$ (using temporal gauge for simplicity) cannot be zero. (This explanation is not necessary for explaining this phenomenon, though.)
  4. The Jefimenko's equation for ${\bf E}$ contains a term that in this case simplifies to $-\iint_S d^2 {\bf r}'\ \dot{\bf K}({\bf r}', t_r)/|{\bf r} - {\bf r}'|$, where ${\bf K}$ is the surface current at the solenoid and $t_r$ is the retarded time. Therefore the electric field can be thought of as being sourced by a changing current, and in this case the changing current through the solenoid induces an external electric field. (Note that Jefimenko's equations don't apply exactly in the quasistatic regime in which this effect was derived, but presumably a similar term is found in the quasistatic equivalent to Jefimenko's equations, which I suspect would be quite challenging to derive.)

The fourth explanation is the most physically direct (although the most mathematically complex) and so possibly the most satisfying, although making it completely rigorous would require a challenging calculation.

$\endgroup$
10
  • $\begingroup$ can you expand a bit? If there’s no $\vec B$ what is the origin of this $\vec E$ field? $\endgroup$ Sep 4, 2021 at 13:23
  • $\begingroup$ @ZeroTheHero Expanded. $\endgroup$
    – tparker
    Sep 4, 2021 at 16:32
  • $\begingroup$ @ZeroTheHero Does this address your question? $\endgroup$
    – tparker
    Sep 7, 2021 at 2:42
  • $\begingroup$ Not sure. I found some Am.J.Phys papers by Griffiths which go in the direction of 2. I still have to read them. $\endgroup$ Sep 7, 2021 at 5:03
  • $\begingroup$ @ZeroTheHero Oh wait, I think I now understand the point of your question. Are you trying to understand how Faraday's law (which can be expressed in a local way) can allow an electric field to be induced in a region (outside of the solenoid) where there's no magnetic field at all? This question was considered in aapt.scitation.org/doi/10.1119/1.18033, with some interpretational discussion at aapt.scitation.org/doi/10.1119/1.18379. $\endgroup$
    – tparker
    Sep 8, 2021 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.