# Simple explanation to the induction from the slowly changing $\vec B$ of a solenoid in the region of $0$ magnetic field

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. 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?

• The induced E-field extends outside the solenoid according to Faraday's Law. – K_inverse Mar 18 '19 at 2:54
• 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... – hyportnex Mar 18 '19 at 12:11
• 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 $\rho$ – hyportnex Mar 18 '19 at 12:16
• @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. – ZeroTheHero Mar 18 '19 at 12:36
• 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. – hyportnex Mar 18 '19 at 12:47

## 1 Answer

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.

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?

• 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. – ZeroTheHero Mar 18 '19 at 15:36
• @ZeroTheHero Would not the magnetic field lines being loops mean that there would still be a (very small) magnetic field outside the solenoid? – Farcher Mar 18 '19 at 15:40
• 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. – ZeroTheHero Mar 18 '19 at 17:13
• @ZeroTheHero I think that the Dasgupta paper deals with an infinite solenoid? – Farcher Mar 18 '19 at 17:26
• 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). – dmckee --- ex-moderator kitten Mar 19 '19 at 0:30