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I came across an example in my book where it has the changing flux from a solenoid passing through a larger ring at the end.

Here's a picture: enter image description here

How does the large loop (or radius $r_1$) even “know” what the flux is, while we were told to assume that the $B$ field due to the solenoid was negligible outside of the smaller cross section (when $r > r_2$), there is no field actually going through the material of the larger conducting loop.

If I understand correctly, it can be said that an electromotive force (emf) is induced due to the Lorentz force on the charge carries, but this would require the electrons to be inside a varying magnetic field which isn't the case here, $B(r = r_1) \approx 0$.

All we have is the Faraday's law, $$ -\int \frac{d \mathbf B}{dt} \cdot d \mathbf s = -\frac{dB}{dt} \pi r_2 ^2 = \mathcal{E}_{EMF} \, ; $$ nowhere does this depend on the size of our loop ($r_1$) in which the emf is being induced. It seems to me there is no way for the loop to “feel” the magnetic flux.

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    $\begingroup$ Ho boy, just wait until you learn about the Aharanov-Bohm effect! $\endgroup$ – DanielSank Dec 1 '15 at 2:46
  • $\begingroup$ related physics.stackexchange.com/questions/156726/… $\endgroup$ – Paul Dec 1 '15 at 3:39
  • $\begingroup$ I'm not sure if it's really possible to answer this without starting from the Coulomb law for electrostatic fields of a charge and then showing how all the forces work when you take special relativity into account. Astrum, have you read Purcell's E&M book? $\endgroup$ – DanielSank Dec 1 '15 at 9:21
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    $\begingroup$ What's now known as the Aharonov-Bohm effect was actually predicted in Ehrenberg and Siday's 1949 semi-classical paper The Refractive Index in Electron Optics and the Principles of Dynamics. See figure 3 in this image. $\endgroup$ – John Duffield Dec 6 '15 at 23:11
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    $\begingroup$ Such a problem would arise in understanding all action at a distance theories where you don't take in the field view behind the theory into account. $\endgroup$ – Viesr Dec 10 '15 at 6:34
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Suppose you have an infinitely long solenoid made of "stacked" loops of radius $a$. The magnetic field is then truely 0 outside, and all the field lines are confined inside the solenoid. Then how can an exterior loop (of radius $b \ge a$) encircling the solenoid could "know" that there's actually a changing magnetic field inside the solenoid, since it is completely confined ?

First, the magnetic flux is defined by this expression : \begin{equation}\tag{1} \Phi_B \equiv \int_{\mathcal{S}_{\text{sol}}} \vec{B} \cdot d\vec{S}, \end{equation} Where $\mathcal{S}_{\text{sol}}$ is the solenoid transverse aera. Since the field vanishes outside the solenoid, you can use the exterior loop aera $S_{\text{loop}}$ instead, and express the magnetic field in terms of the magnetic vector-potential : $\vec{B} = \vec{\nabla} \times \vec{A}$ : \begin{equation}\tag{2} \Phi_B = \int_{\mathcal{S}_{\text{loop}}} \vec{B} \cdot d\vec{S} = \int_{\mathcal{S}_{\text{loop}}} (\vec{\nabla} \times \vec{A}) \cdot d\vec{S}. \end{equation} By Stokes theorem, you then have the magnetic flux expressed as a line integral around the loop : \begin{equation}\tag{3} \Phi_B \equiv \oint_{\mathcal{C}_{\text{loop}}} \vec{A}_{\text{outside}} \cdot d\vec{\ell}. \end{equation} The vector-potential doesn't vanish outside the solenoid (it must be continuous across the solenoid's boundary) : \begin{align}\tag{4} \vec{A}_{\text{inside}} &= \frac{1}{2} \; \vec{B} \times \vec{r}, &\vec{A}_{\text{outside}} &= \frac{a^2}{2 \, \rho^2} \; \vec{B} \times \vec{r}, \end{align} where $a$ is the radius of the solenoid and $\rho$ is the cylindrical variable. $\vec{r}$ is the vector position of any point in space, and $b \ge a$ is the loop radius. Using this vector-potential, it is very easy to verify that \begin{align} \vec{\nabla} \times \vec{A}_{\text{inside}} &= \vec{B}, &\vec{\nabla} \times \vec{A}_{\text{outside}} &= 0, \end{align} and expression (3) gives $\Phi_B = \pi B \, a^2$.

So, the loop doesn't feel the magnetic field itself, but it can interact with the vector-potential. The e.m.f is the time derivative of the flux : \begin{equation}\tag{5} \mathscr{E} = -\: \frac{d \Phi_B}{d t} = -\: \frac{d}{d t}(\pi B \, a^2) = -\: \pi \dot{B} \, a^2. \end{equation} Now, the e.m.f itself is defined as the line integral of the electric field induced on the loop by the time-varying magnetic field inside the solenoid : \begin{equation}\tag{6} \mathscr{E} \equiv \oint_{\mathcal{C}_{\text{loop}}} \vec{E} \cdot d\vec{\ell} = \pm \, E \; 2 \pi b, \end{equation} Then we get $E(t) = \frac{a^2}{2 \, b} \; |\, \dot{B} \,|$ on the loop, or \begin{align}\tag{7} \vec{E}_{\text{inside}}(t, \, \vec{r}) &= -\: \frac{\partial }{\partial t} \, \vec{A}_{\text{inside}} = -\: \frac{1}{2} \; \frac{\partial \vec{B}}{\partial t} \times \vec{r}, \\[18pt] \vec{E}_{\text{outside}}(t, \, \vec{r}) &= -\: \frac{\partial }{\partial t} \, \vec{A}_{\text{outside}} = -\: \frac{a^2}{2 \, \rho^2} \; \frac{\partial \vec{B}}{\partial t} \times \vec{r}, \tag{8} \end{align} which agrees with Maxwell's equation : \begin{align}\tag{9} \vec{\nabla} \times \vec{E}_{\text{inside}} &= -\: \frac{\partial \vec{B}}{\partial t}, \\[18pt] \vec{\nabla} \times \vec{E}_{\text{outside}} &= 0. \tag{10} \end{align} Take note that $\vec{\nabla} \cdot \vec{E} = 0$ everywhere (do all the detailled calculations to verify this !). So the conclusion is that the loop do indirectly feel the magnetic field of the solenoid with the help of its vector-potential, outside the solenoid.


Complement : Take note that $\vec{B}$ must be varying very slowly, or vary linearly with $t$, or else there will be some electromagnetic waves outside the solenoid ! We have \begin{equation}\tag{11} \vec{\nabla} \times \vec{B}_{\text{outside}} = 0 = \mu_0 \, \vec{J}_{\text{outside}} + \frac{1}{c^2} \, \frac{\partial}{\partial t} \, \vec{E}_{\text{outside}}. \end{equation} The current density $\vec{J}$ vanishes inside and outside the solenoid (and it is singular on its boundary !). Then equation (11) and expression (8) give \begin{equation}\tag{12} 0 = \frac{1}{c^2} \, \frac{\partial}{\partial t} \, \vec{E}_{\text{outside}} \propto \frac{1}{c^2} \, \frac{\partial^2 \, \vec{B}}{\partial t^2}. \end{equation} This is also true inside the solenoid.

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  • $\begingroup$ Very nice and illustrative solution. $\endgroup$ – AlQuemist Dec 10 '15 at 20:16
  • $\begingroup$ Why $\pm$ sign appears on RHS of Eq. 6, $\pm 2 \pi b E$? $\endgroup$ – AlQuemist Dec 10 '15 at 20:16
  • $\begingroup$ The sign depends on the way you do the line integral around $\mathcal{C}_{\text{loop}}$. For an open surface (the transverse area of the solenoid), the orientation of the vector $d\vec{S}$ is arbitrary (unless you use the current on the solenoid boundary and the right hand rule). So by Stokes theorem, the line integral on $\mathcal{C}_{\text{loop}}$ may be in the same sense or in the reverse sense of the induced electric field. Since $E \equiv || \vec{E} ||$ is the modulus of a vector, it must be positive. $\endgroup$ – Cham Dec 10 '15 at 20:28
  • $\begingroup$ Also, don't forget that the expression $\dot{B}$ is positive if the magnetic field is increasing inside the solenoid (then $\mathscr{E} < 0$), and $\dot{B}$ is negative if the magnetic field is decreasing (then $\mathscr{E} > 0$). So we must be careful about the signs in each expression. $\endgroup$ – Cham Dec 10 '15 at 20:41
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Based on the Faraday’s induction law: $$ \mathcal{E}_{EMF} = \oint_{C_1} E \cdot dl = -\frac{d \Phi}{d t} ~, $$ where $E$ is the induced electric field and $$ \frac{d \Phi}{d t} = \frac{d B}{d t}\pi r_2^2. $$ Then left-hand-side of this equation, assuming that the integral is over a circular curve $C_1$ with a radius $r = r_1$ (because you are calculating the field at that distance), and that $E$ is uniform over that curve, we have: $$ E \times \text{circumference of a circle of radius } r_1 = E \times 2\pi r_1 ~. $$ so that $$ \begin{align} E\times 2\pi r_1 &= -\frac{dB}{dt}\pi r_2^2 \\ \implies E &= -\frac{dB}{dt} \left(\frac{r_2^2}{2r_1}\right) \end{align} $$ Therefore, with the right calculations you see that an electric field is induced by the change of $B$ in the solenoid, this electric field depends on the shape of both the ring and the solenoid ($r_1$ and $r_2$) and causes the electrons to move on the larger ringe; hence, an electric current is induced in that ring.

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  • $\begingroup$ Thats what I thought.I don't know where OP is confused. $\endgroup$ – Paul Dec 5 '15 at 12:17
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    $\begingroup$ Oh yeah, huh. For some reason, that never "clicked" in my brain until now. Probably because a lot of the problems I was doing focused primarily on the emf. That really makes sense when you look at Amperes law in differential form. So I guess the next question is why does it do this? From what I remember from Purcell, he doesn't talk much about how/why the fields transform from fields removed from their sources. $\endgroup$ – Astrum Dec 5 '15 at 15:25
  • $\begingroup$ "emf" does not know about the flux. Before the flux change there is no "emf". All I can say is that this is how nature behaves. Maybe in special relativity you could find something deeper. $\endgroup$ – Amin R. Dec 5 '15 at 15:41
  • $\begingroup$ This answer provides the correct analytical solution to the problem, but does not answer the essential part of the question: “It seems ... there is no way for the [larger] loop to “feel” the [variation of] magnetic flux [in the smaller loop].” So it is still incomplete. $\endgroup$ – AlQuemist Dec 5 '15 at 16:44
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    $\begingroup$ There's an error in the derivation; should multiply by the circumference, not the area. Then you get the expected $1/r_1$ dependence on $E$. $\endgroup$ – elifino Dec 6 '15 at 7:17
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The analytical solution to the problem of two loops (a solenoid inside a conducting ring) is given by Armin R (here) which shows how the induced electric field in the larger ring depends on the parametres of the problem (the radii).

Here, I attempt to extend Armin R’s contribution by answering the essential (more interesting) part of the question,

How does the large loop even “know” what the flux is, while we assume that the magnetic field $B$ due to the solenoid is negligible outside of the smaller cross section of the solenoid, so that there is no field actually going through the material of the larger conducting loop. ... It seems there is no way for the loop to “feel” the magnetic flux.

To understand how the larger ring “feels” the changing magnetic flux, one should remind that the Maxwell equations which describe the dynamics of electromagnetic fields, predict that any change in the electromagnetic field at some point would propagate with the velocity of light throughout the space. This happens, e.g., in an antenna where the oscillation of electric charge produces waves which we can “feel” (detect) from far away.

The same thing happens in this case, but such ‘details’ or complexities are eliminated by the simplified scenario of the problem; namely, the problem is given in the context of steady-state electrodynamics where the transient dynamics has disappeared and one does not consider how the effect of the change in the localized magnetic field $B$ propagates in space-time to reach the larger ring. This becomes clear by noticing that in the analytical solution by Armin R, everything is instantaneous, $$ E(\color{red}{t}) \propto -\frac{d B(\color{red}{t})}{d t} ~; $$ that is, changing $B$ changes $E$ instantaneously (with no time delay) and this apparently $^\ast$ contradicts the relativistic nature (Lorentz invariance) of Maxwell's equations – the finite time needed for the changes to propagate. So, the actual (full) scenario is that when one changes the localized field $B$, this effect propagates (like waves) in space-time with the velocity of light and ultimately, after $\Delta t \sim \frac{|r_2 - r_1|}{c}$, reaches the charge carriers (electrons) in the larger ring. In this way, the ring “feels” the effect of the change in the magnetic field of the solenoid (despite the magnetic field being localized).

I think considering the Liénard–Wiechert potentials that describe the classical electromagnetic effect of a moving electric point charge, in terms of relativistic, time-varying electromagnetic fields will be enlightening regarding the current question.


$^\ast$ This is only an apparent (not actual) contradiction, as noted in a comment by John Duffield: “one should properly speak of the electromagnetic field $F_{\mu \nu}$ rather than $E$ or $B$ separately” [$\S$ 11.10 of Jackson’s “Classical Electrodynamics”]. Here, I intend to emphasize the instantaneity.

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    $\begingroup$ It's instantaneous because $\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$ means the curl of E is the rate of change of B. See section 11.10 of Jackson's Classical Electrodynamics where he says "one should properly speak of the electromagnetic field Fμν rather than E or B separately". Also see Jefimenko's equations. $\endgroup$ – John Duffield Dec 6 '15 at 23:15
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    $\begingroup$ Your comment is absolutely true; but I wanted to emphasize that instantaneity. Here, the missing part of the picture is the the way which the change in the localized field $B$ reaches the outer ring. The reason one does not see this (in the first glance) is that the full electrodynamics, esp. the wave propagation, is absent in the scenario due to the reduced and simplified nature of the problem. As I mentioned in the answer, the essence of the question is very general; i.e., not specific to this particular setting of two loops. @JohnDuffield $\endgroup$ – AlQuemist Dec 7 '15 at 10:20
  • $\begingroup$ Philosophiæ Naturalis : all points noted. You may be interested in some of the answers I've given previously on the electromagnetic field, such as this. $\endgroup$ – John Duffield Dec 8 '15 at 13:53
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    $\begingroup$ @user36790: I have different view of this; namely, the special relativity tells us that the electric and magnetic fields are two aspects of the same thing (field): the electromagnetic field. It is not like that an electric field produces magnetic field or vice versa. They are always present both together. $\endgroup$ – AlQuemist Dec 21 '15 at 10:33
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    $\begingroup$ Yes, I mean that indeed. @user36790 $\endgroup$ – AlQuemist Dec 21 '15 at 11:50
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Field outside the loop is not negligible at all. It is huge and almost 1/4 of field inside solenoid is experienced by ring in this case. Field due to two consecutive wounds on solenoid is zero outside solenoid which is said to be neglected. As a result only axial part of field exist inside solenoid. This axial field leaves solenoid from north pole with almost half intensity and bends toward south pole which is actually felt by ring with , let's say, 1/4 intensity. So now you understand negligible field is actually said to that part, which is due to two consecutive wounds in solenoid, which cancel in pair due to opposite direction field outside solenoid.

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