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This question comes from this Walter Lewin video at 35:00 where he says "I'm going to confuse you even more" (and I suppose he somewhat succeeded). Walter Lewin claims there is an electric field around the inductor (when the current is changing).

Ok, I will explain the setup myself below, so feel free to either watch the video from the 35 minute mark or read my writing. In my scenario, I will simplify things and talk about an RL circuit instead of an RLC circuit. Afterwards, I will ask my question.

Please note: This post is not about the debate between Kirchhoff's voltage law vs Faraday's law. I consider that debate fully resolved by this PSE post, this PSE post and this video at 41:00, so please do not talk about it here.

Context and Setup for the Question

Consider an RL circuit as depicted below.

enter image description here

Suppose I want to apply Faraday's law ($\oint \vec{E}\cdot d\vec{l} = -\dot{\Phi}_{B}$) to find the equation governing the current of the setup. One way to do this is to take a loop going clockwise along the wire itself. Call this loop 1. See the image below for an attempt to sketch it along with the surface that goes with it.

enter image description here

As you go along the loop, the total electric field is only present in the battery and resistor (the inductor consists of an ideal conductive wire so there is no electric field there): $\oint_{\text{loop 1}} \vec{E}\cdot d\vec{l} = -V + IR$. On the other hand, $-\dot{\Phi}_{B} = -L\dot{I}$ due to the magnetic field in the inductor solenoid. Thus we get $-V + IR = -L\dot{I}$ or rather $-V + IR + L\dot{I} = 0$.

Now take a loop identical to the previous one except it takes a detour to avoid the inductor coil. Call it loop 2 and see the image below.

enter image description here

My chosen surface is now completely flat, and it cuts halfway through the inductor solenoid. Given that the magnetic field is never perpendicular to the flat surface, we have $-\dot{\Phi}_{B} = 0$. The inductor term no longer appears in the magnetic flux integral. However, the inductor term should now appear in the electric field loop integral for the following reason: Take the piece of the path going along the inductor coils and paste it with the (reverse of) the piece of the detour path. See the image below.

enter image description here

For a changing current through the solenoid, there should be a changing magnetic field and this means the associated surface will have $-\dot{\Phi}_{B} = -L\dot{I}$. However by Faraday's law this must equal $\oint \vec{E}\cdot d\vec{l}$ which in this case is $$ \underbrace{\int_{A\rightarrow B\text{ by inductor}} \vec{E}\cdot d\vec{l}}_{\text{this is } 0} + \int_{B\rightarrow A\text{ by detour}} \vec{E}\cdot d\vec{l}. $$ Thus $$ \int_{B\rightarrow A\text{ by detour}} \vec{E}\cdot d\vec{l} = -L\dot{I} $$ and $$ \int_{A\rightarrow B\text{ by detour}} \vec{E}\cdot d\vec{l} = +L\dot{I}. $$ Therefore, when we go back to image #3, we should find $$ \oint_{\text{loop 2}} \vec{E}\cdot d\vec{l} = -V + IR + L\dot{I} = 0. $$ This shows there must be an electric field around the inductor and it must contribute to the detour part of loop 2.

Question(s)

My problem is, I agree with the reasoning above, but as far as I understand the $E$-field around the solenoid has to be azimuthal around the solenoid, so it will always be perpendicular to the "detour path." Hence we should have $\vec{E}\cdot d\vec{l} = 0$ around the "detour path." This clearly contradicts the above reasoning. Where am I going wrong?

I've broken my main question into sub-questions, if it helps.

  1. Where is this electric field coming from? I assume the change in magnetic field (as the current changes) induces an electric field around it via $\nabla\times\vec{E} = -\dot{\vec{B}}$. Is this correct?
  2. Shouldn't the electric field around the solenoid be azimuthal? If so, this would mean $\vec{E}$ is always perpendicular to the "detour path" and $\vec{E}\cdot d\vec{l} = 0$. This is clearly contradicting the above reasoning, so where am I wrong?
  3. How do I correctly visualize the electric field around the solenoid?
  4. Is there an approximate formula for the $E$-field around the finite solenoid? I know the formula for an infinite solenoid but this is a finite one.
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  • $\begingroup$ what's the E-field for an infinite solenoid? $\endgroup$
    – jim
    Commented Aug 22, 2023 at 9:54
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    $\begingroup$ @jim If we say the solenoid of radius $R$ and along the $z$-axis, then $\vec{E} = -\dot{B}r/2\cdot\vec{e}_{\phi}$ if $r < R$ and $\vec{E} = -\dot{B}R^{2}/(2r)\cdot\vec{e}_{\phi}$ if $r\ge R$ where $\vec{e}_{\phi}$ is the unit vector along the azimuthal angle. Maybe I'm mistaken or missing something obvious. $\endgroup$ Commented Aug 22, 2023 at 10:15
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    $\begingroup$ thanks for enlightening me $\endgroup$
    – jim
    Commented Aug 22, 2023 at 10:17

1 Answer 1

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but as far as I understand the $E$-field around the solenoid has to be azimuthal around the solenoid, so it will always be perpendicular to the "detour path."

This is not the case. Total electric field $\mathbf E$ can be expressed as sum of two components:

$$ \mathbf E = \mathbf E_i + \mathbf E_C, $$

where $\mathbf E_i$ is the induced electric field due to accelerated current in the coil (this field's lines indeed go around the solenoid in circles, and are thus perpendicular to the detour path) and $\mathbf E_C$ is conservative component of the total electric field due to all charges on surface of conductors in the circuit (this field's lines go from one terminal of the inductor to the other, like electrostatic field of a dipole, so they go in similar direction as the detour path). Since $\mathbf E_C$ is conservative field, line integral of this field depends only on endpoints, not exact shape of the path, so line integral of $\mathbf E_C$ over the detour path $A\to B$ gives $L\dot{I}$, regardless of its exact shape.

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  • $\begingroup$ Good answer. I will see if I can simulate the charge distribution (on the surface of the conductive wire) along with $\mathbf E_C$ in some sort of toy model, and then post that as my own answer. I'll likely get around to doing it in the future however. $\endgroup$ Commented Aug 25, 2023 at 7:14

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