Differential equation in RL-circuit

I am self-studying electromagnetism right now (by reading University Physics 13th edition) and for some reason I always want to understand things in a crystalclear way and in depth. Now look at this simple RL-circuit: I am assuming the wire is superconducting and suppose at $t=0$ the switch is turned on. My problem is that I want to come up with this differential equation for this circuit: $V-I\cdot R=-L\cdot \frac{dI}{dt}$. I know that I can't use Kirchoff's voltage law $(\oint \vec{E}\cdot d\vec{l}=0)$, because there is a non-conservative electric field near the inductor. Therefore I use Farayday's law of induction: \begin{equation} \mathcal{E}=\oint_{\Gamma} \vec{E}\cdot d\vec{l}=-\frac{d}{dt} \int_{S} \vec{B} \cdot \hat{n}\,dA \end{equation}

where $\Gamma$ is a closed loop and $S$ is the open surface attached to the loop. Now, my textbook says:

In general, the total field $\vec{E}$ at a point in space can be the superposition of an electrostatic field $\vec{E_c}$ caused by a distribution of charges at rest and a magnetically induced, nonelectrostatic field $\vec{E_n}$. That is $\vec{E}=\vec{E_c}+\vec{E_n}$.

Inserting this in Faraday's law I get:

$$\oint_{\Gamma} \vec{E}\cdot d\vec{l}=\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}+\oint_{\Gamma} \vec{E_n}\cdot d\vec{l}=-L\cdot \frac{dI}{dt}$$ Now by definition: $\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}=0$ since $\vec{E_c}$ is conservative. This gives: $\oint_{\Gamma} \vec{E_n}\cdot d\vec{l}=-L\cdot \frac{dI}{dt}$, which is perfectly fine and consistent with my textbook. However it also gives (going counterclockwise in circuit): $\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}=-V+I\cdot R=0$ which it shouldn't. I think my flaw is here, but I've tried to think about this this whole day with no succes.

I have my inspiration from Walter Lewin's video: http://youtu.be/LzT_YZ0xCFY?t=26m31s but I still can't get the equation by Faraday's law.

Any help is greatly appreciated.

• Just to make this clear: a circuit diagram is not a visualization of a field geometry, therefor trying to integrate over some assumed area or path depicted by the diagram doesn't make much sense (if that is what you are trying to do here). You can calculate the behavior of the inductor all by itself and then use Kirchhoff and that will give the correct answer. Having said that, I have been in physics courses where the professor did just that... for whatever reason. – CuriousOne Jun 30 '16 at 0:12
• I see your point, and it seems that I actually am integrating over an assumed area. My closed path is in the counterclockwise direction of the circuit and the surface I integrate over is bounded to that path. Nevertheless, does it make any difference which surface I attach to this closed loop in this situation? Or am I just too narrow-minded? In addition, I don't quite understand how you can calculate the behavior of the inductor all by 'itself' - can you explain this further? – Ziko Humlesen Jun 30 '16 at 0:34
• The best you can achieve by reading a circuit diagram geometrically is the derivation of Kirchhoff's laws, again, but you basically mentioned the problem with that, already. A better way to look at a circuit diagram is to see it as a topological/graph description of a system for which Kirchhoff's laws are valid. The main problem then is to get the signs of the currents and voltages right. If all else fails, check energy conservation explicitly. If the circuit is magically generating exponentially growing voltages/currents, then you know that you got a sign wrong, somewhere. :-) – CuriousOne Jun 30 '16 at 0:52

When dealing with inductors Sears & Zemansky state that "we need to develop a general principle analogous to Kirchhoff's loop rule".

With an inductor present in the circuit they state that there is a non-conservative electric field within the coils $\vec E_n$ as well a conservative electric field $\vec E_c$.
Assuming that the inductor has negligible resistance they then state that the net electric field within the inductor is zero ie $\vec E_c + \vec E_n =0$.

If the end of the inductor are labelled $a$ and $b$ they then state that instead of writing $\oint \vec E_n \cdot d \vec l = - L \frac {dI}{dt}$ for the complete circuit they can just consider that path within the coil as that is where the magnetic flux is changing which then gives $\int_b^a \vec E_n \cdot d \vec l = - L \frac {dI}{dt}$.

So $\int_b^a \vec E_c \cdot d \vec l = + L \frac {dI}{dt} = V_{ab}$ which is the potential of point $a$ relative to point $b$.

Their final statement is "we conclude that there is a genuine potential difference between the terminals of the inductor, associated with conservative, electrostatic forces, despite the fact that the electric field associated with the magnetic induction effect.

So their aim was to be able to use $\oint \vec E_c \cdot d \vec l = 0$ for a complete circuit with no changing magnetic flux through it which is the essence of what Walter Lewin discusses towards the end of his video with the potential difference across the inductor as part of the line integral.

You might find this Yale video useful in which Ramamurti Shankar starts with a mutual inductance with the secondary open circuited.

So where did you go wrong?

However it also gives (going counterclockwise in circuit): $\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}=-V+I\cdot R=0$ which it shouldn't.

Look at this equation:

$$\oint_{\Gamma} \vec{E}\cdot d\vec{l}=\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}+\oint_{\Gamma} \vec{E_n}\cdot d\vec{l}=-L \frac{dI}{dt}$$

Split it term by term:

$$\oint_{\Gamma} \vec{E}\cdot d\vec{l}=\int_{\text{battery}} \vec E_c \cdot d \vec l + \int_{\text{resistor}} \vec E_c \cdot d \vec l + \int_{\text{inductor}} (\vec E_c + \vec E_n) \cdot d \vec l = -L \frac{dI}{dt}$$

$$\Rightarrow \oint_{\Gamma} \vec{E}\cdot d\vec{l}=\int_{\text{battery+resistor+inductor}} \vec E_c \cdot d \vec l + \int_{\text{inductor}} \vec E_n \cdot d \vec l = -L \frac{dI}{dt}$$

$$\Rightarrow \oint_{\Gamma} \vec{E}\cdot d\vec{l}=\oint_{\Gamma} \vec E_c \cdot d \vec l + \int_{\text{inductor}} \vec E_n \cdot d \vec l = -L \frac{dI}{dt}$$

Using your statement: "Now by definition: $\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}=0$"

$$\oint_{\Gamma} \vec{E_c}\cdot d\vec{l}=-V+I\cdot R + L \frac {dI}{dt}=0$$

Going around clockwise or anticlockwise does not matter the only difference is a change of sign for each term.

• Thank you for you answer and the video, but I still don't understand how can $\vec{E_c}$ be nonzero in the conducter, when the wire has negligible resistance? And since it is nonzero doesn't it contradict Walter Lewin's statement about the term $L\cdot \frac{dI}{dt}$ not being a part of the closed loop integral (youtu.be/LzT_YZ0xCFY?t=30m4s)? I can think of a much easier way to derive the differential equation, but it shouldn't matter how I approach this problem and I really want to understand my method, so I would be very grateful if you can answer my questions. – Ziko Humlesen Jun 30 '16 at 15:46
• I can understand Sears & Zemansky's method now thanks to you, nevertheless it is still not clear how $\vec{E_c}$ is nonzero and why it doesn't contradict Walter Lewin. :-( – Ziko Humlesen Jun 30 '16 at 15:52
• In the Lewin video you will note that towards the end when he makes part of the loop bypass the inductor he states that there is no magnetic flux change through the loop and so Kirchhoff's voltage law can be used. His L dI/dt term then moves to the left hand side of the equation and becomes positive. Sears and Zemansky produce an approach which does not worry about a part of the loop bypassing the inductor and are not as clear as to where the loop is. They do however state that the electric field inside the inductor is zero so the $\oint E \cdot dl$ term is zero. – Farcher Jun 30 '16 at 19:54
• I understand it now!!! :D - Wow! Thank you very much Farcher!! – Ziko Humlesen Jun 30 '16 at 22:35