8.02x - Lect 20 - Inductance, RL Circuits, Magnetic Field Energy
Why Physics Stack Exchange recommends it
Physics Stack Exchange answerers and askers point to it for Lewin's treatment of Faraday's law, potential differences across inductors, and RL circuit behaviour, including the ring launcher demo.
Recommended by 5 people across 5 different months, 2014 to 2022.
What it is
Walter Lewin's MIT 8.02 lecture 20 covers inductance, RL circuits and the energy stored in magnetic fields, with live demonstrations. He works through Faraday's law around a circuit loop and shows why Kirchhoff's loop rule needs care when an inductor is present.
What people said
Recommends Monoomni · 2014-11 · Physics Stack Exchange · on “Is the standard explanation for the ring launcher incomplete?”
There is a 90 degree phase difference between the input current and the induced voltage in the ring because of Lenz's law. Then, there is a 90 degree phase difference between this induced voltage and the induced current because of the self-inductance of the ring. This amounts to a 180 degree phase difference between the input current and the induced current which ensures the magnetic fields are always repulsive. If you account for resistance, the phase difference is not quite 180 degrees, but enough that the repulsive forces are greater on average. Dr. Lewin explains it towards the end of the MIT 8.02 lecture #20 about 40 minutes into it (the "levitating woman" demonstration is in lecture 19
Recommends Dang Khai Phan · 2018-08 · Physics Stack Exchange · on “Conservative $E$-field and Kirchoff rule in practice”
I found this lecture online: [this video] at 8:25 and 9:30 he stated something interesting that I find relevant to my question. The lecturer stated that Kirchhoff isn't applicable to this situation which confuse me quite a bit.
Recommends Ofek Gillon · 2020-02 · Physics Stack Exchange · on “Clean link between electromotive force and difference of potential: potential difference around a resistor or inductance”
This is an excellent and non trivial question. Walter Lewin addresses it in his lectures about EM. I suggest you watch the following video from minute 35: [LINK] and more specifically, he talks about the case in an inductor as you asked in this video from 5:45 to 13:45 : [this video]
Recommends user243016 · 2020-05 · Physics Stack Exchange · on “Confusion regarding a few sign conventions in appyling faraday's law to inductive circuits”
In the mit ocw lecture by Prof. Lewin on EMI, He quotes a few statements from here to couple of seconds [this video] of the lecture. I am confused why the sign of $L\frac{dI}{dt}$ changes, when we go around and evaluate the closed loop integral in the opposite direction of the current. By faraday's law , we have $$\oint_{+\partial \Sigma} \vec{E} \cdot d\vec{\ell} = - \frac{d}{dt} \int_{+\Sigma} \vec{B}\cdot d\vec{A}$$ This seemingly suggest that the negative sign would be only affected by the dot product $\vec{B}\cdot d\vec{A}$ , and indeed, the direction of the $d\vec{A}$ (Area vector) is only affected by the sense of path chosen for $d\vec{\ell}$ . Then, what might be the role of current ( not
Recommends Farcher · 2022-07 · Physics Stack Exchange · on “Kirchoff analysis in faraday law for RL circuit”
This is a still from Walter Lewin's video 8.02x - Lect 20 - Inductance, RL Circuits, Magnetic Field Energy [this video] . He evaluates $\displaystyle \int \vec{E} \cdot d\vec{l}$ for each of the three circuit elements starting at the dot to the left of the current arrow at the top of the circuit digram, moving in a clockwise direction which is the same as the current direction label. For the inductor since $\vec E=0$ the integral is zero. For the resistor the direction of the path taken $d\vec \ell$ is the same as the direction of the electric field so the integral is positive $+IR$ . For the cell the direction of the electric field is opposite to that of the path taken and so the integral is neg
Cites An Ignorant Wanderer · 2019-08 · Physics Stack Exchange · on “Can phase angle be larger than 90 for a simple AC RL circuit?”
In this [this video] lecture, Walter Lewin provides the equation for the phase angle in a simple $RL$ circuit: $$\tan(\phi) = {\omega L\over R}$$ Looking at this equation, it seems to me that no matter how large $\omega$ is, the phase angle cannot exceed $\dfrac{\pi}{2}$ . So my take on this is that the current cannot lag the voltage by more than $\dfrac{\pi}{2}$ . I'm pretty sure what I'm saying is wrong, but I was wondering if someone could explain what I'm missing, or confirm what I'm saying and explain the intuition behind it.
Cites StarBucK · 2020-02 · Physics Stack Exchange · on “Clean link between electromotive force and difference of potential: potential difference around a resistor or inductance”
[edit]: I figured out that my explanation is consistent with: from 5:45 to 13:45 : [this video] link suggested by @Ofek Gillon in the comment of the answer. I will try to put a rigorous derivation of R.L circuit without using at all any results that comes from conservative fields. In practice I will never make use of the notion of electric potential which mainly makes sense for electrostatics because field is conservative. I will only work with electromotive forces. [LINK] I compute the electromotive force on the full circuit after having chosen an orientation as given in my image. $$e_{tot}=\oint \mathbf{E}.\mathbf{dl} $$ The inductance is here to model the flux going to the full circuit. In pre