8.02x - Lect 16 - Electromagnetic Induction, Faraday's Law, Lenz Law, SUPER DEMO

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Recommends user16307 · 2017-06 · Electrical Engineering Stack Exchange · on “Circuit yields a peculiar contradiction between KCL, KVL and Faraday's law”

I don't know if this particular circuit/loop is covered in another question but I stumbled upon a video where a peculiar consequence occurs for the following circuit: [LINK] For the above circuit loop according to Faraday's law of induction one can write: EMF = -dΦ/dt And from the basic electric circuit theory for the current one can also write: I = EMF/(R1+R2) But since the same current passes through the resistors( KCL ), something peculiar happens here. Imagine the magnetic flux Φ starts increasing with a constant slope(which means EMF=-dΦ/dt is a constant); and during this time if we observe the voltage V1 across R1 by a scope between the point A and B, according to the logic the voltage

Recommends Farcher · 2017-11 · Physics Stack Exchange · on “What‘s the potential difference between two points of rectangle circuit in a changing magnetic field?”

The question about a voltmeter reading as stated has no answer unless you include a statement about the position of the voltmeter and the leads that connect it to points $A$ and $C$. By connecting a voltmeter into the circuit you have introduced extra loops which have emf induced in them. [LINK] It is not a matter of "real" potential difference it is to do with the reading on a voltmeter. In the diagram above all three voltmeter readings will be different. For more about this I would suggest that you watch Walter Lewin's lecture from about 33 minutes [this video] and the notes that he produced [LINK] . Also Ramamurti Shankar's lecture from about 38 minutes [LINK] is worth watching?

Recommends jonk · 2018-04 · Electrical Engineering Stack Exchange · on “How step-up / step-down transformers work considering that you cannot create or destroy electrons?”

Watch this demonstration of Faraday's law [this video] . Go completely through the entire video starting around 5 minutes into it.

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: [this video] and more specifically, he talks about the case in an inductor as you asked in this video from 5:45 to 13:45 : [LINK]

Recommends Slaven Tojić · 2020-06 · Electrical Engineering Stack Exchange · on “Can two voltmeters connected to the same terminals show different values? Circuit with induced EMF”

The circuit shown below consists of two resistors connected in series that form a closed loop. No electrical sources are in that loop.Let’s assume that the resistance of the wires that connect the resistors and form the loop is zero.Inside the loop is an external changing magnetic field that induces inside that loop a constant EMF of 1 V. [LINK] [LINK] The current inside the loop is calculated using Ohm’s Law : [LINK] The problem was introduced by former MIT professor Walter Lewin. The problem was introduced in this video (starting at 35th min of the video): 8.02x - Lect 16 - Electromagnetic Induction, Faraday's Law, Lenz Law, SUPER DEMO [this video] In the video is shown that two voltmeters conn

Recommends naturallyInconsistent · 2025-05 · Physics Stack Exchange · on “Electrostatic potential difference across a self-inductor”

EDIT: My answer below is wrong! In the comments, we have uncovered your true confusion: The $\int\vec E\cdot\vec{\mathrm d\ell}$ following the inside of the inductor is NOT zero. Go and watch Lewin's lecture here: [this video] If you are in a time-crunch, skip to 33:40 for the main event.

Recommends naturallyInconsistent · 2025-05 · Physics Stack Exchange · on “Electrostatic potential difference across a self-inductor”

Then we have uncovered your true confusion: The line integral of the electric field following the inside of the inductor is **NOT** zero. Go and watch Lewin's lecture here: [this video]

Cites abc · 2017-06 · Electrical Engineering Stack Exchange · on “Energy from pickup coil”

[this video] states the current is only affected by coil resistance (doesn't mention inductance), so I guess V=IR applies

Cites Hey Hey · 2018-11 · Electrical Engineering Stack Exchange · on “Does this experiment show that Kirchhoff's Law hold when there's a changing magnetic field involved in a circuit?”

In this video [LINK] , the electrical engineer and youtuber Mehdi Sadaghdar (ElectroBOOM) disagrees with another video [this video] from professor Walter Lewin. Basically, professor Lewin shows in an experiment that if we have two different resistances connected in a closed loop, and if we generate a changing magnetic field using a coil, the voltage at the endpoints of the two resistances will be different, contrary to the expectations from Kirchhoff's Voltage Law (KVL). simulate this circuit [LINK] – Schematic created using CircuitLab [LINK] According to the experiment, the left voltmeter VM1 shows a voltage different from the second voltmeter VM2. Lewin then concludes that KVL does not hold whe

Cites Sredni Vashtar · 2021-03 · Electrical Engineering Stack Exchange · on “What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field?”

If the varying magnetic field is confined inside the triangle, the (high impedance) voltmeter placed outside it, as you depicted it, will measure a voltage equal to R * I, where I is the current flowing in the loop I = emf / (R + R + R). And it will consistently read that value no matter how the probes are placed, twisted, untwisted, curled, long or short - as long as the probes stay outside of the variable magnetic field region (that is outside of the triangle), do not run circles around it and there are no variable magnetic fields outside of the region inside the triangle. To illustrate why it is so, I will use without loss of generality a simpler circuit with only two resistors that I wil