Recent content by SDL

Not a problem, A.I. rules. You can turn subtitles on and click pause anytime.

Maybe this reference would help: https://web.mit.edu/6.013_book/www/chapter10/10.1.html See "Example 10.1.2. Electric Field of a One-Turn Solenoid". The reference above is a chapter from this book: Herman A. Haus, James R. Melcher. Electromagnetic Fields and Energy...

Ok, something like that. But what about this: ? The field can't stay unchanged: first, it becomes non-uniform to satisfy boundary conditions and, second, the exterior field is now the sum of the original one and the field produced by the induced charges.

It actually does, don't forget you conductor sample is finite. The external electric field is always normal to a conductor's surface. If you insert a rectangular conductor into a uniform field, like it's shown on your drawing, what this field at the upper and the lower surfaces would be?

Yes, this is exactly what I intended to show. Almost true except "voltage" instead of "power", which is a significantly different quantity. This is what capacitance is for: it depends on capacitor geometry, dielectric used, technology and not on electrical parameters like voltage and current...

Let's perform some calculations. We have a coductive plate of size 10x10cm and 1mm thick. Its volume is ##(10^{-1})^2 \cdot 10^{-3}=10^{-5}~\rm{m^3}##. Taking the "typical" size of an atom of ##0.1~\rm{nm}=10^{-10}~\rm{m}## hence its volume ##10^{-30}~\rm{m^3}##, the plate contains ##10^{-5} /...

Think, it was a challenge :wink:

This is true, but the problem isn't here. Assume the constant equals to zero. Angles 0 and ##2 \pi## are actually the same. At angle 0 V is 0 and at ##2 \pi## it equals to ##\epsilon##. Thus the potential is not defined. If we force V to be 0 at angle 0 anyway, then there would be a leap of this...

The last equation ##V=\epsilon \frac \phi {2 \pi}+const## is obviously false, because here V for angle zero is either undefined or experiances a leap. But any scalar potential function has to be continuous.

This is what I've been trying to express all the thread long. As far as I understand the idea, he tries to decompose the resulting E-field into the solenoidal one having a constant value along a circle arond the flux (due to symmetry) and the conservative one, produced by induced charges, to...

On your diagram, I'd call this into question: The EMF is proportional to the angle (this is to be proven separately). As shown on the picture above, angle ##\varphi## is less than ##\pi/2##, so ##\epsilon<0.25V##. This is what I was talking about: The solution does depend on geometry, lumped...

What I asked you to prove is why the EMFs along the two wires are the same and equal to 0.25V. That is to provide calculations instead of "locical reasoning". Can you solve your own circuit using EM laws?

If we subtract the solenoidal part, the conserative remainder would have well-observed voltage drops. The language of physics is mathematics and we all should speak it. Otherwise we take risks to get fantastic results. Under certain boundary and initial conditions Maxwell's equations have a...

This is because the line integral of E along a closed path around the magnetic flux in this circuit is non-zero, while due to KVL it is zero. Certainly I mean the resulting E-field ("non-split"). On this diagram, there are two EMFs along two wires ##\epsilon=0.25V## each. For a perfectly...

First, one has to agree that your examples do not belong to those the circuit theory deals with, because KVL is not true here. Despite of it they can be modelled with lumped elements. I suppose you have been inspired by that paper about a circuit paradox or maybe by some discussions at other...