Talk:Faraday paradox

message: Looking at the references I added to the article, it appears that this subject is still somewhat controversial. Are there situations where only the Lorentz law can be used? Note that the Scanlon reference appears to be the source of the explanation based on Faraday's Law given in this article.--J S Lundeen 02:59, 23 January 2006 (UTC)
thread_title:  Is this still controversial? 

message: I didn't really studied the paradox, so I ;ight be wrong, but a situation that looks close to this paradox is easily solved with special relativity, it is about the magnetic lines moving with the magnet. For example a wire perpendicular to an (almost) infinite wire with current, that moves in a parallel motion in respect of the wire, experiences a lorentz force. But if the infinite wire moves with the same velocity, there exists a restframe were nothing moves. The problem is solved by applying lorentz tranformation to the quadri vector potential, or the maxwell tensor, hence there appears an electric field that exactly cancel the induction. Electromagnetism is stupid, you have the fields, you apply the lawsKlinfran (talk) 11:26, 31 May 2010 (UTC)
thread_title:  paradox 

message: In my opinion, Faraday's disc has nothing to do with electromagnetic induction at all. From the view of the laboratory system, the magnetic field is constant in time (\partial \vec B / \partial t =0). Thus the induction law \text{curl} ~\vec E = -\partial \vec B / \partial t equals zeros at all places and all times, so that the electrical field is a conservative field with some charges being separated in space by the Lorentz force. 
The flux rule
: 
\oint_{C}{{{\vec{E}}'}}\text{d}\vec{s}\approx -\int_{A}{\frac{\partial \vec{B}}{\partial t}}\text{d}\vec{A}+\oint_{C}{{\vec{v}}}\times \vec{B}\ \text{d}\vec{s}=-\frac{\text{d}}{\text{d}t}\int_{A}{{\vec{B}}}\cdot \text{d}\vec{A} 
that is commonly used to explain why there is a rotational electric field, has no explanatory power concerning whether there is induction or not, because E' is measured by an observer in the (moving) system of the path \text{d}\vec s (that means: not in one common reference system). Furthermore, it is only an approximation for non-relativistic (that means: small) velocities.
It is better to describe all variables from a common reference system (e. g. the laboratory system) and use the "true" induction law 
:\oint_{C}{{\vec{E}}}\ \text{d}\vec{s}=-\int_{A}{\frac{\partial \vec{B}}{\partial t}}\ \text{d}\vec{A}
observed in an arbitrary but common inertial system. Note that the "true" induction law does not consider any change of the borders of contour C.
By using the Lorentz transformation for the electromagnetical field (shown here), it is possible to change from one system to another. Doing this, the integration path can easily be completed by a direct line between the two electric contacts (that means by a straight unmoved line through the disc). The electric field \vec E'=0 of the ideally conducting disc (measured in he moved system) will transform to an electric field \vec E \approx \vec E'-\vec v \times \vec B=-\vec v \times \vec B measured in the laboratory system. By doing this, it can be shown that the voltage displayed by the voltmeter is the same voltage that is comprised in the rotating disc. This shows, again, that the field is not rotational, and there is no induction. Michael Lenz (talk) 00:04, 28 May 2011 (UTC)
thread_title:  There is no induction at all 

message: --Wtshymanski (talk) 18:37, 8 July 2011 (UTC)
thread_title: Moved from article, needs rewrite for tone and references 

message: This appears to be a private individual's web site.  While I have no reason to doubt its veracity, it does not constitute a reliable source.Constant314 (talk) 02:01, 4 June 2014 (UTC)
thread_title:  Vanja Janežič 

message: This describes an elegant experiment.  But it is not one of Feynman's examples.  If it is based on Feynman's example then it is WP:SYN.  A reliable source is needed.Constant314 (talk) 02:04, 4 June 2014 (UTC)
thread_title:  Inapplicability of Faraday's law 

message: Good picture, but you do not need the math.  When the switch is flipped, one circuit is destroyed and another created.  During the time that either circuit exists, the flux enclosed by the circuit does not change.  Hence Faraday's law would predict 0 EMF. Constant314 (talk) 04:15, 4 June 2014 (UTC)
thread_title:  An additional rule 

message: This new section seems to probably violate Wikipedia rules on own work WP:OR and synthesis WP:SYN.  It also reaches the non-sensible conclusion that "the induced electric field cannot be described by the Maxwell–Faraday equation".  Constant314 (talk) 03:48, 6 December 2015 (UTC)
replies: text: The Maxwell–Faraday equation is zero-valued when the electric field E is curl-free and the magnetic flux density B is constant. talk2siNkarma86—Expert Sectioneer of  Wikipedia 18:00, 6 December 2015 (UTC)
replies: text: Perhaps you can elaborate.  I am presuming you are discussing the case where there is zero actual EMF which would agree completely with a zero-value E field.Constant314 (talk) 18:28, 7 December 2015 (UTC)
thread_title:  Using the relativistically-correct electric field 

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thread_title:  External links modified 

message: After reading through the article, it is clear that sections 3, 4, and 5 are actually subsections of section 2.  I'm going to fix that.  It will be ugly, but it will actually reflect the structure of the article. Constant314 (talk) 01:59, 7 February 2019 (UTC)
replies: text: It has been done. Constant314 (talk) 02:04, 7 February 2019 (UTC)
thread_title:  Indention of subsections needs correction. 

message: This section starts with the equation
:: \mathbf{F}_{21} = \frac {\mu_0} {4 \pi} I_1 I_2 \oint_{C_1} \oint_{C_2} \frac {d \mathbf{l_1}\ \mathbf{ \times} \ (d  \mathbf{l_2} \ \mathbf{ \times } \ \hat{\mathbf{r}}_{21} )} {r_{21}^2} \ 
However, for the apparatus shown, both currents are zero so this equation reduces to zero.  The entire derivation is unnecessary.Constant314 (talk) 22:00, 30 March 2019 (UTC)
thread_title:  "An additional rule", unnecessary math  

message: The sliding contact at the rim of the disc is a "third party". Actually there are 8 configurations.
# nothing rotates
# only the disc rotates
# only the magnet rotates
# only the contact rotates
# only the disc rests 
# only the magnet rests 
# only the contact rests
# all three rotate
What will happen? At first sight I guess: 
:1 and 8 are equivalent, no emk
as are
:2 and 5, induced emk
:3 and 6, no emk
:4 and 7, induced emk
Madyno (talk) 21:55, 10 July 2022 (UTC)
thread_title:  Sliding contact