From Newsgroup: sci.physics

The Progressing Electric Field Model
Abstract: FaradayrCOs law is empirically derived and, as such, may be
subject to limitations. Notably, it appears to violate the law of
conservation of energy in certain contexts. To establish a more robust formulation, it is necessary to derive the law from first principles. In
this article, we theoretically derive FaradayrCOs law using only
CoulombrCOs law and special relativity. We present the first stage of this derivation: the construction of the 'Progressing Electric Field Model.'
This model determines the curl of the electric field produced by moving charges and calculates the electric potential induced in a wire loop
within that field.

Introduction
In electromagnetism, Faraday's law defines the electromotive force (EMF) induced in a wire loop by a varying magnetic field. Although considered a cornerstone of classical electromagnetism, it remains an empirical law; consequently, it may be incomplete regarding phenomena not yet captured by experimental observation. To illustrate this, consider the following experimental setup: suppose two coils, A and B, are positioned side by
side, with coil B connected to a resistor R, as shown in Figure 1.

Let the current in coil A, denoted as Ia, vary as follows: Ia increases linearly from zero to Imax, then decreases linearly back to zero. The
duration of each phase is +ot. According to Faraday's law, voltages are induced in coils A and B, which we label Va and Vb, respectively. Since Ia varies linearly during each phase, Va and Vb remain constant throughout
those intervals. Within resistor R, the voltage Vb generates a current Ib
and dissipates electric power equal to |VbIb|, both of which are constant
in each phase. Consequently, the total work performed in R after both
phases is 2|VbIb|NUat.

Since Ib is constant, it does not induce a voltage in coil A; therefore,
the value of Va remains unchanged regardless of whether Ib is positive, negative, or zerorCojust as if coil B were not present. When Ia increases,
the voltage in coil A (Va) is positive, and the electrical work performed
in A is given by re2_o^+otruApCuV_a I_a dtpCu. Conversely, when Ia
decreases, the voltage in A becomes -Va, and the work equals -re2_o^+otruApCuV_a I_a pCu dt. Consequently, the total energy consumption
of coil A after both phases equals zero.

Since the energy consumption in coil A is zero, A does not transfer any
energy to coil B. We therefore encounter a case where B performs work
equal to 2|VbIb|NUat while receiving no energy from A. This implies that
the system consisting of coils A and B performs work without any energy
input, which violates the law of conservation of energy.

The cause of this violation is that Faraday's law predicts zero voltage in
A when the current in coil B is constant. By theoretically deriving FaradayrCOs law from more fundamental laws, we can not only resolve this inconsistency but also uncover new phenomena and achieve a deeper understanding of nature.

An example of such a phenomenon is demonstrated in my experiment, which reveals a tangential electromotive force not predicted by FaradayrCOs law.
You can view the experiment in this video on YouTube: https://www.youtube.com/watch?v=P33Hgj68G9M

To begin the theoretical derivation of FaradayrCOs law, we must first
examine the electric field of a moving electron.
rCa

For more detail, please read -2 A Derivation of Faraday's law from
Coulomb's Law and Relativity / 1.The Progressing Electric Field Model -+ https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html https://www.academia.edu/146009113/A_Derivation_of_Faradays_law_from_Coulombs_Law_and_Relativity_1_The_Progressing_Electric_Field_Model

Discussion
This article presents the first stage of a derivation of FaradayrCOs law
based solely on CoulombrCOs law and special relativity. We demonstrate
that the retarded electric field of a moving electronrCowhich we term the 'Progressing Electric Field'rCois a non-conservative vector field. We have derived the mathematical expression for its curl and have found that
several properties of this field are analogous to those of the magnetic
field.

We have derived the electric potential induced in a wire loop by a
progressing electric field, which shares certain properties with the electromotive force defined by FaradayrCOs law. However, because this potential is not yet proportional to the rate of change of the magnetic
field, it cannot be classified as electromotive force at this stage.
Several additional steps of theoretical derivation are required to fully arrive at FaradayrCOs law.

The essential steps of the derivation presented in this article are as follows:
Propagation at the speed of light: In accordance with special relativity,
the electric field propagates at the speed of light, c.
Iso-intensity circles: The electric field of a moving electron radiates
in 'iso-intensity circles.' The centers of these circles correspond to the retarded positions of the electron.
Application of CoulombrCOs law: The intensity of the electric field on an
iso-intensity circle is defined by CoulombrCOs law relative to the
circle's center.
Non-conservative nature: The 'Progressing Electric Field' is a deformation of the static electric field and is inherently
non-conservative.
Instantaneous curl calculation: The Progressing Electric Field is analyzed within a single temporal snapshot, allowing its curl to be
computed using instantaneous values.
Superposition of charges: The curl of the Progressing Electric Field for
a steady current is derived by integrating the fields of individual moving charges.
Induction and LenzrCOs Law: The Progressing Electric Field induces an electric field within a wire loop; the calculated average value of this
field is shown to be consistent with LenzrCOs law.

We have constructed our theory upon the Progressing Electric Field, a
concept that may initially appear to lack direct experimental
verification. One might ask: what if this field is merely a theoretical construct? In reality, experimental evidence for its existence already
exists; however, it has historically been overlooked or reinterpreted
through the lens of well-established theories.

By applying the 'Progressing Electric Field Model,' we will provide a new
and comprehensive explanation for this experimental evidence in a
forthcoming article.

In summary, the 'Progressing Electric Field Model' demonstrates that the induction phenomena traditionally attributed to Faraday's law can be
derived from the relativistic motion of electric charges. By accounting
for the propagation delay of the field at speed c, we establish that the resulting electric field is non-conservative and possesses a non-zero
curl. While this initial stage of the derivation produces results
consistent with LenzrCOs law, further refinement is required to achieve
full mathematical proportionality with the rate of change of magnetic
flux. This theoretical framework not only aligns with the principle of
energy conservation but also opens the door to predicting electromagnetic phenomena beyond the reach of classical empirical laws.

For more detail, please read -2 A Derivation of Faraday's law from
Coulomb's Law and Relativity / 1.The Progressing Electric Field Model -+ https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html https://www.academia.edu/146009113/A_Derivation_of_Faradays_law_from_Coulombs_Law_and_Relativity_1_The_Progressing_Electric_Field_Model

1. Introduction 1
2. The Electric Field of a Moving Electron 2
a) The Static Case: The Immobile Electron 2
b) The Dynamic Case: The Moving Electron and the Progressing Field 3
3. Geometry of the Wire Loop and Iso-intensity Circles 1
a) Partitioning the Wire Loop into Sectors 4
b) Determining the Lengths of Segments AB and CD 5
c) Calculation of the Surface Area of Sector S 6
4. Potential within the Progressing Electric Field 4
a) Definition of Potential 7
b) Calculation of Potential around the Boundary of a Sector 7
c) Demonstration of Null Potential Variation on Arcs 8
d) Resultant Potential Variation for a Complete Sector 9
5. Potential and Field within the Wire Loop 7
a) Summation of Individual Sector Influences 11
b) Characterization of the Electric Field in the Wire Loop 12
c) Application of Stokes' Theorem to the Induced Field 13
6. The Influence of Electric Current 13
7. The Curl of the Progressing Electric Field 13
a) Calculation of the Curl for a Single Moving Charge 15
b) Calculation of the Curl due to a Macroscopic Current 16
c) Connection to the BiotrCoSavart Law 17
d) Connection to the Lorentz Force 17
8. Implications for a Theoretical Faraday's Law 18
9. Discussion 19
Letter to the Readers 20


For more detail, please read -2 A Derivation of Faraday's law from
Coulomb's Law and Relativity / 1.The Progressing Electric Field Model -+ https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html https://www.academia.edu/146009113/A_Derivation_of_Faradays_law_from_Coulombs_Law_and_Relativity_1_The_Progressing_Electric_Field_Model

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From Newsgroup: sci.physics

Kuan Peng wrote:

The Progressing Electric Field Model
[...] https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html

[blogspot.com, where many crackpots are self-publishing.]

On page 6, you write:

| Because v/c re- 1, (l/r_e)^2 is neglected in (4).

ISTM that that is precisely what you MUST NOT do if you want to arrive at a proper relativistic formulation of electromagnetism.

Incidentally, though, that train has already left the station; that ship has already sailed: Quantum electrodynamics is the best special-relativistic formulation of electrodynamics that we have to date; its predictions agreed with experiment to 10 decimal places in 2014 (probably more now). It is how you can even write this and post this as modern computer technology is based
on it.

Since the rest of your work is apparently based on this approximation, unfortunately it is useless (*iff* correct, it does NOT show anything
*new*) as it does NOT consider special-relativistic effects.

| In physics, we can use the sign "=" when a really small quantity is
| neglected.

That is just not true. Physics is an *exact* science. What you can do is
an approximation using the symbol "ree", or, more precisely, make a Maclaurin series approximation and signify the minimum degree of polynomials using the O-notation and then declare that one can neglect them if the variable of the polynomial is close to 0. (Einstein did that with the actual kinetic energy
to derive "E_0 = m c^2".)


https://www.academia.edu/146009113/A_Derivation_of_Faradays_law_from_Coulombs_Law_and_Relativity_1_The_Progressing_Electric_Field_Model

academia."edu", where most crackpots are self-publishing.

| In 1997, I discovered that the Lorentz force occurs because the density of
| a moving electric charge increases due to length contraction.

Yeah, well, it doesn't. The idea of a point-like object that somehow
carries an electric charge does not really work especially when one
considers special relativity; which is why we need quantum field theory to describe it properly.

But you can derive the Lorentz force law from the principle of stationary action in Minkowski space if you only consider the spatial components of the four-vectors; if I had time, I would post it here (it would be not my idea,
but from our Classical Field Theory lecture notes; maybe I will do it later).

See also: <https://en.wikipedia.org/wiki/Crackpot_index>
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

On 1/15/26 00:20, Thomas 'PointedEars' Lahn wrote:

Kuan Peng wrote:

The Progressing Electric Field Model
[...]
https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html

[blogspot.com, where many crackpots are self-publishing.]

On page 6, you write:

| Because v/c re- 1, (l/r_e)^2 is neglected in (4).

ISTM that that is precisely what you MUST NOT do if you want to arrive at a proper relativistic formulation of electromagnetism.

Incidentally, though, that train has already left the station; that ship has already sailed: Quantum electrodynamics is the best special-relativistic formulation of electrodynamics that we have to date; its predictions agreed with experiment to 10 decimal places in 2014 (probably more now). It is how you can even write this and post this as modern computer technology is based on it.

Since the rest of your work is apparently based on this approximation, unfortunately it is useless (*iff* correct, it does NOT show anything
*new*) as it does NOT consider special-relativistic effects.

| In physics, we can use the sign "=" when a really small quantity is
| neglected.

That is just not true. Physics is an *exact* science.

There is something called 'experiment', and it is difficult
to say if that can be 'exact'.

I suggest that one consult with another alien
called the 'Rael'.

I am thinking that 'real' in French is something like 'reel'.
But I am thinking that the 'Rael' is called the 'Rael' because
they consider that the 'Rael' has the right to interchange
the 'ae' with an 'ea' in English. You are utterly disconnected
from 'reality' unless you get approval from the 'Rael'. Your
pointy ears mean nothing.

I am thinking the article is OK if they just delete the words 'and Relativity'.How do you contact the 'Rael' to see if you are living in 'reality'? I do not know. Here is a link.

https://en.wikipedia.org/w/index.php?title=Claude_Maurice_Marcel_Vorilhon


What you can do is

an approximation using the symbol "ree", or, more precisely, make a Maclaurin series approximation and signify the minimum degree of polynomials using the O-notation and then declare that one can neglect them if the variable of the polynomial is close to 0. (Einstein did that with the actual kinetic energy to derive "E_0 = m c^2".)

https://www.academia.edu/146009113/A_Derivation_of_Faradays_law_from_Coulombs_Law_and_Relativity_1_The_Progressing_Electric_Field_Model

academia."edu", where most crackpots are self-publishing.

| In 1997, I discovered that the Lorentz force occurs because the density of | a moving electric charge increases due to length contraction.

Yeah, well, it doesn't. The idea of a point-like object that somehow
carries an electric charge does not really work especially when one
considers special relativity; which is why we need quantum field theory to describe it properly.

But you can derive the Lorentz force law from the principle of stationary action in Minkowski space if you only consider the spatial components of the four-vectors; if I had time, I would post it here (it would be not my idea, but from our Classical Field Theory lecture notes; maybe I will do it later).

See also: <https://en.wikipedia.org/wiki/Crackpot_index>

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From Newsgroup: sci.physics

Thank you for reading my paper and commenting.

What do you think about the violation of the law of conservation of energy
by FaradayrCOs law?

Kuan Peng

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From Newsgroup: sci.physics

Kuan Peng wrote:

Thank you for reading my paper and commenting.

You are welcome.

What do you think about the violation of the law of conservation of energy by FaradayrCOs law?

There are several (at least 3) Faraday's laws. I presume you mean Faraday's law _of induction_:

<https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction>

If so, why do you think that the law of the conservation of _total_ energy would be violated by it?
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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From Newsgroup: sci.physics

I have explained this violation of the law of conservation of energy in
the introduction.

To illustrate this, consider the following experimental setup: suppose two coils, A and B, are positioned side by side, with coil B connected to a resistor R, as shown in Figure 1.

Let the current in coil A, denoted as Ia, vary as follows: Ia increases linearly from zero to Imax, then decreases linearly back to zero. The
duration of each phase is +ot. According to Faraday's law, voltages are induced in coils A and B, which we label Va and Vb, respectively. Since Ia varies linearly during each phase, Va and Vb remain constant throughout
those intervals. Within resistor R, the voltage Vb generates a current Ib
and dissipates electric power equal to |VbIb|, both of which are constant
in each phase. Consequently, the total work performed in R after both
phases is 2|VbIb|NUat.

Since Ib is constant, it does not induce a voltage in coil A; therefore,
the value of Va remains unchanged regardless of whether Ib is positive, negative, or zerorCojust as if coil B were not present. When Ia increases,
the voltage in coil A (Va) is positive, and the electrical work performed
in A is given by the integral of VaIa . Conversely, when Ia decreases, the voltage in A becomes -Va, and the work equals the integral of -VaIa . Consequently, the total energy consumption of coil A after both phases
equals zero.

Since the energy consumption in coil A is zero, A does not transfer any
energy to coil B. We therefore encounter a case where B performs work
equal to 2|VbIb|NUat while receiving no energy from A. This implies that
the system consisting of coils A and B performs work without any energy
input, which violates the law of conservation of energy.

The cause of this violation is that Faraday's law predicts zero voltage in
A when the current in coil B is constant.

Kuan Peng

16/01/2026 |a 08:22, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Thank you for reading my paper and commenting.

You are welcome.

What do you think about the violation of the law of conservation of energy >> by FaradayrCOs law?

There are several (at least 3) Faraday's laws. I presume you mean Faraday's law _of induction_:

<https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction>

If so, why do you think that the law of the conservation of _total_ energy would be violated by it?

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Den 16.01.2026 12:50, skrev Kuan Peng:

I have explained this violation of the law of conservation of energy in
the introduction.

https://pengkuanem.blogspot.com/2026/01/a-derivation-of-faradays-law-from.html

To illustrate this, consider the following experimental setup: suppose
two coils, A and B, are positioned side by side, with coil B connected
to a resistor R, as shown in Figure 1.
Let the current in coil A, denoted as Ia, vary as follows: Ia increases linearly from zero to Imax, then decreases linearly back to zero. The duration of each phase is +ot. According to Faraday's law, voltages are induced in coils A and B, which we label Va and Vb, respectively. Since
Ia varies linearly during each phase, Va and Vb remain constant
throughout those intervals.

Let us first consider coil A only. Let its inductance be L.

(Just summing up, I expect you to agree)

When the current increases from zero to Imax,
the voltage on the current source is:
Va = LriadI/dt = Lria(Imax/+ot)
Note that both Va and Ia are positive.

The energy W stored in the magnetic field
at the time when the current is Imax is:
Wa = Lria(Imax)-#/2

When the current decreases from Imax to zero,
the voltage on the current source is:
Va = LriadI/dt = - Lria(Imax/+ot)
Note that Va is negative while Ia is positive.

That means that energy is delivered back to
the current source.

The energy W released from in the magnetic field
at the time when the current is zero is:
W = - Lria(Imax)-#/2 = -Wa

While the current is increasing, the current source will
deliver energy to the magnetic field,
when the current is decreasing, the magnetic field
will deliver the stored energy back to the current source.

(If the resistance in the coil is different from zero,
some energy will be lost as heat. Some energy will also
be lost as em-radiation. This will be strongly dependent
on +ot. We ignore these losses.)

Now let us consider what will happen in coil B.

Within resistor R, the voltage Vb generates
a current Ib and dissipates electric power equal to |VbIb|, both of
which are constant in each phase. Consequently, the total work performed
in R after both phases is 2|VbIb|NUat.

This is correct.

When the current in A is increasing, the voltage over
the resistor R will be constant Vb and the current Ib = Vb/R.

When the current in A is decreasing, the voltage over
the resistor R will be constant -Vb and the current -Vb/R = -Ib.
Energy loss Wb = 2|VbIb|ria+ot

Now let us consider how this will affect coil A.

Since Ib is constant, it does not induce a voltage in coil A; therefore,
the value of Va remains unchanged regardless of whether Ib is positive, negative, or zerorCojust as if coil B were not present. When Ia increases, the voltage in coil A (Va) is positive, and the electrical work
performed in A is given by the integral of VaIa . Conversely, when Ia decreases, the voltage in A becomes -Va, and the work equals the
integral of -VaIa . Consequently, the total energy consumption of coil A after both phases equals zero.

Let's take it from the beginning:

When the current in A is increasing, the magnetic field in A
will be increasing. Part of this field will go through B,
so there will be an increasing flux through B. This will induce
a constant voltage Vb and current Ib in B. This will give a
constant magnetic flux through B. Part of this flux go through A
and will have the opposite direction of the flux in A.
The result is that the flux in A will increase slower, and the
energy stored in the magnetic field at the time when the current
is Imax will be less than Wa = Lria(Imax)-#/2 . Let's call it Wa'.

When the current in A is decreasing, the magnetic field in A
will be decreasing. Part of this field will go through B,
so there will be an decreasing flux through B. This will induce
a constant voltage -Vb and current -Ib in B. This will give a
constant magnetic flux through B. Part of this flux go through A
and will have the same direction of the flux in A.
The result is that the flux in A will decrease slower, and the
energy stored in the magnetic field at the time when the current
is zero will be W = 0.

The net result is that the stored energy in A will increase from
zero to Wa', and then decrease back to zero.

This means that the current source has delivered the energy Wa
to the field, but has only got Wa' back.

Wa-Wa' = Wb

Since the energy consumption in coil A is zero, A does not transfer any energy to coil B.

Strange conclusion.

The energy consumption in coil A is zero.
A transfers energy to B.
This energy is supplied by the the current source in A.

We therefore encounter a case where B performs work
equal to 2|VbIb|NUat while receiving no energy from A. This implies that
the system consisting of coils A and B performs work without any energy input, which violates the law of conservation of energy.

The cause of this violation is that Faraday's law predicts zero voltage
in A when the current in coil B is constant.

--
Paul

https://paulba.no/
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Le 16/01/2026 |a 22:15, "Paul B. Andersen" a |-crit :

Let us first consider coil A only. Let its inductance be L.

(Just summing up, I expect you to agree)

When the current increases from zero to Imax,
the voltage on the current source is:
Va = LriadI/dt = Lria(Imax/+ot)

The energy W stored in the magnetic field
at the time when the current is Imax is:
Wa = Lria(Imax)-#/2

Yes. In accordance with FaradayrCOs law

When the current decreases from Imax to zero,
the voltage on the current source is:
Va = LriadI/dt = - Lria(Imax/+ot)

Yes. In accordance with FaradayrCOs law

That means that energy is delivered back to
the current source.

The energy W released from in the magnetic field
at the time when the current is zero is:
W = - Lria(Imax)-#/2 = -Wa

Yes. In accordance with FaradayrCOs law

While the current is increasing, the current source will
deliver energy to the magnetic field,
when the current is decreasing, the magnetic field
will deliver the stored energy back to the current source.

(If the resistance in the coil is different from zero,
some energy will be lost as heat.

Yes. In accordance with FaradayrCOs law

Now let us consider what will happen in coil B.

When the current in A is increasing, the voltage over
the resistor R will be constant Vb and the current Ib = Vb/R.

When the current in A is decreasing, the voltage over
the resistor R will be constant -Vb and the current -Vb/R = -Ib.
Energy loss Wb = 2|VbIb|ria+ot

Yes. In accordance with FaradayrCOs law

Now let us consider how this will affect coil A.

Let's take it from the beginning:

When the current in A is increasing, the magnetic field in A
will be increasing. Part of this field will go through B,
so there will be an increasing flux through B. This will induce
a constant voltage Vb and current Ib in B. This will give a
constant magnetic flux through B. Part of this flux go through A
and will have the opposite direction of the flux in A.
The result is that the flux in A will increase slower, and the
energy stored in the magnetic field at the time when the current
is Imax will be less than Wa = Lria(Imax)-#/2 . Let's call it Wa'.

With what law is computed Wa or Wa'? is it FaradayrCOs law?

This will give a constant magnetic flux through B.
Part of this flux go through A and will have the opposite direction of the flux
in A.

This part is constant in time I suppose. So, d(part)/dt =0 and Va=
LriadI/dt with no d(Bb)/dt =/=0.
This is FaradayrCOs law.

When the current in A is decreasing, the magnetic field in A
will be decreasing. Part of this field will go through B,
so there will be an decreasing flux through B. This will induce
a constant voltage -Vb and current -Ib in B. This will give a
constant magnetic flux through B. Part of this flux go through A
and will have the same direction of the flux in A.
The result is that the flux in A will decrease slower, and the
energy stored in the magnetic field at the time when the current
is zero will be W = 0.

With what law is computed W? is it FaradayrCOs law?

The net result is that the stored energy in A will increase from
zero to Wa', and then decrease back to zero.

This means that the current source has delivered the energy Wa
to the field, but has only got Wa' back.

Wa-Wa' = Wb

Are Wa , Wa' and Wb computed with FaradayrCOs law ? Or are they not
computed but only imagined?

Since the energy consumption in coil A is zero, A does not transfer any
energy to coil B.

Strange conclusion.

isn't it?

The energy consumption in coil A is zero.

Yes. In accordance with FaradayrCOs law

A transfers energy to B.

Yes. In accordance with FaradayrCOs law

This energy is supplied by the current source in A.

Can you compute the supplied energy with FaradayrCOs law or other
MaxwellrCOs equation?

Kuan Peng

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Kuan Peng wrote:

I have explained this violation of the law of conservation of energy in
the introduction.

The total energy is conserved in electrodynamics as well, which can be proved:-|

The energy density of the electromagnetic field is in Gaussian units

u(X, t) = (1/8-C) [|E(X, t)|^2 + |B(X, t)|^2],

where X is 3-position, and E and B are the electric and magnetic (flux
density) field vectors, respectively. We can also write this

u(X, t) = (1/8-C) [E(X, t) ria E(X, t) + B(X, t) ria B(X, t)],

where "ria" indicates the standard real scalar product.

Taking the partial derivative with respect to time, we find by the product rule

ree_t u(X, t) = (1/8-C) [ree_t E(X, t) ria E(X, t) + E(X, t) ria ree_t E(X, t)
+ ree_t B(X, t) ria B(X, t) + B(X, t) ria ree_t B(X, t)]

= (1/8-C) [2 E(X, t) ria ree_t E(X, t) + 2 B(X, t) ria ree_t B(X, t)]
= (1/4-C) (E ria ree_t E + B ria ree_t B),

where for simplicity in the last step I have not written the coordinate dependence of the fields anymore.

Using two of Maxwell's equations in Gaussian units,

rec |u E = -(1/c) ree_t B [Faraday's Law of Induction]
rec |u B = (4-C/c) J + (1/c) ree_t E [Amp|?re--Maxwell Circuital Law],

where J is the electric current density, we can write

ree_t u = (1/4-C) {E ria [c rec |u B - 4-C J] - c B ria (rec |u E)}
= (1/4-C) {c E ria (rec |u B) - 4-C E ria J - c B ria (rec |u E)}
= (c/4-C) {E ria (rec |u B) - B ria (rec |u E)} - E ria J.

We can use the product rule

rec ria (A |u B) = B ria (rec |u A) - A ria (rec |u B)

to write

ree_t u(X, t) = -(c/4-C) rec ria (E |u B) - E ria J.

one introduces the Poynting vector

S = (c/4-C) (E |u B),

to write the continuity equation

ree_t u(X, t) + rec ria S + J ria E = 0

rec ria S is basically the energy transported away by electromagnetic waves (which, following vector calculus, propagate in a direction perpendicular
to the electric and magnetic field lines).

J ria E is the work done by the electromagnetic field (actually, only the electric field) per unit volume and unit time:

The electromagnetic force is given by

F = q [E + V |u B),

where q is the electric charge of a particle that is moving in an electromagnetic field with the velocity V(R) = dR/dt. The work done on it
by the EM field along a curve C is

W = re2_C dR ria F(R)
= re2 dt dR/dt ria F(t)
= re2 dt V(t) ria q [E + V|u B)
= re2 dt q [V ria E + V ria (V |u B)]
= re2 dt q V ria E,

[V |u B is perpendicular to V, so V ria (V |u B) = 0], so the work done per unit
time is

dW/dt = q V ria E = q dR/dt ria E = dq/dt R ria E = I ria E,

where I is the directed electric current. The work done by the
electromagnetic field per unit volume and unit time is

dw/dt = J ria E. reA

___
-| Based on lecture notes for the course "Classical Electrodynamics" (autumn
semester 2023) by Uwe-Jens Wiese, University of Bern

To illustrate this, consider the following experimental setup: suppose two coils, A and B, are positioned side by side, with coil B connected to a resistor R, as shown in Figure 1.

Let the current in coil A, denoted as Ia, vary as follows: Ia increases linearly from zero to Imax, then decreases linearly back to zero.

For the current to *change*, work has to be done, thus energy is transferred
to or away from the coil. Thus the energy stored in the coil is not
conserved, but nobody claimed that it would be then; the law is, and this is what happens here, too, that the *total* energy stored in the coil *and* of
its environment is conserved.

The duration of each phase is +ot. According to Faraday's law, voltages are induced in coils A and B,

No, that is NOT what Faraday's law of induction is about. It states that an electric _current_ is induced by a changing _magnetic field_:

rec |u E = -(1/c) reeB/reet,

where E is the electric field vector, B is the magnetic flux density field vector, and t is time.

But you already presume a changing *current*.

The cause of this violation is that Faraday's law predicts zero voltage in
A when the current in coil B is constant.

No, the reason of this *seeming* violation is that you are confusing
yourself by assuming something that Faraday's law is not about, and
you have not considered all the facts.
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PointedEars

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Le 17/01/2026 |a 15:13, Thomas 'PointedEars' Lahn a |-crit :

The total energy is conserved:-|

Energy is always conserved.

All the formulas you have provided below are derived from MaxwellrCOs equations. So, they depend on Faraday's Law of Induction. If Faraday's Law
of Induction is flawed, then the reasoning based on these formulas cannot prove that Faraday's Law of Induction is not flawed.

u(X, t) = (1/8-C) [|E(X, t)|^2 + |B(X, t)|^2],
u(X, t) = (1/8-C) [E(X, t) ria E(X, t) + B(X, t) ria B(X, t)],
ree_t u(X, t) = (1/8-C) [ree_t E(X, t) ria E(X, t) + E(X, t) ria ree_t E(X, t)
+ ree_t B(X, t) ria B(X, t) + B(X, t) ria ree_t B(X, t)]
= (1/8-C) [2 E(X, t) ria ree_t E(X, t) + 2 B(X, t) ria ree_t B(X,
t)]
= (1/4-C) (E ria ree_t E + B ria ree_t B),
rec |u E = -(1/c) ree_t B [Faraday's Law of Induction]
rec |u B = (4-C/c) J + (1/c) ree_t E [Amp|?re--Maxwell Circuital Law],
ree_t u = (1/4-C) {E ria [c rec |u B - 4-C J] - c B ria (rec |u E)}
= (1/4-C) {c E ria (rec |u B) - 4-C E ria J - c B ria (rec |u E)}
= (c/4-C) {E ria (rec |u B) - B ria (rec |u E)} - E ria J.
rec ria (A |u B) = B ria (rec |u A) - A ria (rec |u B)
ree_t u(X, t) = -(c/4-C) rec ria (E |u B) - E ria J.
S = (c/4-C) (E |u B),
ree_t u(X, t) + rec ria S + J ria E = 0
F = q [E + V |u B),
W = re2_C dR ria F(R)
= re2 dt dR/dt ria F(t)
= re2 dt V(t) ria q [E + V|u B)
= re2 dt q [V ria E + V ria (V |u B)]
= re2 dt q V ria E,
dW/dt = q V ria E = q dR/dt ria E = dq/dt R ria E = I ria E,

dw/dt = J ria E. reA

For the current to *change*, work has to be done, thus energy is transferred to or away from the coil. Thus the energy stored in the coil is not conserved, but nobody claimed that it would be then; the law is, and this is what happens here, too, that the *total* energy stored in the coil *and* of its environment is conserved.

I absolutely agree with the principle of the law of conservation of
energy. So, the voltages and currents in the coils A and B must respect
this law. That is, the energy consumption in A equals the energy
consumption in B which equals the dissipated energy in the resistor which
is lost definitively.

But, because Faraday's Law of Induction determines that the voltages in
the coil A is exactly opposite in the two phases, the coil A gets back
100% the energy it radiated in the first phase, and finally the coil A
does not lose the dissipated energy in the resistor. It is the voltages determined by Faraday's Law of Induction that violates the law of
conservation of energy, not the coils A and B in a real experiment.

In a real experiment, the voltages and currents in the coils A and B
respect the equality: consumption in A equals consumption in B which
equals dissipated energy in the resistor.

So, the voltages predicted by Faraday's Law of Induction do not correspond
the voltages and currents measured in a real experiment. That is,
Faraday's Law of Induction has a flaw.

This is all what I say: real experiment respects the law of conservation
of energy, the voltages predicted by Faraday's Law of Induction does not.
The Law that human derived, Faraday's Law of Induction is incomplete. But, nature is always complete .

No, that is NOT what Faraday's law of induction is about. It states that an electric _current_ is induced by a changing _magnetic field_:

rec |u E = -(1/c) reeB/reet,

where E is the electric field vector, B is the magnetic flux density field vector, and t is time.

Sorry, but Faraday's law of induction states that an electric _
rCLvoltagerCY _ is induced by a changing _magnetic field , not current.

But you already presume a changing * *.

Current changes.

No, the reason of this *seeming* violation is that you are confusing
yourself by assuming something that Faraday's law is not about, and
you have not considered all the facts.

Please explain.

Kuan Peng

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Thomas 'PointedEars' Lahn wrote:

Kuan Peng wrote:

To illustrate this, consider the following experimental setup: suppose two >> coils, A and B, are positioned side by side, with coil B connected to a
resistor R, as shown in Figure 1.

Let the current in coil A, denoted as Ia, vary as follows: Ia increases
linearly from zero to Imax, then decreases linearly back to zero.

For the current to *change*, work has to be done, thus energy is transferred to or away from the coil. Thus the energy stored in the coil is not conserved, but nobody claimed that it would be then; the law is, and this is what happens here, too, that the *total* energy stored in the coil *and* of its environment is conserved.

The duration of each phase is +ot. According to Faraday's law, voltages are

induced in coils A and B,

No, that is NOT what Faraday's law of induction is about. It states that an electric _current_ is induced by a changing _magnetic field_:

rec |u E = -(1/c) reeB/reet,

where E is the electric field vector, B is the magnetic flux density field vector, and t is time.

But you already presume a changing *current*.

To elaborate:

When an electric current is flowing through coil A, that produces a non-zero magnetic field around the wire of the coil:

rec |u B = (4-C/c) J + (1/c) ree_t E. [Amp|?re--Maxwell Circuital Law]

Assuming a constant external electric field, this reduces to

rec |u B = (4-C/c) J.

The magnetic field can be calculated using either the Biot--Savart Law in general, or (as here) the Kelvin--Stokes theorem considering the cylindrical symmetry of the problem (here: for a cylindrical conducting wire):

re4_A dA ria (rec |u B) = (4-C/c) re4_A dA ria J
<==> re<_C dL ria B = (4-C/c) I(r)
<==> re2_0^{2-C} d-a r B(r) = (4-C/c) I(r)
<==> 2-C r B(r) = (4-C/c) I(r)
<==> B(r) = 2 I(r)/(c r),

where C is a circular path with radius r centered on the axis of symmetry of the wire, and dL is an infinitesimal piece of it:

R = r [cos(-a), sin(-a)]^T,
dL = dR/d-a = r [-sin(-a), cos(-a)]^T = r E_-a,
B(r) = B E_-a,

where by B the magnetic (flux density) field strength is meant, and E_-a is a unit vector in the -a-direction of cylindrical coordinates. [The actual magnetic field is H = B/++reC. The direction of the magnetic field lines, parallel to E_-a, i.e. counter-clockwise when looked at towards -E_z when the current is flowing parallel to E_z, is represented by Amp|?re's right-hand grip/(cork)screw rule.]

Thus you can see that the strength of the magnetic field *around* a wire is proportional to the electric current flowing in it.

In order to change the current in coil A, somebody or something has to do
work: Energy is added to or removed from the system. [Typically the current changes by changing direction: it is an alternating current. Thus the
polarity of the magnetic field changes as well.]

When the current changes, so does the magnetic field.

A changing magnetic field B induces a non-zero electric field E:

rec |u E = -(1/c) ree_t B, [Faraday's Law of Induction]

where by E and B we mean the vector fields again.

This electric field produces a force F = q E, thus an electric current in
the nearby coil B (which will in turn produce a non-zero magnetic field
around it that will to some extent influence coil A [self-induction].)

So you can see that *in total*, no energy is lost and none is gained.
--
PointedEars

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Kuan Peng wrote:

I absolutely agree with the principle of the law of conservation of
energy. So, the voltages and currents in the coils A and B must respect
this law. That is, the energy consumption in A equals the energy
consumption in B

There is no such thing as "energy consumption"; that there would be, is a fundamental misconception (among laypeople). Energy is not consumed, it is converted from one or more forms to one or more other forms.

which equals the dissipated energy in the resistor which is lost definitively.

No, energy is NOT lost. When the circuit is closed, energy is partially converted to the energy of the electromagnetic field whose electric field component E produces the force that produces the electric current, also in
coil A. From there is is partially converted to the energy of the magnetic field B around coil A. From the energy of the magnetic field around coil A
it is partially converted to the energy of the electric field E near coil B, inducing a current there (J ria E). In order to do that, energy is converted to that of electromagnetic waves which are radiated away (rec ria S).

There is really just one field, the electromagnetic field, which stores that energy, with the energy density

u(X, t) = (1/8-C) (||E(X, t)||^2 + ||B(X, t)||^2)

which changes according to

ree_t u(X, t) = -rec ria S - J ria E

as I showed before.

It is also partially converted to heat in all electric elements which are becoming warmer. So for actual conductors, ISTM that there would have to be
an additional term in the continuity equation above that is describing that.

See also:

Veritasium (2021): The Big Misconception About Electricity <https://www.youtube.com/watch?v=bHIhgxav9LY>

Veritasium (2022): How Electricity Actually Works <https://www.youtube.com/watch?v=oI_X2cMHNe0>

No, that is NOT what Faraday's law of induction is about. It states that an >> electric _current_ is induced by a changing _magnetic field_:

rec |u E = -(1/c) reeB/reet,

where E is the electric field vector, B is the magnetic flux density field >> vector, and t is time.

Sorry, but Faraday's law of induction states that an electric _ rCLvoltagerCY _ is induced by a changing _magnetic field , not current.

Wrong:

,-<https://en.wikipedia.org/w/index.php?title=Faraday%27s_law_of_induction&oldid=1332760723>
|
| In electromagnetism, Faraday's law of induction describes how a changing
| magnetic field can induce an electric current in a circuit. [...]
--
PointedEars

Twitter: @PointedEars2
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Kuan Peng <titang78@gmail.com> wrote or quoted:

Since the energy consumption in coil A is zero, A does not transfer any >energy to coil B.

Electromagnetism, especially the part with coils, isn't exactly
my strong suit! But if I had to take a stab at it, I'd say the
current in coil B creates a field that pushes back against the
change in current in coil A (effectively Lenz's law). So you
end up having to put in extra energy to keep the current rising
linearly, and that's the energy that gets dissipated.

Lenz's Law (1834):

|An induced current is always in such a direction as to oppose
|the motion or change causing it.

. Web:

|Lenz's Law ensures that induced currents flow in a direction
|that opposes the change in magnetic flux, promoting energy
|conservation.

.


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Le 18/01/2026 |a 05:28, Thomas 'PointedEars' Lahn a |-crit :

There is no such thing as "energy consumption"; that there would be, is a fundamental misconception (among laypeople). Energy is not consumed, it is converted from one or more forms to one or more other forms.

OK

It is also partially converted to heat in all electric elements which are becoming warmer. So for actual conductors, ISTM that there would have to be an additional term in the continuity equation above that is describing that.

Electric energy is converted to heat in the resistor.

This electric energy comes from the coil B.

The coil B gets the energy from the coil A.

The coil A gets the energy from a battery.

See also:
Wrong:

,-<https://en.wikipedia.org/w/index.php?title=Faraday%27s_law_of_induction&oldid=1332760723>
|
| In electromagnetism, Faraday's law of induction describes how a changing
| magnetic field can induce an electric current in a circuit. [...]

What happens when the electric circuit has a cut?

Kuan Peng

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From Newsgroup: sci.physics

Kuan Peng wrote:

Le 18/01/2026 |a 05:28, Thomas 'PointedEars' Lahn a |-crit :

It is also partially converted to heat in all electric elements which are
becoming warmer. So for actual conductors, ISTM that there would have to be >> an additional term in the continuity equation above that is describing that.

Electric energy is converted to heat in the resistor.

Yes, but not only there.

This electric energy comes from the coil B.

The coil B gets the energy from the coil A.

The coil A gets the energy from a battery.

Put *very* simply, yes. But I have explained before what happens in detail.

If you have watched the video, certainly the second one, you should now understand that all circuit elements receive electromagnetic energy from the battery via the electromagnetic field through the air (or whatever is
between them and the battery) *and* via the electromagnetic field on the surface of and outside the conducting wire. (NOT from the electrons which
are drifting very slowly, and with an alternating current, even back and
forth, instead.)

As a result, a light bulb as special kind of resistor turns on after the
time that an electromagnetic wave needs to travel from the battery to it *directly*, d/c, where d is the Euclidean distance.

It would even turn on when it is not at all connected to the circuit if it
were close enough to it and the field would be strong enough where it is
(there is an inverse-square law; such experiments have been made).

[Nikola Tesla tried to make this work over long distances (Wardenclyffe
Tower), but -- being a crackpot inventor, not a scientist -- he failed
because basically he did not consider the inverse-square law for EM waves. Today we are using it successfully in wireless charging, but your mobile
device has to be close enough to the charging station for that to work.]

In any case, even with this simplified explanation you can see that energy flows, is converted, and is not ever lost. Even if you cannot accept the dissipation of electromagnetic energy as given by the pointing vector, after all, work is being done.

,-<https://en.wikipedia.org/w/index.php?title=Faraday%27s_law_of_induction&oldid=1332760723>
|
| In electromagnetism, Faraday's law of induction describes how a changing >> | magnetic field can induce an electric current in a circuit. [...]

What happens when the electric circuit has a cut?

If you watch the second video, you can see that this is the key point here:
It does not matter. That is, it is not required that there is an electrical conductor connecting the coils. It is the changing magnetic field around
one coil, due to the changing current through it, that induces the current
in the other.
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PointedEars

Twitter: @PointedEars2
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From Newsgroup: sci.physics

Kuan Peng wrote:

Le 18/01/2026 |a 05:28, Thomas 'PointedEars' Lahn a |-crit :

It is also partially converted to heat in all electric elements which are
becoming warmer. So for actual conductors, ISTM that there would have to be >> an additional term in the continuity equation above that is describing that.

Electric energy is converted to heat in the resistor.

Yes, but not only there.

This electric energy comes from the coil B.

The coil B gets the energy from the coil A.

The coil A gets the energy from a battery.

Put *very* simply, yes. But I have explained before what happens in detail.

If you have watched the video, certainly the second one, you should now understand that all circuit elements receive electromagnetic energy from the battery via the electromagnetic field through the air (or whatever is
between them and the battery) *and* via the electromagnetic field on the surface of and outside the conducting wire. (NOT from the electrons which
are drifting very slowly, and with an alternating current, even back and
forth, instead.)

As a result, a light bulb as special kind of resistor turns on after the
time that an electromagnetic wave needs to travel from the battery to it *directly*, d/c, where d is the Euclidean distance.

It would even turn on when it is not at all connected to the circuit if it
were close enough to it and the field would be strong enough where it is
(there is an inverse-square law; such experiments have been made).

[Nikola Tesla tried to make this work over long distances (Wardenclyffe
Tower), but -- being a crackpot inventor, not a scientist -- he failed
because basically he did not consider the inverse-square law for EM waves. Today we are using it successfully in wireless charging, but your mobile
device has to be close enough to the charging station for that to work.]

In any case, even with this simplified explanation you can see that energy flows, is converted, and is not ever lost. Even if you cannot accept the dissipation of electromagnetic energy as given by the Poynting vector, after all, work is being done.

,-<https://en.wikipedia.org/w/index.php?title=Faraday%27s_law_of_induction&oldid=1332760723>
|
| In electromagnetism, Faraday's law of induction describes how a changing >> | magnetic field can induce an electric current in a circuit. [...]

What happens when the electric circuit has a cut?

If you watch the second video, you can see that this is the key point here:
It does not matter. That is, it is not required that there is an electrical conductor connecting the coils. It is the changing magnetic field around
one coil, due to the changing current through it, that induces the current
in the other.
--
PointedEars

Twitter: @PointedEars2
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Le 18/01/2026 |a 23:50, Thomas 'PointedEars' Lahn a |-crit :

In any case, even with this simplified explanation you can see that energy flows, is converted, and is not ever lost. Even if you cannot accept the dissipation of electromagnetic energy as given by the Poynting vector, after all, work is being done.

Yes, electromagnetic energy is dissipated.

What happens when the electric circuit has a cut?

If you watch the second video, you can see that this is the key point here: It does not matter. That is, it is not required that there is an electrical conductor connecting the coils. It is the changing magnetic field around
one coil, due to the changing current through it, that induces the current
in the other.

Yes, there is changing magnetic field in the cut.

But, do you think that Faraday's law of induction is correct?

Kuan Peng


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Kuan Peng wrote:

But, do you think that Faraday's law of induction is correct?

It has to be correct to high precision -- otherwise you could not read this.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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Le 19/01/2026 |a 22:03, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

But, do you think that Faraday's law of induction is correct?

It has to be correct to high precision -- otherwise you could not read this.

So, for you there cannot be any improvement to Faraday's law of induction.


Kuan Peng

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Le 18/01/2026 |a 19:46, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Since the energy consumption in coil A is zero, A does not transfer any >>energy to coil B.

Electromagnetism, especially the part with coils, isn't exactly
my strong suit! But if I had to take a stab at it, I'd say the
current in coil B creates a field that pushes back against the
change in current in coil A (effectively Lenz's law). So you
end up having to put in extra energy to keep the current rising
linearly, and that's the energy that gets dissipated.

Yes. you are absolutely right. This is how energy gets balanced in coils A
and B in real experiment .

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the
change in current in coil A

And there is no law in electromagnetism that defines a rCLfield that
pushes back rCY . So, we need to correct FaradayrCOs law or create a new
law to define the rCLfield that pushes back rCY

Kuan Peng

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Kuan Peng <titang78@gmail.com> wrote or quoted:

And there is no law in electromagnetism that defines a rCLfield that
pushes back rCY . So, we need to correct FaradayrCOs law or create a new
law to define the rCLfield that pushes back rCY

The back-reaction of the second coil on the power follows from Ohm's
law and Maxwell's equations, but not from Faraday's law alone.

The second coil has an emf acting on it by Faraday's law.
By Ohm's law a current appears in the second coil. According
to the Ampere-Maxwell equation (part of Maxwell's equations)

rot B = ++_0 J + ++_0 e_0 dE/dt,

a field arises from this current (here you can use the
magnetostatic approximation

rot B = ++_0 J

). J here is the current in the second coil.

A portion of this field creates a magnetic flux through
the first coil, which leads to an EMF in the first coil
by Faraday's law, increasing the load on the power supply.

Unicode:

EYcU |u EYEU = ++reC EYEe + ++reC -|reC reeEYEa/reet

EYcU |u EYEU = ++reC EYEe (approximation)

Summary of some laws:

Faraday's law states that a time-varying magnetic field induces
a circling electric field.

The Amp|?re-Maxwell law states that currents and time-varying
electric fields produce circling magnetic fields.

Ohm's law states that the current density through a conductor
is proportional to the electric field.

Lenz's law states that an induced current flows in a direction
such that the magnetic field it produces opposes the change
in magnetic flux that induced it. It can be derived from
Faraday's law, the Amp|?rerCoMaxwell law, Ohm's law, and the
conservation of energy (or the Lorentz force law).


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Kuan Peng writes:

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the
change in current in coil A

Coil B is a coil with current in it. Faraday's law predicts that
it will generate a field which opposes that generated by coil A.
--
John Hasler
john@sugarbit.com
Dancing Horse Hill
Elmwood, WI USA
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Kuan Peng wrote:

Le 19/01/2026 |a 22:03, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

But, do you think that Faraday's law of induction is correct?

It has to be correct to high precision -- otherwise you could not read this.

So, for you there cannot be any improvement to Faraday's law of induction.

I did not say that. In fact, I indicated the contrary.
--
PointedEars

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ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

The back-reaction of the second coil on the power follows from Ohm's
law and Maxwell's equations, but not from Faraday's law alone.

It may be instructive to say that Maxwell's laws

Gauss's law (electricity): div E = rho / epsilon_0

Gauss's law (magnetism): div B = 0

Faraday's law: curl E = - dB/dt

Ampere-Maxwell law: curl B = mu_0 J + mu_0 epsilon_0 dE/dt

can be written as just

d F = 0, d *F = J

after E and B are united to the two-form F (d being the exterior
differential, * the Hodge dual, and J the current 3-Form).

Even if one is not familiar with the exterior differential, this
show the unity of Maxwell's equations. They are just one single
law. So, this might make it clear that while "Faray's law" was
found before Maxwell's equations historically, one cannot actually
just split Faraday's law away and consider it in isolation always.

(I'm kinda tempted to define an operator "D" as the pairing
(d,d*) and a pairing j as (0,J) and then write "DF=j". Well,
you can see it this way, but that's my non-standard notation.)


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ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

Even if one is not familiar with the exterior differential, this
show the unity of Maxwell's equations. They are just one single
law. So, this might make it clear that while "Faray's law" was
found before Maxwell's equations historically, one cannot actually
just split Faraday's law away and consider it in isolation always.

Sorry, there were some typos above!

Michael Faraday (1791 - 1867), born to a blacksmith in a London
slum, devoured science from books he bound as a teen apprentice,
meticulously transcribing Humphry Davy's Royal Institution lectures
and boldly presenting himself as a lab assistant in 1813 - Davy,
awed by the young autodidact's zeal, hired him despite his wife
Jane's persistent snobbery and mistreatment of the lowly upstart.

Faraday's genius erupted in 1831 with his seminal discovery of
electromagnetic induction - now immortalized as Faraday's law
(o-int E dl = -d(Phi_B)/dt) - achieved by rigging coils around
an iron ring until a twitching galvanometer revealed that
a changing magnetic field births an electric current, powering
the first dynamo prototype where a spinning copper disc atop
a horseshoe magnet generated ceaseless electricity, all while
he dazzled Friday crowds with fireworks-like demos, liquefied
chlorine for fridges, and twisted light rays with magnets [1]
in notebook eureka moments, twice rejecting the Royal Society
presidency amid lifelong humility.

[1] The Faraday effect causes a polarization rotation which is
proportional to the projection of the magnetic field along the
direction of the light propagation.


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ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

d F = 0, d *F = J

I looked it up in Thirring (A Course in Mathematical Physics II)
and he actually writes, "d *F = - *J".

There is the concept of the codifferential delta, that allows
to write this as, "delta F = J", in Unicode, "+|F=J".


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Kuan Peng wrote:

Le 18/01/2026 |a 19:46, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Since the energy consumption in coil A is zero, A does not transfer any >>> energy to coil B.

Electromagnetism, especially the part with coils, isn't exactly
my strong suit! But if I had to take a stab at it, I'd say the
current in coil B creates a field that pushes back against the
change in current in coil A (effectively Lenz's law). So you
end up having to put in extra energy to keep the current rising
linearly, and that's the energy that gets dissipated.

Yes. you are absolutely right. This is how energy gets balanced in coils A and B in real experiment .

There is no "balancing of energy".

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the
change in current in coil A

Because that is not what that law is about. However, with the
Amp|?re--Maxwell Law (Maxwell's refinement of Amp|?re's Circuital Law-|), you can see that the induced electric field (which produces the current) induces "a(nother)" magnetic field-#.

We begin with FaradayrCOs law _of induction_ (AISB, there are _several_ "Faraday's laws"). In differential form and SI units, it is

rec |u E = reAreeBreoreet. (1)

The Amp|?re--Maxwell Law is in differential form and SI units

rec |u B = ++reC J + (1/c-#) reeEreoreet, (2)

where the capital letters mean vector fields.

But when we are thinking about "a (magnetic) (B) field that pushes back
against the change in current in coil A", that field is *not the same*
B field as in eq. (1). So we should label it differently, for example

rec |u B' = ++reC J + (1/c-#) reeEreoreet. (3)

[I am not sure yet if this E is the same E as in eq. (1), or the E in
eq. (1) only produces the J here. If it is not the same E, it should
be labeled differently as well, e.g. "E'".]

[Without proof yet:]

Eventually, the resulting magnetic field is the superposition of B and B';
the polarity of B' is opposite that of B, so it weakens the B field, so to speak, thus its change reduces the electric current produced by the change
of the B-field.-#

And there is no law in electromagnetism that defines a rCLfield that
pushes back rCY .

There is; it is called Lenz's Law:

<https://en.wikipedia.org/wiki/Lenz%27s_law>

It can probably be derived from the equations above; maybe I will do it when
I have more time.

So, we need to correct FaradayrCOs law or create a new
law to define the rCLfield that pushes back rCY

/Ex falso quodlibet./

___
-| Maxwell discovered that the total electric charge would only be conserved
if he modified Amp|?re's Circuital Law by adding a -- what he called --
"displacement current". [Since I have done the derivation again anyway
in an attempt to derive Lenz's Law, I might as well post it :'-)]

If we calculate the divergence of the left-hand side and right-hand side
of eq. (2), we obtain

rec ria (rec |u B') = rec ria (++reC J + +|reC ++reC reeEreoreet)
<==> 0 = ++reC (rec ria J) + +|reC ++reC reereoreet (rec ria E)

because the divergence of a curl field is zero ("a field with closed
field lines has no sources"), and partial derivatives of twice
differentiable functions commute (Schwarz--Clairaut Theorem). But we
also have Gauss' Law:

rec ria E = -U/+|reC,

where -U is the electric charge density. So

0 = ++reC (rec ria J) + +|reC/+|reC ++reC ree-Ureoreet
<==> 0 = rec ria J + ree-Ureoreet.
<==> rec ria J = -ree-Ureoreet.

This is the (non-relativistic) *continuity equation* for classical
electrodynamics. In words, it means: For an electric current to flow out
of a volume of space, the electric charge density in that volume must
decrease.

In other words, the total electric charge is conserved: When the electric
charge decreases in one volume, it must increase in another (the
adjacent) one. Or, if the charge density in a volume remains constant,
either there is no electric current passing through that volume, or as
much electric charge flows into it as out of it.

In yet other words, it is not possible to produce electric charge out
of nowhere (you have to take away the required amount of opposite
charge, which is what we actually mean by "charging"), or to destroy it
without neutralizing it.

-# There is only one magnetic field, actually only one electromagnetic
field. But it is useful to speak of the contributions to either by
different processes as different fields. This is allowed by the
principle of superposition which applies here because Maxwell's
equations are *linear* differential equations, and so the sum of
two solutions is also a solution. For example, if

rec |u E_1 = reAreereoreet B_1
rec |u E_2 = reAreereoreet B_2,

i.e. E_1 and B_1, and E_2 and B_2, are pairwise solutions of this
equation, then

rec |u E_1 + rec |u E_2 = reAreereoreet B_1 reA reereoreet B_2
<==> rec |u (E_1 + E_2) = reAreereoreet (B_1 + B_2),

so E_1 + E_2 and B_1 + B_2 are solutions as well. reA
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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Le 20/01/2026 |a 15:08, John Hasler a |-crit :

Kuan Peng writes:

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the
change in current in coil A

Coil B is a coil with current in it. Faraday's law predicts that
it will generate a field which opposes that generated by coil A.

What if the current in B is constant?
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Le 20/01/2026 |a 15:08, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Le 19/01/2026 |a 22:03, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

But, do you think that Faraday's law of induction is correct?

It has to be correct to high precision -- otherwise you could not read this.

So, for you there cannot be any improvement to Faraday's law of induction.

I did not say that. In fact, I indicated the contrary.

So, You mean that there can be correction to Faraday's law of induction .
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Le 20/01/2026 |a 14:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

The back-reaction of the second coil on the power follows from Ohm's
law and Maxwell's equations, but not from Faraday's law alone.

The second coil has an emf acting on it by Faraday's law.
By Ohm's law a current appears in the second coil. According
to the Ampere-Maxwell equation (part of Maxwell's equations)

The second coil has an emf acting on it by Faraday's law. This EMF is constant. So, the current in the second coil is constant.

a field arises from this current

The field from this current is constant.

A portion of this field creates a magnetic flux through the first coil,

This magnetic flux is constant because the current in the second coil is constant.

which leads to an EMF in the first coil by Faraday's law,

Constant magnetic flux does not change, so " which leads to an EMF " which
is zero in the first coil.

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Le 20/01/2026 |a 17:11, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

However, FaradayrCOs law does not define :

We begin with FaradayrCOs law _of induction_ (AISB, there are _several_ "Faraday's laws"). In differential form and SI units, it is

And there is no law in electromagnetism that defines a rCLfield that
pushes back rCY .

There is; it is called Lenz's Law:

If at the beginning the current in A is constant, do you think that there
is a current in the coil B?
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Kuan Peng <titang78@gmail.com> wrote or quoted:

Le 20/01/2026 |a 14:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :
The second coil has an emf acting on it by Faraday's law. This EMF is >constant. So, the current in the second coil is constant.

Ok.

a field arises from this current

The field from this current is constant.

Ok.

A portion of this field creates a magnetic flux through the first coil,

This magnetic flux is constant because the current in the second coil is >constant.

Ok.

which leads to an EMF in the first coil by Faraday's law,

Constant magnetic flux does not change, so " which leads to an EMF " which >is zero in the first coil.

I see that I have not addressed this point before. Let me give
it a try!

The (increasing) current I1(t) in the first coil creates a flux

u11(t) = L1 I1(t)

through the first loop, where L1 is the self-inductance of
the first loop (by the definition of inductance). The "(t)"
is intended to indicate the time dependency.

The (constant) current I2 in the second coil creates a flux

u12 = M I2

through the first loop where M is the mutual inductance of the
loops (by the definition of the mutual inductance).

The total flux through the first loop is

u1(t) = u11(t) + u12
= L1 I1(t) + M I2.

The sign of I2 is opposite that of I1 by Lenz's law.

So one can write: u1(t) = L1 |I1(t)| - M |I2|.

The flux u1(t) is reduced by M |I2| even if I2 is constant.

Thus, to get the same flux as without the other coil, |I1(t)|
must be greater, which requires more energy than without the
other coil.


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ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

through the first loop, where L1 is the self-inductance of

PS: "loop" is intended to have the same meaning as "coil" here.


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Kuan Peng wrote:

Le 20/01/2026 |a 15:08, John Hasler a |-crit :

Kuan Peng writes:

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the
change in current in coil A

Coil B is a coil with current in it. Faraday's law predicts that
it will generate a field which opposes that generated by coil A.

What if the current in B is constant?

[Those are vector fields that can depend both on time *and* position. Therefore, it is important to state precisely with respect to what a field
is constant: Does it not vary over time, or does it perhaps vary over time,
but not in space?]

Then, if the external electric field is constant over time or the external electric field strength is constantly increasing over time, too, the
resulting magnetic field is constant over time, too, as you can see from the Amp|?re--Maxwell Law

rec |u B = ++reC J + (1/c-#) reeEreoreet.

Take the (partial) derivative with respect to time, then

ree/reet (rec |u B) = ++reC reeJ/reet + (1/c-#) ree-#Ereoreet-#.

If I = const., then ||J|| = dI/dA = const., and reeJ/reet = 0, so

ree/reet (rec |u B) = rec |u reeB/reet = (1/c-#) ree-#Ereoreet-#.

Calculating the surface integral over some cross-sectional area A, one finds
by application of the Kelvin--Stokes Theorem (K/S):

re4_A dA ria (rec |u reeB/reet) = (1/c-#) ree-#Ereoreet-# re4_A dA

K/S
<==> re<_C dL ria reeB/reet = (1/c-#) ree-#Ereoreet-# re4_A dA,

where C is the delimiting curve, so eventually

reeB/reet ~ (1/c-#) ree-#Ereoreet-# re4_A dA,

and unless ree-#Ereoreet-# rea 0, then reeB/reet = 0. ree-#Ereoreet-# = 0 if either reeEreoreet = 0,
i.e. E = const. (wrt. time), or reeEreoreet = const. (wrt. time) rea 0. reA
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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Kuan Peng wrote:

Le 20/01/2026 |a 17:11, Thomas 'PointedEars' Lahn a |-crit :

There is; it is called Lenz's Law:

If at the beginning the current in A is constant,

What exactly do you mean by that?

do you think that there is a current in the coil B?

Unless coil B is in the vicinity of a voltage source, or if the external electric field is not constant, then there should not be:

rec |u E = reAreeBreoreet,
rec |u B = ++reC J + (1/c-#) reeEreoreet

and

J = -a E,

where -a is conductivity, and

I = re4 dA ria J,

so there is a current when the magnetic field changes, or there is a
non-zero electric field (that is produced by the voltage source).

But one has to be careful here. A *measured* current is rarely exactly 0
all the time; for example, there is interference from external source, and
the phenomenon and practical problem of a /Kriechstrom/ (literally: crawling current). Experimental evidence that a measured current is not zero is therefore insufficient to confirm the claim that Faraday's law of induction would be wrong or needed refinement. Much of it depends on your
experimental setup.
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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Le 21/01/2026 |a 08:29, Thomas 'PointedEars' Lahn a |-crit :

If at the beginning the current in A is constant,

What exactly do you mean by that?

do you think that there is a current in the coil B?

Unless coil B is in the vicinity of a voltage source, or if the external electric field is not constant, then there should not be:

We discuss the phenomenon in ideal condition with no field other than
those from A and B.
If you agree that
1. When the current in A is constant, the induced voltage and current in B
are zero.
Then,
2. When the current in B is constant, the induced voltage and current in A
are zero.

3. If the current in B is induced by the current in A, the current in B is
constant. Then, the voltage and current in A induced by the current in B
are zero, which is our case 2.


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Le 21/01/2026 |a 02:12, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Le 20/01/2026 |a 15:08, John Hasler a |-crit :

Kuan Peng writes:

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the >>>>> change in current in coil A

Coil B is a coil with current in it. Faraday's law predicts that
it will generate a field which opposes that generated by coil A.

What if the current in B is constant?

[Those are vector fields that can depend both on time *and* position. Therefore, it is important to state precisely with respect to what a field
is constant: Does it not vary over time, or does it perhaps vary over time, but not in space?]

The current in A varies linearly. The current in B is constant.
So, the magnetic field of A+B varies linearly.
A and B are both in this magnetic field which varies linearly.
According to FaradayrCOs law, the induced voltages in A and B are
proportional to dB/dt which is constant.
The induced voltage and current in A with or without the presence of B is
the same.
So, the energy dissipation is zero with or without the presence of B.

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Le 20/01/2026 |a 21:58, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Thus, to get the same flux as without the other coil, |I1(t)|
must be greater, which requires more energy than without the
other coil.

Great. But FaradayrCOs law does not specify how much more energy than
without the other coil. This is the missing term of FaradayrCOs law.


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Kuan Peng <titang78@gmail.com> wrote or quoted:

Great. But FaradayrCOs law does not specify how much more energy than >without the other coil. This is the missing term of FaradayrCOs law.

Faraday's law states that a time-varying magnetic field induces
a circling electric field.

It does not give that energy.

But that does not mean that terms need to be added to Faraday's
law because Faraday's law is not meant to describe everything
in the world when it is taken in isolation.

That energy? It can be calculated using a combination of several laws.

Faraday's law gives the induced emf E2(t) in the second loop from the
time rate of change of magnetic flux due to the first coil's current.

E2(t) = - M * dI1/dt

where M is the mutual inductance and I1(t) is the current in the
first coil.

With the loop 2 resistance R known, Ohm's law gives the induced
current

I2(t) = E2(t) / R.

The power dissipated as heat in the second loop is then

P2(t) = E2(t) * I2(t) = [E2(t)]^2 / R

The extra energy delivered to the second loop over some time
interval [t0, t1] is

W2 = integral from t0 to t1 of P2(t) dt
= integral from t0 to t1 of [E2(t)]^2 / R dt

Thus, Faraday's law provides E2(t); combined with the known
resistance R (and, if needed, the mutual inductance M and
primary current I1(t)), it determines the additional power
and energy absorbed in the second coil, without adding any
new term to Faraday's law itself.


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ram@zedat.fu-berlin.de (Stefan Ram) wrote or quoted:

rot B = ++_0 J + ++_0 e_0 dE/dt,

Oops! "rot" should be "curl" above.


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Le 21/01/2026 |a 22:20, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Great. But FaradayrCOs law does not specify how much more energy than >>without the other coil. This is the missing term of FaradayrCOs law.

Faraday's law states that a time-varying magnetic field induces
a circling electric field.

It does not give that energy.

But that does not mean that terms need to be added to Faraday's
law because Faraday's law is not meant to describe everything
in the world when it is taken in isolation.

That energy? It can be calculated using a combination of several laws.

Faraday's law gives the induced emf E2(t) in the second loop from the
time rate of change of magnetic flux due to the first coil's current.

E2(t) = - M * dI1/dt

where M is the mutual inductance and I1(t) is the current in the
first coil.

With the loop 2 resistance R known, Ohm's law gives the induced
current

I see. FaradayrCOs law is a tool. As a tool, it does not have to respect
the law of conservation of energy
PoyntingrCO theorem is a tool and does not have to respect the law of conservation of energy
Ohm's law is a tool and does not have to respect the law of conservation
of energy
We have to combine several laws to fabricate a global solution that
respects the law of conservation of energy

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From Newsgroup: sci.physics

Kuan Peng wrote:

Le 21/01/2026 |a 22:20, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Great. But FaradayrCOs law does not specify how much more energy than
without the other coil.

Gibberish.

This is the missing term of FaradayrCOs law.

Faraday's law states that a time-varying magnetic field induces
a circling electric field.

It does not give that energy.

But that does not mean that terms need to be added to Faraday's
law because Faraday's law is not meant to describe everything
in the world when it is taken in isolation.

That energy? It can be calculated using a combination of several laws.

Faraday's law gives the induced emf E2(t) in the second loop from the
time rate of change of magnetic flux due to the first coil's current.

E2(t) = - M * dI1/dt

where M is the mutual inductance and I1(t) is the current in the
first coil.

With the loop 2 resistance R known, Ohm's law gives the induced
current

I see. FaradayrCOs law is a tool.

No, Faraday's law _of induction_ (how many more times do I have to tell
you?) is an empirically confirmed physical law.

AISB, you could not read this if it were fundamentally wrong: most electric appliances, certainly electronic devices, include transformers (or require power adapters which include them to transform high voltage to low voltage) which are working based on that law:

<https://en.wikipedia.org/wiki/Transformer>

As a tool, it does not have to respect the law of conservation of energy

No, you simply have no clue what you are talking about.

PoyntingrCO theorem is a tool and does not have to respect the law of conservation of energy

No, you simply have no clue what you are talking about.

Ohm's law is a tool and does not have to respect the law of conservation
of energy

No, you simply have no clue what you are talking about.

We have to combine several laws to fabricate a global solution that
respects the law of conservation of energy

No; as I have proved to you in the very beginning, when we calculate the
change of the energy density of the electromagnetic field with respect to
time, the continuity equation for (classical) electrodynamics including
the Poynting vector and the work done by the elctromagnetic field, results *naturally*.

What I have not shown is how to calculate the energy density itself because
it is a rather lengthy calculation. But it can be shown by integrating the Lorentz force

F = q (E + V |u B)

over the path of a charged particle, which in turn can be derived from the special-relativistic (Lorentz-covariant) Lagrangian for a charged particle coupled to the electromagnetic field (given by the Maxwell tensor with components F_ab = ree_a A_b reA ree_b A_a, where A_a = [reA-o/c, A] is the four-potential, where -o is the electric potential in E = reArec-o, and A is the
magnetic vector potential in B = rec |u A).
--
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.
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Thomas 'PointedEars' Lahn wrote:

What I have not shown is how to calculate the energy density itself because it is a rather lengthy calculation.

By that I meant how to _derive the formula_ for the energy density of the EM field in Gaussian units,

u(X, t) = 1/(8-C) [E-#(X, t) + B-#(X, T)].

Once you have that formula, its value is rather trivial to calculate, of course.

But it can be shown by integrating the Lorentz force

F = q (E + V |u B)

over the path of a charged particle,

Hint: The absolute value of the work done by the electromagnetic field on a charged particle,

W = re2_P dS ria F = re2_P dS ria q (E + V |u B)

is equal to the difference between the energy that was stored in the field before it did that work and after that. Notice that if dS is an
infinitesimal line element of the particle's spatial path P, then dS || V
which means that the magnetic field does not do work on the particle:
dS ria q (V |u B) = 0. (But that does not mean that no energy is stored in it.)

which in turn can be derived from the special-relativistic (Lorentz-covariant)
Lagrangian for a charged particle coupled to the electromagnetic field (given by the Maxwell tensor with components F_ab = ree_a A_b reA ree_b A_a, where A_a = [reA-o/c, A] is the four-potential, where -o is the electric potential in
E = reArec-o, and A is the magnetic vector potential in B = rec |u A).

Hint: The total action is the sum of the action for a free particle,

S_0[x] = -m c^2 re2_W d-a,

where -a is proper time, and the action for the interaction with the EM field,

S_i[x; A] = q re2_W d-a A_a dx^a/d-a,

where W is a section of the (timelike) worldline of the particle. As usual, one can find the Euler--Lagrange equations by calculating the variation of
the total action, here

S[x; A] = S_0[x] + S_I[x; A].
--
PointedEars

Twitter: @PointedEars2
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[Supersedes because of too many typos]

Thomas 'PointedEars' Lahn wrote:

What I have not shown is how to calculate the energy density itself because it is a rather lengthy calculation.

By that I meant how to _derive the formula_ for the energy density of the EM field in Gaussian units,

u(X, t) = 1/(8-C) [E-#(X, t) + B-#(X, t)].

Once you have that formula, its value is rather trivial to calculate, of course.

But it can be shown by integrating the Lorentz force

F = q (E + V |u B)

over the path of a charged particle,

Hint: The absolute value of the work done by the electromagnetic field on a charged particle,

W = re2_P dS ria F = re2_P dS ria q (E + V |u B)

is equal to the difference between the energy that was stored in the field before it did that work and after that. Notice that if dS is an
infinitesimal line element of the particle's spatial path P, then dS || V
which means that the magnetic field does not do work on the particle:
dS ria q (V |u B) = 0. (But that does not mean that no energy is stored in it.)

which in turn can be derived from the special-relativistic (Lorentz-covariant)
Lagrangian for a charged particle coupled to the electromagnetic field (given by the Maxwell tensor with components F_ab = ree_a A_b reA ree_b A_a, where A_a = [reA-o/c, A] is the four-potential, where -o is the electric potential in
E = reArec-o, and A is the magnetic vector potential in B = rec |u A).

Hint: The total action is the sum of the action for a free particle,

S_0[x] = -m c^2 re2_W d-a,

where W is a section of the (timelike) worldline of the particle, -a is
proper time, and the action for the interaction with the EM field,

S_I[x; A] = q re2_W d-a A_a dx^a/d-a.

As usual, one can find the Euler--Lagrange equations by calculating the variation of the total action, here

S[x; A] = S_0[x] + S_I[x; A].

Because the variation is a *linear* differential operation, and the Euler--Lagrange equations are *linear* differential equations, knowing the
free relativistic Lagrangian, it suffices to vary S_I[x; A] to obtain the equations of motion for the interaction and add the non-interacting ones to obtain the Lorentz force equations.
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Le 23/01/2026 |a 13:19, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Le 21/01/2026 |a 22:20, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Great. But FaradayrCOs law does not specify how much more energy than >>>> without the other coil.

Gibberish.

This is the missing term of FaradayrCOs law.

Faraday's law states that a time-varying magnetic field induces
a circling electric field.

It does not give that energy.

But that does not mean that terms need to be added to Faraday's
law because Faraday's law is not meant to describe everything
in the world when it is taken in isolation.

That energy? It can be calculated using a combination of several laws. >>>
Faraday's law gives the induced emf E2(t) in the second loop from the
time rate of change of magnetic flux due to the first coil's current.

E2(t) = - M * dI1/dt

where M is the mutual inductance and I1(t) is the current in the
first coil.

With the loop 2 resistance R known, Ohm's law gives the induced
current

I see. FaradayrCOs law is a tool.

No, Faraday's law _of induction_ (how many more times do I have to tell
you?) is an empirically confirmed physical law.

AISB, you could not read this if it were fundamentally wrong: most electric appliances, certainly electronic devices, include transformers (or require power adapters which include them to transform high voltage to low voltage) which are working based on that law:

<https://en.wikipedia.org/wiki/Transformer>

As a tool, it does not have to respect the law of conservation of energy

No, you simply have no clue what you are talking about.

PoyntingrCO theorem is a tool and does not have to respect the law of
conservation of energy

No, you simply have no clue what you are talking about.

Ohm's law is a tool and does not have to respect the law of conservation
of energy

No, you simply have no clue what you are talking about.

We have to combine several laws to fabricate a global solution that
respects the law of conservation of energy

No; as I have proved to you in the very beginning, when we calculate the change of the energy density of the electromagnetic field with respect to time, the continuity equation for (classical) electrodynamics including
the Poynting vector and the work done by the elctromagnetic field, results *naturally*.

What I have not shown is how to calculate the energy density itself because it is a rather lengthy calculation. But it can be shown by integrating the Lorentz force

F = q (E + V |u B)

over the path of a charged particle, which in turn can be derived from the special-relativistic (Lorentz-covariant) Lagrangian for a charged particle coupled to the electromagnetic field (given by the Maxwell tensor with components F_ab = ree_a A_b reA ree_b A_a, where A_a = [reA-o/c, A] is the four-potential, where -o is the electric potential in E = reArec-o, and A is the
magnetic vector potential in B = rec |u A).

FaradayrCOs law is the same for all. I see now the difference of the understanding between you and me.

Thank you.

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Please trim your quotes to the relevant minimum.

Kuan Peng wrote:

FaradayrCOs law is the same for all.

What do you mean by that?

I see now the difference of the understanding between you and me.

I am not sure that you have understood me properly; anyhow:

Thank you.

You are welcome.
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From Newsgroup: sci.physics

Kuan Peng wrote:

Le 21/01/2026 |a 02:12, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Le 20/01/2026 |a 15:08, John Hasler a |-crit :

Kuan Peng writes:

However, FaradayrCOs law does not define :

the current in coil B creates a field that pushes back against the >>>>>> change in current in coil A

Coil B is a coil with current in it. Faraday's law predicts that
it will generate a field which opposes that generated by coil A.

What if the current in B is constant?

[Those are vector fields that can depend both on time *and* position.
Therefore, it is important to state precisely with respect to what a field >> is constant: Does it not vary over time, or does it perhaps vary over time, >> but not in space?]

The current in A varies linearly. The current in B is constant.
So, the magnetic field of A+B varies linearly.

I wonder how there is a current in (the secondary) coil B at all *before* electromagnetic induction. Usually there is not, i.e. the secondary coil
is NOT connected to a voltage source, but to some electric appliance:

<https://en.wikipedia.org/wiki/Electromagnetic_induction#History>

In any case, your third statement is only superficially true because the
change of the current in coil A changes the magnetic field in and around
coil A (which is also the magnetic field in and around around coil B). The change of the magnetic field induces another current which counteracts the
one in coil A and perhaps even coil B (Lenz's Law).

That, in a sense, another current is induced by that change follows from Faraday's law of induction (eq. 2 below) that was used to induce a current
in coil B in the first place; but its direction is not obvious (to me).

In vacuum:

rec |u B = ++reC (J + +|reC ++reC reeE/reet) = ++reC J + (1/c-#) reeE/reet, (1)
rec |u E' = -reeB'/reet, (2)

where (IIUC) E' is now the contribution to the electric field that is
induced by the change in the magnetic field B' due to the induced current,
that produces Lenz's opposing current ':-)

A and B are both in this magnetic field which varies linearly.

It is not, for the reason explained above.

According to FaradayrCOs law,

_of induction_ (but I guess we can drop that to simplify this discussion)

the induced voltages in A and B are proportional to dB/dt

Yes, that follows (sort of) from the differential form above, since the
induced _electric field_ gives rise to a spatial difference of electric potential, i.e. a voltage.

which is constant.

It is not.

The induced voltage and current in A with or without the presence of B is the same.

It is not.

So, the energy dissipation is zero with or without the presence of B.

/Ex falso quodlibet./
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Kuan Peng wrote:

Le 21/01/2026 |a 08:29, Thomas 'PointedEars' Lahn a |-crit :

If at the beginning the current in A is constant,

What exactly do you mean by that?

You have not answered my question.

do you think that there is a current in the coil B?

Unless coil B is in the vicinity of a voltage source, or if the external
electric field is not constant, then there should not be:

We discuss the phenomenon in ideal condition with no field other than
those from A and B.

Technically, the fields are not "from A and B", but exist in all of
spacetime, and just have different values at different locations and
times; but OK.

If you agree that
1. When the current in A is constant, the induced voltage and current in B
are zero.

I agree provisionally: *voltages* are _never_ induced, _currents_ are, due
to electric fields.

Then,
2. When the current in B is constant, the induced voltage and current in A
are zero.

Same as above.

3. If the current in B is induced by the current in A, the current in B is
constant.
Then, the voltage and current in A induced by the current in B
are zero, which is our case 2.

I disagree with both statements because of Lenz's Law.
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Thomas 'PointedEars' Lahn wrote:

In vacuum:

rec |u B = ++reC (J + +|reC ++reC reeE/reet) = ++reC J + (1/c-#) reeE/reet, (1)

The careful reader has probably noticed it: There is an extra ++reC between the parentheses :'-)

rec |u E' = -reeB'/reet, (2)

where (IIUC) E' is now the contribution to the electric field that is
induced by the change in the magnetic field B' due to the induced current, that produces Lenz's opposing current ':-)

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Le 27/01/2026 |a 22:45, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

What exactly do you mean by that?

You have not answered my question.

I mean that the law

rec |u E' = -reeB'/reet, (2)

is the same for all physicists. We understand it as that the curl of the electric field equals minus the rate of change of the magnetic field.

But each physicist attribute it different sense in his head and it is his proper interpretation that makes that different physicist when predicting
the outcome of a same experiment using the same formula rec |u E' = -reeB'/reet can give different result.

Le 27/01/2026 |a 22:37, Thomas 'PointedEars' Lahn a |-crit :

I wonder how there is a current in (the secondary) coil B at all *before* electromagnetic induction. Usually there is not, i.e. the secondary coil
is NOT connected to a voltage source, but to some electric appliance:

1. Before the current circulates in coil A, there is not a current in coil
B .
2. After the current circulates in coil A, a current is induced in coil B. 3. Now, we have a current in coil A and a current in coil B.
4. Because the current in coil A increases linearly, the magnetic field in
coil B increases linearly and its rate of change is -reeB'/reet which is constant.
5. Because-reeB'/reet is constant, rec |u E' = -reeB'/reet is constant. So, E' is constant.
6. Because E' is constant, the induced voltage in coil B is constant. Let Vb be this voltage
7. The resistor R is connected to the coil B and the voltage Vb is across R, the current in R equals Ib = Vb / R
8. Because Vb is constant, Ib = Vb / R is constant.
9. Now we have the current Ia in coil A and Ib in coil B. Ia generates the
magnetic field Ba and Ib generates Bb in space.
10. Now the total magnetic field in space is total B= Ba+ Bb
11. Because Ba increases linearly and Bb is constant, total B increases linearly with the same rate of change than Ba
12. Now the coil A and B are in the magnetic field total B which increases linearly and ree( total B)/reet is constant
13. Because rec |u E' = -ree( total B)/reet, E' is constant in the coil A and B. So, the induced voltage in the coil A and B, Va and Vb , are both constant
14. So, when the current Ia in coil A increases linearly, Va>0, Vb>0 when the current Ia in coil A decreases linearly, Va<0, Vb<0 .
15. The dissipation in coil B equals 2Vb*Vb/R*t>0
16. The dissipation in coil A equals time integral of (Va-Va) Ia*dt=0


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Kuan Peng wrote:

Le 27/01/2026 |a 22:45, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

What exactly do you mean by that?

You have not answered my question.

I mean that the law

rec |u E' = -reeB'/reet, (2)

is the same for all physicists.

Yes, but you confused the questions. I asked about this:

If at the beginning the current in A is constant,

What exactly do you mean by that?

We understand it as that the curl of the electric field equals minus
the rate of change of the magnetic field.

Yes.

But each physicist attribute it different sense in his head and it is his proper interpretation that makes that different physicist when predicting the outcome of a same experiment using the same formula rec |u E' = -reeB'/reet can give different result.

No.

Le 27/01/2026 |a 22:37, Thomas 'PointedEars' Lahn a |-crit :

If you had only replied to one posting as you should have, you would
probably not have confused the questions.

I wonder how there is a current in (the secondary) coil B at all *before*
electromagnetic induction. Usually there is not, i.e. the secondary coil
is NOT connected to a voltage source, but to some electric appliance:>

1. Before the current circulates in coil A, there is not a current in coil
B .
2. After the current circulates in coil A, a current is induced in coil B.

In practice there is no "circulating current" as it is an "alternating"
current to maximize the induced current.

Also, in general you should not think of electricity as electrons (or worse, positive charges) flowing through a conductor from one end of a circuit to
the other, like flowing water. That is NOT how it works:

Veritasium: The Big Misconception About Electricity <https://youtu.be/bHIhgxav9LY>

3. Now, we have a current in coil A and a current in coil B.

So far we are (partially) in agreement, then.

4. Because the current in coil A increases linearly, the magnetic field in
coil B increases linearly

This could only be said of the _average strength_ of the magnetic field (but
it would be wrong regardless, see below). The magnetic *field* is a
*vector* *field*; it does not make sense to say that a field increases, especially not a vector field.

and its rate of change is -reeB'/reet which is constant.

No. You have to consider that any change of the magnetic field also induces
a current that flows opposite the current that produced the non-zero field values -- Lenz's Law -- which I had indicated by putting primes in the
*second* induction equation.

5. Because-reeB'/reet is constant, rec |u E' = -reeB'/reet is constant. [...]

/Ex falso quodlibet./
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Kuan Peng <titang78@gmail.com> wrote or quoted:

But each physicist attribute it different sense in his head and it is his >proper interpretation that makes that different physicist when predicting >the outcome of a same experiment using the same formula rec |u E' = >-reeB'/reet can give different result.

Well, with that formula there really isn't much of an issue.
There's one part in electrodynamics that's kind of murky,
and that's the stuff tied to the radiation reaction.

|The Abraham-Lorentz formula has disturbing implications,
|which are not entirely understood a century after the law
|was first proposed.
from "Radiation Reaction" in "Introduction to Electrodynamics"
(2013) - David J. Griffiths


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Le 29/01/2026 |a 12:54, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

But each physicist attribute it different sense in his head and it is his >>proper interpretation that makes that different physicist when predicting >>the outcome of a same experiment using the same formula rec |u E' = >>-reeB'/reet can give different result.

Well, with that formula there really isn't much of an issue.
There's one part in electrodynamics that's kind of murky,
and that's the stuff tied to the radiation reaction.

|The Abraham-Lorentz formula has disturbing implications,
|which are not entirely understood a century after the law
|was first proposed.
from "Radiation Reaction" in "Introduction to Electrodynamics"
(2013) - David J. Griffiths

Electromagnetism has many paradoxes. For example, Lorentz force violates NewtonrCOs third law.


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Kuan Peng <titang78@gmail.com> wrote or quoted:

You think the current to be "alternating", I think it to be direct,
although increasing current.

Maxwell's equations hold in any case.

Is electron beam an electric current?

Yes. The current is the charge per unit time passing a given point.
Negative charges count as positive charges in the opposite direction.

Do we use MaxwellrCOs equations to
describe its behavior?

The behavior (acceleration) of each electron is given by
Newton's second law, F=ma, where F = F_em + F_other, and the
electromagnetic force F_em = Q[E + (v x B)] (when radiation
reaction [the recoil due to the electron's own electromagnetic
radiation] is negligible, which it usually is).


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From Newsgroup: sci.physics

Kuan Peng <titang78@gmail.com> wrote or quoted:

Electromagnetism has many paradoxes. For example, Lorentz force violates >NewtonrCOs third law.

This is true. But it just means that Newton's laws have a limited
scope. I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.

|In electrostatics and magnetostatics the third law holds,
|but in electrodynamics it does not.
"NewtonrCOs Third Law in Electrodynamics" in "Introduction to
electrodynamics" (2013) by David J. Griffiths.


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Kuan Peng writes:

You think the current to be "alternating", I think it to be direct,
although increasing current.

It is a superposition of a DC component and a triangle wave. We can
ignore the DC component in the steady state.
--
John Hasler
john@sugarbit.com
Dancing Horse Hill
Elmwood, WI USA
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From Newsgroup: sci.physics

Kuan Peng wrote:

Le 29/01/2026 |a 04:19, Thomas 'PointedEars' Lahn a |-crit :

Le 27/01/2026 |a 22:37, Thomas 'PointedEars' Lahn a |-crit :

I wonder how there is a current in (the secondary) coil B at all *before* >>>> electromagnetic induction. Usually there is not, i.e. the secondary coil >>>> is NOT connected to a voltage source, but to some electric appliance:>

1. Before the current circulates in coil A, there is not a current in coil
B .
2. After the current circulates in coil A, a current is induced in coil B. >>

In practice there is no "circulating current" as it is an "alternating"
current to maximize the induced current.

By " Before the current circulates in coil A " , I mean " Before the current occurs in coil A "

You think the current to be "alternating",

It is alternating.

I think it to be direct,

Merely an academic possibility. In real life, a transformer transforms high voltage to low voltage or vice-versa because the current is an alternating current.

Also, in general you should not think of electricity as electrons (or worse, >> positive charges) flowing through a conductor from one end of a circuit to >> the other, like flowing water. That is NOT how it works:

Is electron beam an electric current?

Yes.

Do we use MaxwellrCOs equations to describe its behavior?

Yes.

Your point being?

Veritasium: The Big Misconception About Electricity
https://youtu.be/bHIhgxav9LY

I have viewed this video a while ago. I think he is not a specialist of electromagnetism.

You are not in a position to make an informed judgement because evidently
you have never studied physics. He has, and so have I; he is correct.

Here too, we see the same video but have different interpretation.

I am not interpreting, I *know* because I have studied it. *You* are interpreting because you do NOT know. Big difference.

4. Because the current in coil A increases linearly, the magnetic field in
coil B increases linearly

This could only be said of the _average strength_ of the magnetic field (but >> it would be wrong regardless, see below). The magnetic *field* is a
*vector* *field*; it does not make sense to say that a field increases,
especially not a vector field.

The flux of magnetic field is a scalar and can increase.

Yes.

The induced voltage

Voltages are not induced.

is proportional to the rate of increase of the flux.

Only because of the electric field that is induced by the change of the magnetic (flux density) field. We have been over this already.

and its rate of change is -reeB'/reet which is constant.

No. You have to consider that any change of the magnetic field also induces >> a current that flows opposite the current that produced the non-zero field >> values -- Lenz's Law -- which I had indicated by putting primes in the
*second* induction equation.

The rate of change of the flux is constant, then the induced voltage is constant.

Read again what I wrote. *facepalm*
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John Hasler wrote:

Kuan Peng writes:

You think the current to be "alternating", I think it to be direct,
although increasing current.

It is a superposition of a DC component and a triangle wave. We can
ignore the DC component in the steady state.

Nonsense.
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Kuan Peng wrote:

Electromagnetism has many paradoxes. For example, Lorentz force violates NewtonrCOs third law.

:-D

No, it does NOT.
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Thomas 'PointedEars' Lahn wrote:

Kuan Peng wrote:

1. Before the current circulates in coil A, there is not a current in coil
B .
2. After the current circulates in coil A, a current is induced in coil B.

In practice there is no "circulating current" as it is an "alternating" current to maximize the induced current.

No idea why I put "alternating" in quotes here. An alternating current
changes direction periodically, in practice usually with a frequency of
either 50 Hz or 60 Hz, or a mixture of both (so about 50 to 60 times per second):

<https://en.wikipedia.org/wiki/Alternating_current#AC_power_supply_frequencies> --
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Le 30/01/2026 |a 01:20, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Electromagnetism has many paradoxes. For example, Lorentz force violates
NewtonrCOs third law.

:-D

No, it does NOT.

Please see this:
Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Electromagnetism has many paradoxes. For example, Lorentz force violates >>NewtonrCOs third law.

This is true. But it just means that Newton's laws have a limited
scope. I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.

|In electrostatics and magnetostatics the third law holds,
|but in electrodynamics it does not.
"NewtonrCOs Third Law in Electrodynamics" in "Introduction to electrodynamics" (2013) by David J. Griffiths.

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Le 30/01/2026 |a 01:08, John Hasler a |-crit :

Kuan Peng writes:

You think the current to be "alternating", I think it to be direct,
although increasing current.

It is a superposition of a DC component and a triangle wave. We can
ignore the DC component in the steady state.

We can. But the paradox subsists.



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Kuan Peng <titang78@gmail.com> wrote or quoted:

Please see this:
Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Electromagnetism has many paradoxes. For example, Lorentz force violates >>>NewtonrCOs third law.

This is true. But it just means that Newton's laws have a limited
scope. I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.
|In electrostatics and magnetostatics the third law holds,
|but in electrodynamics it does not.
"NewtonrCOs Third Law in Electrodynamics" in "Introduction to >>electrodynamics" (2013) by David J. Griffiths.

Here's an example for this (taken from Griffiths):

Assume, q1 is moving along the x axis at constant speed,
and q2 has the charge of q1 and is moving along the y axis
at the same speed.

The electric force is repulsive.

The magnetic field of of q1 at q2 points into the page,
so the magnetic force on q2 is towards the /right/.
(The "page" on which the coordinate system is drawn.)

The magnetic field of of q2 at q1 points out of the page,
so the magnetic force on q1 is /upwards/.

So, the force of q1 on q2 is /not/ opposite to the force
of q2 on q1.

(We can assume additional forces guiding the two particles to move
along those two axes at constant speed to fulfill our preconditions.)


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Le 29/01/2026 |a 23:08, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

The behavior (acceleration) of each electron is given by
Newton's second law, F=ma, where F = F_em + F_other, and the
electromagnetic force F_em = Q[E + (v x B)]

Is this correct? or

Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.

This is correct

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Le 30/01/2026 |a 01:18, Thomas 'PointedEars' Lahn a |-crit :

Kuan Peng wrote:

Le 29/01/2026 |a 04:19, Thomas 'PointedEars' Lahn a |-crit :

You think the current to be "alternating",

It is alternating.

I think it to be direct,

Merely an academic possibility. In real life, a transformer transforms high voltage to low voltage or vice-versa because the current is an alternating current.

My coils A and B are not those of a transformer. Just two plain coils. Nevertheless, FaradayrCOs law should work with them with a linearly
increasing current.

Also, in general you should not think of electricity as electrons (or worse,
positive charges) flowing through a conductor from one end of a circuit to >>> the other, like flowing water. That is NOT how it works:

Is electron beam an electric current?

Yes.

Do we use MaxwellrCOs equations to describe its behavior?

Yes.

Your point being?

If an electron beam is a current and MaxwellrCOs equations work for it.
Then, a current in a conductor should be consistent with MaxwellrCOs
equations whether

think of electricity as electrons (or worse, positive charges) flowing through a
conductor
from one end of a circuit to the other,

Or NOT.

You are not in a position to make an informed judgement because evidently
you have never studied physics. He has, and so have I; he is correct.

You are the judge

I am not interpreting, I *know* because I have studied it. *You* are interpreting because you do NOT know. Big difference.

You are the judge

The induced voltage

Voltages are not induced.

You can say it.

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Kuan Peng <titang78@gmail.com> wrote or quoted:

Le 29/01/2026 |a 23:08, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

The behavior (acceleration) of each electron is given by
Newton's second law, F=ma, where F = F_em + F_other, and the >>electromagnetic force F_em = Q[E + (v x B)]

Is this correct? or
Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.

This is correct

Newton's /first/ and /second/ laws are /always/ correct.
Only the third law is not valid in electrodynamics.

Feynman discusses this in Volume II, at the end of 26-2
"The fields of a point charge with a constant velocity".


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Le 30/01/2026 |a 20:04, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Le 29/01/2026 |a 23:08, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

The behavior (acceleration) of each electron is given by
Newton's second law, F=ma, where F = F_em + F_other, and the >>>electromagnetic force F_em = Q[E + (v x B)]

Is this correct? or
Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.

This is correct

Newton's /first/ and /second/ laws are /always/ correct.
Only the third law is not valid in electrodynamics.

Feynman discusses this in Volume II, at the end of 26-2
"The fields of a point charge with a constant velocity".

I do not agree, but OK.
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Le 29/01/2026 |a 04:19, Thomas 'PointedEars' Lahn a |-crit :

Le 27/01/2026 |a 22:37, Thomas 'PointedEars' Lahn a |-crit :

I wonder how there is a current in (the secondary) coil B at all *before* >>> electromagnetic induction. Usually there is not, i.e. the secondary coil >>> is NOT connected to a voltage source, but to some electric appliance:>

1. Before the current circulates in coil A, there is not a current in coil
B .
2. After the current circulates in coil A, a current is induced in coil B.

In practice there is no "circulating current" as it is an "alternating" current to maximize the induced current.

By " Before the current circulates in coil A " , I mean " Before the
current occurs in coil A "

You think the current to be "alternating", I think it to be direct,
although increasing current. This is an example of " each physicist
attribute it different sense in his head ", we are not talking about the
same experiment while talking about the same FaradayrCOs law.

Also, in general you should not think of electricity as electrons (or worse, positive charges) flowing through a conductor from one end of a circuit to the other, like flowing water. That is NOT how it works:

Is electron beam an electric current? Do we use MaxwellrCOs equations to describe its behavior?

Veritasium: The Big Misconception About Electricity https://youtu.be/bHIhgxav9LY

I have viewed this video a while ago. I think he is not a specialist of electromagnetism.
Here too, we see the same video but have different interpretation.

4. Because the current in coil A increases linearly, the magnetic field in
coil B increases linearly

This could only be said of the _average strength_ of the magnetic field (but it would be wrong regardless, see below). The magnetic *field* is a
*vector* *field*; it does not make sense to say that a field increases, especially not a vector field.

The flux of magnetic field is a scalar and can increase. The induced
voltage is proportional to the rate of increase of the flux.

and its rate of change is -reeB'/reet which is constant.

No. You have to consider that any change of the magnetic field also induces a current that flows opposite the current that produced the non-zero field values -- Lenz's Law -- which I had indicated by putting primes in the *second* induction equation.

The rate of change of the flux is constant, then the induced voltage is constant.

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Le 30/01/2026 |a 19:17, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit :

Kuan Peng <titang78@gmail.com> wrote or quoted:

Please see this:
Le 29/01/2026 |a 23:13, ram@zedat.fu-berlin.de (Stefan Ram) a |-crit : >>>Kuan Peng <titang78@gmail.com> wrote or quoted:

Electromagnetism has many paradoxes. For example, Lorentz force violates >>>>NewtonrCOs third law.

This is true. But it just means that Newton's laws have a limited
scope. I do not deem this to be an actual paradox, because we can
clearly see that here electromagnetism has priority over Newton.
|In electrostatics and magnetostatics the third law holds,
|but in electrodynamics it does not.
"NewtonrCOs Third Law in Electrodynamics" in "Introduction to >>>electrodynamics" (2013) by David J. Griffiths.

Here's an example for this (taken from Griffiths):

Assume, q1 is moving along the x axis at constant speed,
and q2 has the charge of q1 and is moving along the y axis
at the same speed.

The electric force is repulsive.

The magnetic field of of q1 at q2 points into the page,
so the magnetic force on q2 is towards the /right/.
(The "page" on which the coordinate system is drawn.)

The magnetic field of of q2 at q1 points out of the page,
so the magnetic force on q1 is /upwards/.

So, the force of q1 on q2 is /not/ opposite to the force
of q2 on q1.

(We can assume additional forces guiding the two particles to move
along those two axes at constant speed to fulfill our preconditions.)

exact
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