Faraday and conservation laws

In summary,I suspect you aren't correctly juggling with infinities. Why don't you try to solve the exercise in the link (hint by @rude man) and see if a limit R##\downarrow ##0 exists at all ?The solution with the nonzero R is trivial:"Induced emf=BulCurrent in the loop I=Bul/RLaplace force=BIl=B^2ul^2/RTo keep velocity constant you apply an opposite external forceThe power of this force is dissipated on the resistor as heat."
  • #1
12
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Hello to all. Second post here so I apologise if I post to the wrong place with wrong altitude etc
We have a schematic of a problem like here:
https://www.physicsforums.com/threa...ver-a-loop-under-influence-of-b-field.750763/

The conducting rod ab shown in the figure (Figure 1) makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field 1 T, perpendicular to the plane of the figure. The rod has initial velocity u=1m/s mass m=1kg. R of every conductor is 0 ohm

Initial kinetic is 0.5J. because of the Faraday law of induction an infinite force will halt the rod instantaneously in something that pretty much resembles a plastic colission with a wall. Final kinetic energy of the rod is 0.
Initial momentum is 1kg m/s and final momentum is 0.
As there are no losses in the circuit 0.5 joule must have gone to radiating electromagnetic field and the photons of this field must have 1kg m/s of momentum.
Let's suppose that all the photons are emitted directional (not true) in the direction of the initial velocity of the rod.
There is 8 orders of magnitude discrepancy based on the momentum energy relation for the photons emitted E=pc
What am I doing wrong?
Thank you for your time.
 
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  • #2
Leontas said:
Hello to all. Second post here so I apologise if I post to the wrong place with wrong altitude etc
Right altitude, wrong forum: should have been the same as the one for the link you gave: Intro Physics Homework
shown in the figure (Figure 1)
Don't see no number 1, but I suppose you mean
1574179129274.png

What am I doing wrong?
I suspect you aren't correctly juggling with infinities. Why don't you try to solve the exercise in the link (hint by @rude man) and see if a limit R##\downarrow ##0 exists at all ?

Furthermore I can't follow the reasoning with the photons. Does it work when R>0 ?
 
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  • #3
In calculations of this kind radiation is completely ignored, i.e., you work in the quasi-magnetostatic approximation. The motion of the rod induces a current due to the magnetic force acting on the electrons in the rod. Energy is dissipated through the ohmic loss, i.e., through heating up the wires and the rod.
 
  • #4
BvU said:
Furthermore I can't follow the reasoning with the photons. Does it work when R>0 ?
As almost always, photons do not help in understanding a question of classical electrodynamics. Even in the context of quantum theory the field paradigm is the right argument, at least qualitatively. Photons are no particles and cannot be understood as such! Massive point particles are already a nuissance in relativity theory, and massless ones are the more suspicious ;-)).
 
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  • #5
BvU said:
Right altitude, wrong forum: should have been the same as the one for the link you gave: Intro Physics Homework
Thank you for your reply. It has nothing to do with homework that's why I posted here. Hopefully, I haven't made a mistake.

BvU said:
Don't see no number 1, but I suppose you mean
https://www.physicsforums.com/attachments/253042

I suspect you aren't correctly juggling with infinities. Why don't you try to solve the exercise in the link (hint by @rude man) and see if a limit R##\downarrow ##0 exists at all ?
The attachment you provide do not link anywhere. What I meant was the same figure of the linked post with R=0
The solution with the nonzero R is trivial:
"Induced emf=Bul
Current in the loop I=Bul/R
Laplace force=BIl=B^2ul^2/R
To keep velocity constant you apply an opposite external force
The power of this force is dissipated on the resistor as heat."

Now in my question the only force acting on the rod is the Laplace force which goes to infinity as R approaches zero. The deceleration is also infinite due to Newton's second law. Resistance is 0 so no heat dissipation. The question is "as the rod halts where does the energy and momentum go?". So my first thought was that the energy goes to the electromagnetic field as the loop acts as an antenna with an infinite pulse of current. So I come to your question

BvU said:
Furthermore, I can't follow the reasoning with the photons. Does it work when R>0 ?

If we have a nonzero R at least the energy is dissipated there (I still do not know about the momentum) so there is no need to discuss about field energy.
To further clarify: If I am correct on energy and momentum radiating (which I most surely am wrong) from the system, even if the direction of the emitted radiation is directional (its the configuration with which you can have the largest momentum with a predetermined energy of photons. If you emit isotropically the total photon momentum is zero) for the energy dissipation requirement of 0.5J photons can only "carry" 0.5/c kgm/s of momentum and not 1kgm/sSecond thoughts on my question
Even if there is zero resistance there must be finite inductance of the loop(otherwise it would not be a loop). The generated loop current induces a secondary magnetic field and the energy is stored there. But the momentum??
In a way, this problem is similar to the two capacitor problem(if you haven't heard please google it. there are some publications on this. Also if you have time please look at sources on flux compression generators where ohmic losses are neglected and most energy is dissipated in a pulse)

My initial intention is at least qualitatively to find a limit (of acceleration of the rod? induced current in the loop?) when the conservation of energy in the context mechanical energy -> heat losses across R do not apply because the radiated energy starts to be progressively important part of the energy budget of the system. The intention is not to find a new theory or prove quantum electrodynamics or relativity wrong :-)

I hope I clarified my question.
Any help will be greatly appreciated
This problem drives me nuts!
Thank you!
 
  • #6
vanhees71 said:
As almost always, photons do not help in understanding a question of classical electrodynamics. Even in the context of quantum theory the field paradigm is the right argument, at least qualitatively. Photons are no particles and cannot be understood as such! Massive point particles are already a nuissance in relativity theory, and massless ones are the more suspicious ;-)).

I know but I hope some people with deep understanding on these topics could help me at least qualitatively.
thank you for your reply!
 
  • #7
A "spin off" question:
In these types of problems when there is a closed circuit and a current flows due to the induced emf in the loop if this current creates a secondary magnetic field comparable in magnitude to the initially applied field do we have the right to not take this into account when we calculate the induced EMF as Bul and not B'ul where B' is the total magnetic field? ( as the rod moves to the right it creates a field that opposes the initially applied field on its left and intensifies the initially applied field on its right - due to the right-hand rule)
I will greatly appreciate your views on this too!
 
  • #8
In principle you are right. What you mean is to take into account the Hall effect, which in fact should be done in a complete relativistic treatment, but for all practical purposes it's negligible. Note that the drift velocity of the electrons in a wire with usual currents is of the order of 1mm/s (sic!). That's really small compared to ##c## ;-)).

For a full relativistic treatment of the straight cylindrical wire (DC case), see my recent Insights article:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/
 
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  • #9
Leontas said:
A "spin off" question:
In these types of problems when there is a closed circuit and a current flows due to the induced emf in the loop if this current creates a secondary magnetic field comparable in magnitude to the initially applied field do we have the right to not take this into account when we calculate the induced EMF as Bul and not B'ul where B' is the total magnetic field? ( as the rod moves to the right it creates a field that opposes the initially applied field on its left and intensifies the initially applied field on its right - due to the right-hand rule)
I will greatly appreciate your views on this too!
Well let me ask you these two questions:

Does the moving rod move in its own magnetic field?
Does the moving rod move in the magnetic field of the rails?An extra question that may be worth pondering: If current in the rod is changing, does the changing current at the left side of the rod induce a voltage on the right side of the rod? I mean left side and right side of a vertical rod.
 
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  • #10
jartsa said:
Well let me ask you these two questions:

A)Does the moving rod move in its own magnetic field?
B)Does the moving rod move in the magnetic field of the rails?C)An extra question that may be worth pondering: If current in the rod is changing, does the changing current at the left side of the rod induce a voltage on the right side of the rod? I mean left side and right side of a vertical rod.
A) I do not know
B)yes and I forgot to mention the added magnetic field of the rails(as a current flows in them) . I think we have to take account for this too in order to calculate the induced voltage on the loop.
C) on the right side of the rod are static rails. Inside the rails there is changing flux but I do not care because we "can"make the rails arbitrary thin.

Thank you for your reply.
I was desperate that nobody would answer. I still have the problem unsolved in my mind however.
 
  • #11
vanhees71 said:
In principle you are right. What you mean is to take into account the Hall effect, which in fact should be done in a complete relativistic treatment, but for all practical purposes it's negligible. Note that the drift velocity of the electrons in a wire with usual currents is of the order of 1mm/s (sic!). That's really small compared to ##c## ;-)).

For a full relativistic treatment of the straight cylindrical wire (DC case), see my recent Insights article:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

Thank you vanhees71. I cannot understand how the hall effect is relevant :-(. As for the relativistic treatment you say that does not apply on this paradigm. However I will read and try to understand :-) your essay as soon as I get some time.
I still cannot understand where the energy and the momentum go. To the best of my thinking the rod will not halt but will start oscillating. Still in dark here. Probably I am a lost cause :-(
 
  • #12
Leontas said:
Initial kinetic is 0.5J. because of the Faraday law of induction an infinite force will halt the rod instantaneously in something that pretty much resembles a plastic colission with a wall. Final kinetic energy of the rod is 0.
At the end, by which I mean after a long time, the rod is moving very slowly, and very small current exists in two very long segments of the two rails. The inductance of the circuit is very large.

Magnetic energy is inductance multiplied by the square of current. Divided by two. ##E=1/2*LI^2##

So the lost kinetic energy has become magnetic energy.At the beginning the moving rod generates an electromotive force: ##U=Bvl##
Voltage U in a circuit with inductance L causes the current to rise at some finite rate. As the current is still small, the force on the rod is small.

The rod is a weak magnet, and there is the strong magnet that creates the 1 Tesla filed. There is a magnetic force between these two magnets. If it's not correct to call the rod a magnet, then we could say there is a force between the electric currents in the rod and the electric currents in the magnet, which conveniently happens to be an electromagnet.What happens between the start and the end, I leave that as an exercise for the reader. :smile: (If I understand railguns corrctly, then the rod experiences a repulsive force from those parts of the rails where there is a current )
 
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  • #13
Leontas said:
Thank you vanhees71. I cannot understand how the hall effect is relevant :-(. As for the relativistic treatment you say that does not apply on this paradigm. However I will read and try to understand :-) your essay as soon as I get some time.
I still cannot understand where the energy and the momentum go. To the best of my thinking the rod will not halt but will start oscillating. Still in dark here. Probably I am a lost cause :-(
Ok, then let's finally do the problem given in #2 and let's be content with the non-relativistic treatment (which is completely sufficient for usual experimental situations in the usual lectures). Let's denote the plane with the rod shown in the figure the ##(xy)## plane of a Cartesian coordinate system. Then the rod's velocity is ##\vec{v}=v \vec{e}_x## and ##\vec{B}=-B \vec{e}_z## (with ##B>0##). Then Faraday's Law in integral form (note that the rod is moving, and thus you need the complete Law!)
$$\mathcal{E}=\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B})=-\frac{1}{c} \mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B},$$
where ##A## is the rectangle along the wires and the rod. Now
$$\vec{v} \times \vec{B}=-v B \vec{e}_x \times \vec{e}_z=v B \vec{e}_y, \quad \mathrm{d}^2 \vec{f} = \mathrm{d} x \mathrm{d} y \vec{e}_3$$
and thus
$$\mathcal{E}=v B L.$$
That leads to a current
$$i=\frac{v B L}{R}$$
and a force on the rod
$$\vec{F}=-L i B \vec{e}_y \times \vec{e}_z=-L i B \vec{e}_x=-\frac{B^2 L^2}{R} v,$$
and thus the equation of motion of the rod is
$$\dot{v}=-\frac{B^2 L^2}{m R} v=-\gamma v.$$
The solution is
$$v(t)=v_0 \exp (-\gamma t).$$
Now let's calculate the mechanical energy loss. Since for ##t \rightarrow \infty## it's
$$\Delta E_{\text{mech}}=E_{\text{kin}}(t \rightarrow \infty)-E_{\text{kin}}(t=0)=-\frac{m}{2} v_0^2.$$
We can also calculate how much Ohmic loss there is. The power consumption is
$$P(t)=R i^2=\frac{B^2 L^2}{R} v^2(t)=m \gamma v_0^2 \exp(-2 \gamma t).$$
And thus the total energy converted into heat through the resistor is
$$E_{\text{therm}}=\int_0^{\infty} \mathrm{d} t P(t)=\frac{m v_0^2}{2}=-\Delta E_{\text{mech}},$$
i.e., energy conservation is fulfilled.
 
  • #14
jartsa said:
At the beginning the moving rod generates an electromotive force: U=BvlU=Bvl
Voltage U in a circuit with inductance L causes the current to rise at some finite rate. As the current is still small, the force on the rod is small.
That does not explain why the current in the circuit does not get ridiculously large. So how about this:

Inductance limits the current at the very beginning. Something else must limit the current later.

So I guess the current stops increasing when the rod is not moving in a magnetic field anymore when the current in the rails is cancelling out the original magnetic field.

Or in other words when the EMF induced by the motion of the rod in the field of the permanent magnet is canceled out by the EMF induced by the motion of the rod in the field of the rails.

Does that sound right?

The magnetic field of the rails looks like this:
https://en.wikipedia.org/wiki/Linear_motor#/media/File:Railgun-1.svg
 

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