Is there a difference between Faraday’s induction experiments?

  • #1
cg0303
31
3
Faraday first demonstrated his law of induction by showing that when he created a current in one wire coil it created a second current in a second coil.

Another experiment showed the same effect. When he moved a magnet through a wire coil, it created a current in the coil.

My question is this: it seems that the two scenarios are different. The second experiment simply demonstrates Maxwell’s third law (named after Faraday) as it simply involves a moving magnet and magnetic field. However, the first experiment involves an accelerating charge, as the charges in the first coil accelerate when a current is created. Accelerating charges create electromagnetic waves, which goes beyond Faraday’s law of induction.
 

Answers and Replies

  • #2
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
35,994
4,715
Faraday first demonstrated his law of induction by showing that when he created a current in one wire coil it created a second current in a second coil.

Another experiment showed the same effect. When he moved a magnet through a wire coil, it created a current in the coil.

My question is this: it seems that the two scenarios are different. The second experiment simply demonstrates Maxwell’s third law (named after Faraday) as it simply involves a moving magnet and magnetic field. However, the first experiment involves an accelerating charge, as the charges in the first coil accelerate when a current is created. Accelerating charges create electromagnetic waves, which goes beyond Faraday’s law of induction.

They are not different. The induced current in the second coil in each scenario is due the changing magnetic flux in the second coil.

Changing magnetic flux can occur by changing (i) magnetic field strength, (ii) area that the magnetic field goes through (iii) the angle that the area and the magnetic field makes.

In the first example, the current in the first coil is increasing, and thus, the magnetic field due to the current is also increasing. The second coil sees an increasing magnetic field (Case (i)). Thus, a changing flux, resulting in an induced current in the second coil.

In the second example, the moving magnet changes the magnetic field strength in the coil, because the magnetic field strength of a magnet typically depends on distance from it. So the coil sees a changing magnetic field as well when the magnet moves. So same change in flux.

They are both demonstrations of the same principle.

Zz.
 
  • Like
Likes etotheipi and vanhees71
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
Faraday's Law of induction nowadays is understood to be the Maxwell equation
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0,$$
and that's the full equation, no approximations made.

The first example is not so much different from the 2nd, because there a time-dependent magnetic field is at the place of the 2nd coil because the current in the first coil implies the existence of a time-dependent magnetic field due to the Ampere-Maxwell Law
$$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}.$$
Of course, you have to take all 4 Maxwell equations + the constitutive equations for the fields in a medium into account to get a complete set of dynamical equations.

The 2nd example provides a time-varying magnetic field due to the motion of the magnet but also an electric field due to this motion. That's of course best understood using a full relativistic description of the em. field (it's anyway a relativistic field theory).

A 3rd variant is to keep the permanent magnet at rest and move the coil, which again is based on the same principles. It's this example which lead Einstein to the discovery of the fully consistent theory of electrodynamics of moving bodies, which is known as the special theory of relativity.
 
  • Like
Likes cg0303 and etotheipi
  • #4
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
Accelerating charges create electromagnetic waves, which goes beyond Faraday’s law of induction.
The fact that accelerating charges create electromagnetic waves is a consequence of Faraday's law of induction and Ampere's-Maxwell law (Ampere's law with the addition of displacement current term by Maxwell). So it's not something far beyond Faraday's law as you might thought at first place.
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
It's not beyond Faraday's Law of induction at all. Faraday's Law of induction is one of the fundmamental Maxwell equations necessary to derive the existence of em. fields!
 
  • #6
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
necessary to derive the existence of em. fields
You mean necessary to derive the wave equation which the E and B field obey.
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
Sure, I should have written: necessary to derive the existence of em. wave solutions.
 
  • #8
cg0303
31
3
Faraday's Law of induction nowadays is understood to be the Maxwell equation
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0,$$
and that's the full equation, no approximations made.

The first example is not so much different from the 2nd, because there a time-dependent magnetic field is at the place of the 2nd coil because the current in the first coil implies the existence of a time-dependent magnetic field due to the Ampere-Maxwell Law
$$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}.$$
Of course, you have to take all 4 Maxwell equations + the constitutive equations for the fields in a medium into account to get a complete set of dynamical equations.

The 2nd example provides a time-varying magnetic field due to the motion of the magnet but also an electric field due to this motion. That's of course best understood using a full relativistic description of the em. field (it's anyway a relativistic field theory).

A 3rd variant is to keep the permanent magnet at rest and move the coil, which again is based on the same principles. It's this example which lead Einstein to the discovery of the fully consistent theory of electrodynamics of moving bodies, which is known as the special theory of relativity.
Ok. Would it be correct to say that, while both examples demonstrate the law of induction, electromagnetic waves exist only in the first?
 
  • #9
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
Ok. Would it be correct to say that, while both examples demonstrate the law of induction, electromagnetic waves exist only in the first?
No, electromagnetic waves are being produced by the moving magnet as well. If the magnet oscillates mechanically back and forth with frequency ##f## then it is the source of electromagnetic waves of the same frequency ##f##. Generally any motion with acceleration of the magnet produces EM waves.
 
  • Like
Likes vanhees71 and cg0303
  • #10
cg0303
31
3
No, electromagnetic waves are being produced by the moving magnet as well. If the magnet oscillates mechanically back and forth with frequency ##f## then it is the source of electromagnetic waves of the same frequency ##f##. Generally any motion with acceleration of the magnet produces EM waves.
In the second experiment, I imagined the magnet moving at a constant velocity, not moving back and forth. In that case, I'm assuming there would not be an EM wave?
 
  • #11
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
In the second experiment, I imagined the magnet moving at a constant velocity, not moving back and forth. In that case, I'm assuming there would not be an EM wave?
Even in this case, I think an EM wave is being produced . That is because even when the magnet is moving with constant velocity, the magnetic field around the magnet is time varying, and according to Maxwell's equations a time varying magnetic field is always accompanied by a time varying electric field and together they form an EM wave.
 
  • #12
cg0303
31
3
Even in this case, I think an EM wave is being produced . That is because even when the magnet is moving with constant velocity, the magnetic field around the magnet is time varying, and according to Maxwell's equations a time varying magnetic field is always accompanied by a time varying electric field and together they form an EM wave.
What determines the frequency of the wave in this case?
 
  • #13
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
What determines the frequency of the wave in this case?
Not sure, i think in this case the em wave will have a spectrum of frequencies and not a single frequency.
 
  • #14
A.T.
Science Advisor
11,672
2,964
Even in this case, I think an EM wave is being produced . That is because even when the magnet is moving with constant velocity, the magnetic field around the magnet is time varying, and according to Maxwell's equations a time varying magnetic field is always accompanied by a time varying electric field and together they form an EM wave.
Just to be clear, are you saying that an isolated magnet at constant velocity in an inertial frame produces EM waves, that propagate away from it? The waves carry energy away, so does the magnet slow down if you don't propel it?

What happens in the rest frame of the magnet? Is there radiation too? If yes, where does the energy for it come from?
 
  • #15
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
For a magnet at constant velocity there are no waves radiated, as is easily seen from transforming back to the rest frame of the magnet, where you have a magnetostatic field, which is not a radiation field. The field in the frame, where the magnet is moving is just the Lorentz-boosted field of the magnetostatic field.
 
  • Like
Likes cg0303, Delta2 and A.T.
  • #16
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
The field in the frame, where the magnet is moving is just the Lorentz-boosted field of the magnetostatic field.
In the frame where the magnet is moving, isn't the magnetic field time varying?
 
  • #17
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
Yes it is. Also the field of a uniformly moving point charge is time-dependent. There's nevertheless no radiation. It's just a boosted Coulomb field.
 
  • #18
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
Yes it is. Also the field of a uniformly moving point charge is time-dependent. There's nevertheless no radiation. It's just a boosted Coulomb field
So ##\frac{\partial \mathbf{B}}{\partial t}\neq 0## but is it ##\frac{\partial \mathbf{B}}{\partial t}=constant## that's why there is no radiation?

What is a Lorentz-boosted field?
 
  • #19
artis
1,330
882
@cg0303 you say these experiments are different because
it seems that the two scenarios are different. The second experiment simply demonstrates Maxwell’s third law (named after Faraday) as it simply involves a moving magnet and magnetic field. However, the first experiment involves an accelerating charge, as the charges in the first coil accelerate when a current is created.

But think about it this way , in the two coil scenario which is a transformer as it is known, does the second coil have any information about what sort of charges or what exactly is happening in the first coil? All the second coils knows is that a magnetic field is increasing and decreasing through it. Imagine you were the second coil, can you see the electrons in the first coil? No you can't. But can you experience a increase/decrease in B field? Yes you can. So you have a time varying field.

Then think of the permanent magnet and coil scenario , again imagine you are the coil, what do you see? Does it even matter what you see, you again feel/register an increase/decrease in field strength.

On the same note you can take a symmetrical permanent cylinder shaped magnet with axial polarization and rotate it around it's axis of polarization and there would be no field change.
Permanent magnet or electromagnet it doesn't matter, the effect on a nearby coil is the same as long as the permanent magnet is moved in the right way near or through the coil.


Let me give another simple example. Take a transformer , probe the secondary winding , you see 50Hz sine wave, then probe a small electrical generator for a 50hz grid that has permanent magnet rotor, what do you see at the stator coils? again 50hz sine wave,

I think we could say the two scenarios are different simply in how the magnetic field was created in the source but not different from the coil's perspective as both cases involve the field lines increasing/decreasing through the coil and the coil couldn't care less what caused the field lines as long as they are changing.
 
  • #20
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
So ##\frac{\partial \mathbf{B}}{\partial t}\neq 0## but is it ##\frac{\partial \mathbf{B}}{\partial t}=constant## that's why there is no radiation?

What is a Lorentz-boosted field?
By definition a radiation field has parts, for which ##\vec{E} \times \vec{B}## go like ##1/r^2## for ##r \rightarrow \infty## and Lorentz-boosted static fields have no such parts, i.e., they go faster to 0 at infinity, and thus the total energy radiated out from any finite spatial region vanishes at infinity, i.e., no energy is transported far out from the sources of the fields.

For the Lorentz transformation of the em. field, see Eqs. (2.3.36) and (2.3.37) in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
 
  • #21
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
By definition a radiation field has parts, for which ##\vec{E} \times \vec{B}## go like ##1/r^2## for ##r \rightarrow \infty## and Lorentz-boosted static fields have no such parts, i.e., they go faster to 0 at infinity, and thus the total energy radiated out from any finite spatial region vanishes at infinity, i.e., no energy is transported far out from the sources of the fields.

For the Lorentz transformation of the em. field, see Eqs. (2.3.36) and (2.3.37) in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
So , something that is important for me personally, there are components of E and B such that the Poynting vector ##\mathbf{E}\times\mathbf{B}## is not zero but it falls faster than ##1/r^2## (it falls like ##1/r^4## for example for Lorentz boosted fields).

I can't seem to be able to locate Eqs (2.3.36) and (2.3.37), it stops at Eq 2.3.18.

But yes, thanks a lot for all and for this, it seems a great text for special relativity (and not only).
 
  • #22
vanhees71
Science Advisor
Insights Author
Gold Member
2022 Award
22,124
13,031
Sorry for my typo. I mean of course Eqs. (3.2.36) and (3.2.37) on page 53. Unfortunately the text is only in its beginnings :-(.
 
  • #23
tech99
Science Advisor
Gold Member
2,621
1,169
Even in this case, I think an EM wave is being produced . That is because even when the magnet is moving with constant velocity, the magnetic field around the magnet is time varying, and according to Maxwell's equations a time varying magnetic field is always accompanied by a time varying electric field and together they form an EM wave.
But in such a case are not the two fields in quadrature rather than in-phase as for an EM wave?
 
  • #24
tech99
Science Advisor
Gold Member
2,621
1,169
No, electromagnetic waves are being produced by the moving magnet as well. If the magnet oscillates mechanically back and forth with frequency ##f## then it is the source of electromagnetic waves of the same frequency ##f##. Generally any motion with acceleration of the magnet produces EM waves.
As a magnet always contains charges, I think it is difficult to know whether it is the magnet or the charges doing the radiating.
 

Suggested for: Is there a difference between Faraday’s induction experiments?

Replies
222
Views
6K
Replies
2
Views
310
Replies
3
Views
527
  • Last Post
Replies
2
Views
585
Replies
3
Views
145
Replies
2
Views
908
  • Last Post
Replies
6
Views
827
  • Last Post
Replies
5
Views
371
  • Last Post
Replies
1
Views
288
Top