Waveform of Classic Electromagnetic Induction

[Paul Colby]

A non uniform magnetic field when sweeping across the coil will also tend to cut the side conductors of the coil (assuming side conductors are of significant length).

I think the point lost is that a time change of the field magnitude alone will cause emf to be generated, no cutting of field lines involved. EMF generated is related to change of flux which may or may not involve lines being cut.

 

[Charles Link]

For instructional purposes, it can be useful to work with a simpler scenario, e.g. a uniform magnetic field with a rotating coil, which is equivalent to a uniform but rotating magnetic field with a stationary coil. The magnetic field from a uniformly magnetized cylindrical magnet adds considerable mathematical complexity to the problem.
Determining the magnetic field from a cylindrical permanent (uniformly magnetized) magnet is a necessary part of this more complex calculation, while the determination of the EMF in the coil is an additional calculation that is most readily done numerically, (computer methods). I would have to believe that in general the waveform will be one with numerous significant Fourier components, besides the fundamental sinusoid.

See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ for computing the magnetic field of a cylindrical magnet.

 

[Paul Colby]

determining the magnetic field from a cylindrical permanent (uniformly magnetized) magnet is a necessary part of this more complex calculation, while the determination of the EMF in the coil is an additional calculation that is most readily done numerically, (computer methods).

Actually, the reciprocity integral relation I gave in #49 can be applied to this problem. One needs to use $J_2=\nabla\times M$ for the current. $E_2$ is then related to the generated voltage at the terminals to the coil. Integration by parts yields the integral of $B_1\cdot M$ over the volume of the magnet. $B_1$ is the field generated by supplying a current $J_1$ to the coil terminals. The only time (or angle) dependence in this integral comes from the angle $M$ makes with $B_1$ provided the both fields are uniform.

 

[b.shahvir]

This isn't helping in resolving my query .

 

[Paul Colby]

This isn't helping in resolving my query .

What would? The calculation I outlined in #73 can be done for uniform fields in closed form. All the principles remain valid without the uniform assumption. Non uniform fields lead to the waveform distortions as many have pointed out. What remains that is unclear?

 

[b.shahvir]

What remains that is unclear?

Contribution of induced emf in coil due to flux cutting.

 

[Paul Colby]

Lacking a definition of flux cutting no one can say. The calculation outlined yields the complete answer. It’s the best I can do.

 

[Delta2]

Contribution of induced emf in coil due to flux cutting.

Can you give us a reference paper where they calculate emf due to flux cutting by a conductor?
If you find anything that has integrals that contain $\mathbf{B} \times \mathbf{v}$ then don't bother, it is equivalent to faraday's law of induction.

 

[b.shahvir]

The equations for faraday's laws of EM are common to both methods of induced emf (rate of change of flux linkage as well as flux cutting). I was wondering if there was a way to distinguish between the two in the rotating magnet case.

 

[Delta2]

There is really only one method of calculating EMF and that's the rate of change of flux. What you refer as flux cutting I interpreted it as calculating integrals that contain $\mathbf{B}\times \mathbf{v}$ and is suitable for systems that contain moving wires with velocity $\mathbf{v}$. But this method is equivalent to the rate of change of flux method.

Here in this example in the rest frame of the rotating magnet the wire of the coil is moving, so you can use the second method but its going to be a hell of more complicated.

 

[b.shahvir]

Here in this example in the rest frame of the rotating magnet the wire of the coil is moving, so you can use the second method

Provided the results are same with both methods.

 

[alan123hk]

I believe that Faraday's law applies to all situations, for example, the magnet is stationary and the coil is moving, or the coil is stationary and the magnet is moving, or both the coil and the magnet are moving, or the mutual inductance of the two coils that move relative to each other, or others.

 

[Delta2]

Provided the results are same with both methods.

They are guaranteed to be the same, check https://en.wikipedia.org/wiki/Faraday's_law_of_induction sections 3 (Proof) and 4 (EMF for non thin wires)

 

[b.shahvir]

I believe that Faraday's law applies to all situations, for example, the magnet is stationary and the coil is moving, or the coil is stationary and the magnet is moving, or both the coil and the magnet are moving, or the mutual inductance of the two coils that move relative to each other, or others.

Absolutely. But in this particular case the contention was which method is best applicable (or both) to the rotating magnet case.
From the arrangement it implies that two situations are applicable to this case;
1) 'Statically' induced emf due to rate of change of flux linkage with coil (transformer simulation)
2) 'Dynamically' induced emf due to cutting of magnetic flux lines by side conductors of the coil as a non uniform magnetic field sweeps across the coil (dynamo simulation)
I believe the above 2 types of EM induction would give rise to individual emfs which may contribute as a whole to the total induced emf value in the coil at a particular instant.

 

[alan123hk]

Absolutely. But in this particular case the contention was which method is best applicable (or both) to the rotating magnet case.
From the arrangement it implies that two situations are applicable to this case;
1) 'Statically' induced emf due to rate of change of flux linkage with coil (transformer simulation)
2) 'Dynamically' induced emf due to cutting of magnetic flux lines by side conductors of the coil as a non uniform magnetic field sweeps across the coil (dynamo simulation)
I believe the above 2 types of EM induction would give rise to individual emfs which may contribute as a whole to the total induced emf value in the coil at a particular instant.

If this is a permanent magnet, so the strength of the magnetic field source itself does not change, then this should be a motion-induced electromotive force. You can use the Lorentz force (the cutting of the magnetic field) or the change in the magnetic flux through the coil to calculate the induced voltage. Both methods should get the same answer, but I don't think the two methods should be used at the same time. As for the method to be used, it depends on the actual situation and each person's choice.

For the example you put forward in #1, because the magnetic field generated by the magnet is very uneven in space, when the magnet rotates, the relative angle of the coil and the magnet is constantly changing, so no matter which method is used, I believe that manually calculating the actual induced voltage is very difficult.

 

[hutchphd]

I believe the above 2 types of EM induction would give rise to individual emfs which may contribute as a whole to the total induced emf value in the coil at a particular instant.

All changes in flux through the coil will result in "cutting" of flux lines. Either the lines move apart or the coil moves to a region of different line density. The results comport with Faraday's Law in either case.
But you cannot calculate a result by endlessly using semantics. Set up a model and calculate.
Edit: I recommend a program called Vizimag which I have used and offers a free 30 day trial I believe. This problem is inherently a 2D problem. Print out some fields (you can choose the line value) and look at how the fields and line cuttings behave. I think your intuition has been pretty good here.
You need to think about line cuttings and line "uncuttings" as well. It is the net change that matters for a loop.

 

[Charles Link]

Suggest you try the experiment if you have access to an oscilloscope. You could spin a cylindrical magnet that is about 10 cm long with an area of 1.0 cm^2. Just some ballpark numbers: The magnetic field at the endface $B \approx 1.0$ W/m^2, so that the flux $\Phi \approx 1.0$E-4 . It should be easy to achieve a $\Delta t=.1$ seconds for a half cycle. With $N=10$ turns, and assuming the entire flux goes through the coils, you get an EMF $\mathcal{E}=10$ mV. I think you are correct when you stated previously that the sinusoid will be distorted with a double "up" hump, and a double "down" hump.

Notice also, to get a stronger signal, you can spin the magnet faster.

 

[Baluncore]

Post #29 specified the dimensions, but not the number of turns on the coil.

 

[b.shahvir]

Suggest you try the experiment if you have access to an oscilloscope.

No I don't, that's why put query on forum.

 

[Charles Link]

No I don't, that's why put query on forum.

I presently don't have access to an oscilloscope either. It would be interesting to see some experimental results.

 

[Charles Link]

What you have here is a small electrical generator. One of the things I find most interesting is the cylindrical magnet and computing the magnetic field from it. See post 72. The calculation can be done by the magnetic pole method or by magnetic surface currents. Both methods get the identical result for the magnetic field.

 

[Tom.G]

Well, I finally ran the experiment.

The hardest part was finding a bar magnet on short notice. Actually I never did find one locally, I did find a horseshoe magnet so I ran two experiments, one with the horseshoe and one with a bar (obtained by sacrificing the horseshoe).

The coil is a spool of 30 AWG wirewrap wire I had around (no idea how many turns, many) and the oscilloscope is a Tektronix 465.

The test with the U (horseshoe) magnet was done by hand-holding the magnet and sweeping it across the diameter of the coil

The string tied around the center of the bar magnet was used for suspension and twisted to supply spin to the magnet upon release. Magnet was approximately on the coil axis.

The funny 'scope traces are because they are a composite of video frames taken with a bargain-store pocket video camera that is many years old.

(There should be 4 images, preview shows 6. I'll try a post-and-edit for clean-up... Arrgh they changed from full size to thumbnails! Oh well.)

Cheers,
Tom

 

[alan123hk]

Thank you very much for sharing the results of your experiment.

I think the result of Bar-composite.png seems similar to my previous prediction of induced voltage waveform.

My prediction is $~+A→-A→0→-A→+A$
But it is actually $~+A~→0→-A→-A→0→+A$

.

 

[Delta2]

Great work @Tom.G thanks for doing this experiment for all of us to confirm some of the theoretical discussion we had here.

 

[Delta2]

Actually the prevailing opinion here is that we will have two positive humps and two negative humps with some zero inbetween

but

if I interpret the oscilloscope screens correct (I am not EE so I don't have experience of working with an oscilloscope) I think we have two positive humps (local maximums) with another positive reversed hump in between (local minimum) and two negative humps with another negative reversed hump in between.

We missed this extra reversed hump..

 

[b.shahvir]

Well, I finally ran the experiment.

Cheers,
Tom

Thanks very much for taking time out for science and helping us with the practical aspect of the discussion. I greatly appreciate your effort and am grateful for it.
The experimental results almost matches the theoretical assumptions related to this arrangement. However, the only contradiction to the assumption is the continuity of the double peaks. In my opinion it should have gone to 0 between the double peaks as below;
0 +A 0 -A 0 -A 0 +A 0 +A 0 -A 0 -A 0 +A and so on...
I wonder why there was no 0 state instead of continuous v curves in between the double peaks.

 

[b.shahvir]

Actually the prevailing opinion here is that we will have two positive humps and two negative humps with some zero inbetween

Exactly! because 0 state in between indicates periodic pole reversals when the magnet rotates. I presume it could be due to oscillations of the string during rotation (causing changing flux linkages) which may be preventing the voltages from dropping to 0 in between pole reversals, but am not sure.

 

[Delta2]

Exactly! because 0 state in between indicates periodic pole reversals when the magnet rotates. I presume it could be due to oscillations of the string during rotation (causing changing flux linkages) which may be preventing the voltages from dropping to 0 in between pole reversals, but am not sure.

I think the voltage drops to 0 between pole reversals, but it does not drop to 0 in between the (positive) humps, instead it drops to a local minimum>0. (or local maximum <0 for the two negative humps).

 

[b.shahvir]

I think the voltage drops to 0 between pole reversals, but it does not drop to 0 in between the humps, instead it drops to a local minimum>0. (or local maximum <0).

In between the double peaks is the stage when the pole reversals take place. Voltage should go to 0 before each pole reversal.

 

[Delta2]

In between the double peaks is the stage when the pole reversals take place. Voltage should go to 0 before each pole reversal.

Yes you are right that in between the double peaks is the pole reversal. But for some reason the voltage doesn't drop to 0, instead it goes to a local minimum.

 

[b.shahvir]

Nope I don't agree with that, the two positive humps is when the north pole is approaching the coil and leaving the coil, and the third hump inbetween is when the north pole is aligned with the coil, that's what I think.
Similarly for the two negative humps and the south pole.

I would politely disagree. On careful analysis you will find the double humps(+ & - ) are actually formed when the poles reverse.

Let's say the the N pole is approaching the coil = 0 to +A
Next, the N pole is perfectly aligned with the axis of the coil = 0 state.
Again N pole is leaving the coil = 0 to -A
Next, the magnet is aligned perpendicular to axis of coil = 0 state
Next, the S pole is approaching the coil = 0 to -A (this is actually the 0 state between the double humps)
And the process continues as above...

 

[Delta2]

Next, the magnet is aligned perpendicular to axis of coil = 0 state

I disagree only with this, we don't get a 0 when this happens instead we get a positive local minimum (or a negative local maximum ).

 

[b.shahvir]

I disagree only with this, we don't get a 0 when this happens instead we get a positive local minimum (or a negative local maximum ).

Why? Could it be due to magnet field geometry or stray induction in oscilloscope probes or due to instability of the string! What prevents the voltage from going to 0 especially when the waveform is not a sinusoid.

 

[b.shahvir]

Is it an air coil or iron core coil? In my opinion the non zero state could be due to output current inductance in case of iron core coil.
Or maybe this is a current waveform. For proper voltage waveform you might require to drop it across a load resistance.

 

[Delta2]

Is it an air coil or iron core coil? In my opinion the non zero state could be due to output current inductance in case of iron core coil.
Or maybe this is a current waveform. For proper voltage waveform you might require to drop it across a load resistance.

Could be due to the self inductance of the coil but I doubt it, i think its because of the magnet field geometry as you said.