Waveform of Classic Electromagnetic Induction

[b.shahvir]

Yes. If the magnet is completely immersed in a uniform magnetic field while rotating. By a uniform magnetic field, I refer to one that is generated by, for example, a Helmholtz coil when the coil is energized by a constant current. In this case, the generated EMF by a spinning magnet is exactly a sinusoid. The generated EMF is related to the volume integral of a constant B field dotted with the magnetization. The only time dependence comes from the direction of the magnetization, which is a pure sinusoid in time.

Do you have a diagram of the set up? Will the results be similar if I replace a bar magnet with a cylindrical dipole magnet in the present rotating magnet case?

 

[Paul Colby]

Do you have a diagram of the set up?

Did you follow the link? If so, picture your rotating bar magnet in between the coils. This is very much the setup I used to measure and compare samarium cobalt magnet strengths.

Will the results be similar if I replace a bar magnet with a cylindrical dipole magnet in the present rotating magnet case?

I'd have to work it out. If one spins a magnet in a Helmholtz coil, the result will be a sinusoid provided the angle between the axis of the coil and the volume integral of the spinning magnetization changes in time.

 

[b.shahvir]

Did you follow the link? If so, picture your rotating bar magnet in between the coils. This is very much the setup I used to measure and compare samarium cobalt magnet strengths.

How do you measure the output voltage, eg. where to connect oscilloscope probes.

 

[Paul Colby]

How do you measure the output voltage, eg. where to connect oscilloscope probes.

On the terminals of the Helmholtz coil. Electrically it's exactly the same as the OP. Only the geometry of the coil is changed between the two cases.

 

[b.shahvir]

On the terminals of the Helmholtz coil. Electrically it's exactly the same as the OP. Only the geometry of the coil is changed between the two cases.

Please refer the output waveforms in post #147 uploaded by Alan. In your opinion, will the output waveform be any different if the experiment was done using Helmholtz coils?

 

[Charles Link]

Can I generate a perfect sinewave if I use a cylindrical dipole magnet as rotor? The experimental results with such an arrangement will be quite interesting.

I can't follow this part or the subsequent responses. The magnet that was being used all along was a cylindrical dipole magnet. Meanwhile a long solenoid makes for a uniform field. Helmholtz coils only gets close to a uniform field in the center of the coils.

 

[Paul Colby]

will the output waveform be any different if the experiment was done using Helmholtz coils

Yes. Be aware I'm using the reciprocity theorem (#49). The reciprocity theorem is an integral relation that relates two physical solutions of the field equation. In this case, the first solution is the B field generated by a current supplied to the coil without the magnet and the second, the EMF generated in the coil by the spinning magnet.

If one were to energize the coil shown in #147 with a current, the generated B field over the volume occupied by the magnet would be highly non-uniform. When one looks at the second case, a spinning magnet, the time dependence of the generated EMF is a complicated function of time as shown in #147 and elsewhere.

If one replaces the coil of #147 with a carefully designed Helmholz coil, the B-field generated by a current is then highly uniform over (the future) location of the spinning magnet. Being a constant allows one to pull the B field out from the integral. The remaining time dependence is sinusoidal.

 

[Charles Link]

If one replaces the coil of #147 with a carefully designed Helmholz coil, the B-field generated by a current is then highly uniform over (the future) location of the spinning magnet. Being a constant allows one to pull the B field out from the integral. The remaining time dependence is sinusoidal.

The explanation is somewhat unclear, but I think I disagree. If this is referring to putting the rotating magnet inside the Helmholtz coils, I believe the EMF generated will be very minimal. (Edit: My mistake here. See post 184.)

 

[b.shahvir]

The magnet that was being used all along was a cylindrical dipole magnet.

No, a bar magnet. Please refer Tom's experiment in earlier posts.

 

[Charles Link]

No, a bar magnet. Please refer Tom's experiment in earlier posts.

The other magnet @Tom.G used in post 92 was a horseshoe magnet. Cylindrical dipole magnet is the same thing as a bar magnet.

 

[b.shahvir]

Cylindrical dipole magnet is the same thing as a bar magnet.

No, the field geometry is quite different as discussed in earlier posts.

 

[Charles Link]

No, the field geometry is quite different as discussed in earlier posts.

Please list the post number you are referring to. I contend the two (bar magnet and cylindrical dipole magnet) are the same. If I have it incorrect, I certainly would want to update my understanding of the subject, but I do think you are in error here.

 

[Paul Colby]

The explanation is somewhat unclear, but I think I disagree.

Okay, let's do it.

The reciprocity theorem relates two physical time-harmonic solutions of Maxwell's equations. The form I'm using is,

$\iiint E_1\cdot J_2 dx^3 = \iiint E_2\cdot J_1 dx^3$

$E_1$ and $J_1$ are the time-harmonic fields generated in all of space by a current $J_1$ supplied to the coil without the spinning magnetization which is current $J_2 = \nabla\times M.$ We assume a form,

$M(x) = M_o \rho(x)$

where $M_o$ is a complex constant 3-vector. When multiplied times $e^{i\omega t}$ will generate a uniformly spinning vector. (I've assumed a vector direction that is spatially constant over the magnet volume. Also, this is not a necessary assumption since a suitable complex vector value would also describe a rotating field at each point.)

$\iiint E_1\cdot M_o\nabla\times \rho(x) dx^3 = -M_o\cdot\iiint \rho(x)\nabla\times E_1 dx^3$

From Maxwell's equations we have,

$\nabla\times E_1 = -i\omega B_1$

From our coil design, $B_1$ is assumed to be constant over the extent of the magnet. Since $\rho(x)$ is nonzero only over the volume of the magnet, the integral is over the magnet volume only.

$\iiint E_2\cdot J_1dx^3 = -i\omega M_o\cdot B_1\iiint\rho(x)dx^3$

Okay, $J_1$ is our current that generates $B_1$ and exists only in the coil wire and the generator. $E_2$ is zero everywhere except across the coil terminals. If we normalize $J_1$ we get,

$V = -i\omega M_o\cdot B_1\iiint\rho(x)dx^3$

Where $V$ is the voltage generated by spinning the magnet at a rate $\omega$.

 

[b.shahvir]

Please list the post number you are referring to. I contend the two (bar magnet and cylindrical dipole magnet) are the same. If I have it incorrect, I certainly would want to update my understanding of the subject, but I do think you are in error here.

In theory, a dipole magnet has negligible length between the 2 opposing poles, it's more like an one turn air coil. The magnetic field geometries of both are supposedly similar.

 

[Charles Link]

Okay, let's do it.

The reciprocity theorem relates two physical time-harmonic solutions of Maxwell's equations. The form I'm using is,

$\iiint E_1\cdot J_2 dx^3 = \iiint E_2\cdot J_1 dx^3$

$E_1$ and $J_1$ are the time-harmonic fields generated in all of space by a current $J_1$ supplied to the coil without the spinning magnetization which is current $J_2 = \nabla\times M.$ We assume a form,

$J_2 = M_o \rho(x)$

where $M_o$ is a complex constant 3-vector. When multiplied times $e^{i\omega t}$ will generate a uniformly spinning vector.

$\iiint E_1\cdot M_o\nabla\times \rho(x) dx^3 = -M_o\cdot\iiint \rho(x)\nabla\times E_1 dx^3$

From Maxwell's equations we have,

$\nabla\times E_1 = -i\omega B_1$

From our coil design, $B_1$ is assumed to be constant over the extent of the magnet. Since $\rho(x)$ is nonzero only over the volume of the magnet, the integral is over the magnet volume only.

$\iiint E_2\cdot J_1dx^3 = -i\omega M_o\cdot B_1\iiint\rho(x)dx^3$

Okay, $J_1$ is our current that generates $B_1$ and exists only in the coil wire and the generator. $E_2$ is zero everywhere except across the coil terminals. If we normalize $J_1$ we get,

$V = -i\omega M_o\cdot B_1\iiint\rho(x)dx^3$

Where $V$ is the voltage generated by spinning the magnet at a rate $\omega$.

I need to study this in detail=I'll need to reply later=busy schedule today, but what I think I may show is that the spinning magnet has a sinusoidal component only because basically $E=-\mu_2 \cdot B_1$ changes in the course of the motion of the magnet. If $B_1$ (the coil relative to the magnet) is uniform, there will be no sinusoidal signal.

 

[hutchphd]

As usual we are in the semantic wasteland after 164 entries.. Here are my definitions

1. A dipole magnet is a permanent or solenoidal magnet having linear form and equal opposite poles on the ends
2. A magnetic dipole is the infinitesimally sized limit of (1)

As previously indicated the double hump (remember the OP ??) requires a finite size bar magnet.

Can we at least agree as to the question?

 

[Paul Colby]

If B1 (the coil relative to the magnet) is uniform, there will be no sinusoidal signal.

Recall, this is an application of reciprocity. $B_1$ is the sinusoidal B-field generated by a sinusoidal current, $J_1$, applied to the coil.

 

[hutchphd]

Recall, this is an application of reciprocity. $B_1$ is the sinusoidal B-field generated by a sinusoidal current, $J_1$, applied to the coil.

I am confused. As I recall the reciprocity lemma applies specifically to linear systems. A rotating ferromagnet is not a linear system. I think I have lost the logical thread here.

 

[Paul Colby]

I am confused. As I recall the reciprocity lemma applies specifically to linear systems. A rotating ferromagnet is not a linear system. I think I have lost the logical thread here.

I routinely underestimate the difficulty of reciprocity arguments for time-harmonic electromagnetic fields. The system we are talking about is very much linear. Gauging by the replies, I get the feeling people don't quite follow the argument, which is very much above the B-level discussion we're having here. $J_1$ and $B_1$ are treated as mathematical devices, though one could well use actual fields and currents to characterize a test setup. $J_2=\nabla\times M$ and the generated field $E_2$ are the fields resulting from a rotating magnetization source, $M$. $J_1$ and $J_2$ are assumed to not have nonlinear effects on the constitutive properties.

 

[b.shahvir]

As usual we are in the semantic wasteland after 164 entries.. Here are my definitions

1. A dipole magnet is a permanent or solenoidal magnet having linear form and equal opposite poles on the ends
2. A magnetic dipole is the infinitesimally sized limit of (1)

Can you provide a link or reference article. I cannot seem to find one indicating a difference between the two. Thanks.

 

[alan123hk]

Please refer to https://www.pengky.cn/zz-generator-...nator/rotating-magnetic-field-alternator.html

It shows a 3D animation of a single-phase rotating field generator, which appears to produce a sine wave EMF output.

In my imagination, if a coil rotates in a uniform magnetic field to produce a sine wave output, as long as the geometric structure is set correctly, the rotating magnet that produces an approximately uniform magnetic field in the coil seems to be equivalent, so we can expect the output EMF to be at least approximately sine wave.

 

[b.shahvir]

Please refer to https://www.pengky.cn/zz-generator-...nator/rotating-magnetic-field-alternator.html

It shows a 3D animation of a single-phase rotating field generator, which appears to produce a sine wave EMF output.

Thanks but the magnet and coil arrangement of the OP is different from this arrangement... and so is the output waveforms.

 

[alan123hk]

Thanks but the magnet and coil arrangement of the OP is different from this arrangement... and so is the output waveforms.

Yes, I am considering the equivalent principle regarding another configuration.

Referring to the magnet and coil arrangement of the OP, if we assume a stationary magnet and a rotating coil, do you think the induced EMF of the coil will be different?
I'm still not sure what the answer is, will it be asymmetric, so is the answer different?

This is another demonstration of the OP's magnet and coil arrangement.

 

[hutchphd]

. J1 and J2 are assumed to not have nonlinear effects on the constitutive properties.

So for this case what are the appropriate linear constituative relations for $J_2$ . Assume it to be a run of the mill NdFeB cylindrical bar magnet magnetized axially.
I feel that you may not follow my argument.

 

[Charles Link]

If the cylindrical magnet has finite length, when it rotates, it does not generate the sinusoidal harmonic field that is being assumed above. Meanwhile, if it surrounded by a long cylindrical solenoidal type coil, (or alternatively placed inside a Helmholtz coil), the EMF will be minimal. The magnet needs to be external to the solenoidal coil. (Edit: My mistake here. See post 184.)

 

[Paul Colby]

So for this case what are the appropriate linear constituative relations for $J_2$ . Assume it to be a run of the mill NdFeB cylindrical bar magnet magnetized axially.
I feel that you may not follow my argument.

$\mu=\mu_o$ is an adequate approximation for a fully saturated magnet. We’re talking about $\mu$ at 10 to 100 hertz, not the static component. All fields and currents have an implicit $e^{i\omega t}$ dependence in the calculations I’ve made.

 

[Paul Colby]

Meanwhile, if it surrounded by a long cylindrical solenoidal type coil, (or alternatively placed inside a Helmholtz coil), the EMF will be minimal. The magnet needs to be external to the solenoidal coil.

I can speak to actual experience. The Helmholtz coil was wound on 5” pvc pipe. 30 or so turns on each end. The magnets were rectangular about 1”x1/4”x1/8” or so. They were magnetized through the narrow dimension. Spinning at 10 or so hertz midway between the windings developed about 1/2 a volt of signal. This was 30 years ago so I’m going by recollections.

I have provided a fairly complete analysis of a configuration not being measured here. Applying the very same analysis with $B_1$ being very non uniform over the magnet will produce exactly the wave forms shown elsewhere.

 

[b.shahvir]

Yes, I am considering the equivalent principle regarding another configuration.

Referring to the magnet and coil arrangement of the OP, if we assume a stationary magnet and a rotating coil, do you think the induced EMF of the coil will be different?
I'm still not sure what the answer is, will it be asymmetric, so is the answer different?

This is another demonstration of the OP's magnet and coil arrangement.

Thanks very much for uploading this video. As a matter of fact this video is the central premise of this thread. Also thankful to Tom for his demo.

 

[hutchphd]

μ=μo is an adequate approximation for a fully saturated magnet. We’re talking about μ at 10 to 100 hertz, not the static component. All fields and currents have an implicit eiωt dependence in the calculations I’ve made.

My apologies I did not understand your argument. I agree with your analysis, and the results you quote are exactly what I would expect. The magnet is saturated and stays saturated. The AC response is $\mu_0$. Thanks

 

[b.shahvir]

The discussions on this thread have been intellectually stimulating and fruitful and I am thankful to all the contributors for making this possible. However, there is still a yearning for more...
It would be interesting to observe the experimental setup for the following 2 cases;
1) Double peaked output waveform with 0 state in between the two peaks
2) A perfectly sinusoidal output waveform

What extent of modifications will be required in the rotating magnet arrangement in order to obtain the above output waveforms?