Waveform of Classic Electromagnetic Induction

[Tom.G]

It is perhaps worth mentioning that @Tom.G is operating his apparatus at a very low frequency, (about 1 or 2 Hz) and that the signal levels ($\mathcal{E}=-\frac{d \Phi}{dt}$ ) would be considerably larger if the frequency were increased, e.g. to 60 Hz. It could be interesting to see some experimental results for higher frequencies, even in the range of 10 Hz.

See the video referenced in post 173,
https://www.physicsforums.com/posts/6503116

Tom

 

[alan123hk]

I believe that in the very uneven 3D magnetic field produced by a rotating magnetic dipole, calculating the rate of change of the magnetic flux passing through the 3D coil is a very difficult task. If it refers to the use of manual calculations, whether it is analytical calculations or numerical calculations, it is even more difficult to imagine and almost impossible to complete.

Maybe I was too exaggerated before. In simple cases, this calculation may not be very difficult. For example, consider only the near field produced by an infinitesimal magnetic dipole and apply a one-turn coil (axial thickness is zero). But unfortunately I found that there is no double humps waveform in this case, so I want to raise a question below.

Is it possible use multiple magnetic dipoles arranged in a long strip to simulate a bar magnet, so to find a magnetic field equivalent to a bar magnet rotating in space ?

If this method works, place a one-turn coil in this magnetic field, and then according to the rate of change of the magnetic flux passing through this one-turn coil, the induced EMF be obtained. I hope this induced EMF waveform is consistent with the experiment, that is, the aforementioned local minimum/maximum will appear between the double peaks.

 

[b.shahvir]

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?

I suppose the above would be more interesting to analyse

 

[b.shahvir]

Or this arrangement for that matter.

 

[Charles Link]

Is it possible use multiple magnetic dipoles arranged in a long strip to simulate a bar magnet, so to find a magnetic field equivalent to a bar magnet rotating in space ?

This is one way to interpret the pole method of magnetism, where all of the opposite matching poles in the material cancel, except at the end faces. In any case, the calculation is one of just two poles. See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ which I mentioned previously in post 72.

Post 1 of the "link" summarizes both the magnetic pole method and the magnetic surface current method of computing the result. Both methods get the identical result for the magnetic field. The pole model is the much simpler one in this case.

Someone who is handy with computer programming could probably somewhat routinely put together a program that will generate the waveform for a specified geometry of the rotating magnet with a single coil. $B(t)$ would need to be computed for the two poles for about 1000 points in the cycle. It would not even be necessary to integrate $\int B \cdot dA$ in a very simplified approach, but doing the integral numerically should not be difficult to program. In any case, $\mathcal{E}=- \frac{d \Phi}{dt}$ is readily computed for each of the 1000 points .

It may be worth mentioning, (I think I may be stating the obvious), that the computation of the magnetic field $B$ for the rotating bar magnet is a completely static type calculation. It is not necessary to account for any motion of the moving magnetic poles. The calculation of the magnetic field $B$ for the rotating bar magnet is rather straightforward.

 

[Charles Link]

I should mention as a follow-on to the above, that the magnetic pole of magnetic surface charge density $\sigma_m=\pm M$ and area $A$ can be assumed to be a point magnetic charge at the center of the end face. This gives point magnetic charges $q_m= \pm \sigma_m A=\pm M A$, on the end faces, and the $H$ field is simply an inverse square analogous to $E$, with $\mu_o$ instead of $\epsilon_o$. Finally $B=\mu_o H$. The north pole is the positive one, and the south pole is the negative one. Only the component of $B$ perpendicular to the plane of the coil is needed for computing the flux.

I presently don't have any computer computing capability on my Chromebook or I would program this up. It should be fairly straightforward to numerically generate the waveform for a bar magnet that is 5 cm long and is 10 cm from a small coil. I do expect the double hump feature will emerge.

 

[b.shahvir]

I think the best option would be a practical demonstration to help us better understand and analyse the results.

 

[hutchphd]

I have done an analytical model and obtained results consistent with my previous posts (and the lovely experiments).. The model consists if a N and a S magnetic monopole on the ends of a rotating stick of length d. A small sensing coil is placed outside the radius of rotation. By varying d (while keeping the dipole moment fixed) one easily reproduces the cos for small d and then flatter then double hump for d approaching the coil. There are no surprises in the model.
Sorry for the tease but I will slog through the LaTeX in the next few days. Too much sh*t on my fan right now!
EDIT: I meant that literally I need to fix my fridge!

 

[alan123hk]

I think the best option would be a practical demonstration to help us better understand and analyse the results.

Experiments and practical demonstrations are of course very important, but scientists and engineers do need to develop some theoretical-based calculation methods to simulate and predict the results of some physical processes, which may be necessary for the design of complex and sophisticated systems.

 

[alan123hk]

I tried to use the magnetic charge model to simulate a rotating bar magnet, and then found an equation for the change of magnetic flux through the stationary coil with time in the simplest case, then I differentiated this equation, and I finally found the equation for the induced EMF of the coil.

The calculated induced EMF is similar to what I had imagined before, but it does have a certain degree of difference.

 

[Charles Link]

Very good @alan123hk :)

The peaks in the EMF are closer than expected to the peaks in the flux, (at which point we get a zero in the EMF), but in hindsight, that isn't too surprising.

What is your ratio of $L/d$, where $L$ is the length of the magnet, and $d$ the distance to the small coil?

 

[Charles Link]

@alan123hk From what I derived, I think an identical expression, (the small print is hard to read), you used $L/2=5$, and $d=20$.

 

[alan123hk]

@alan123hk From what I derived, I think an identical expression, (the small print is hard to read), you used L/2=5, and d=20.

Yes, there are two magnetic poles (N and S), the distance between them is 10, and the distance from the one-turn coil to the center of the two magnetic poles is 20.

Adjusting these distance-related parameters will change the induced EMF waveform.

 

[hutchphd]

I have done an analytical model and obtained results consistent with my previous posts (and the lovely experiments).. The model consists if a N and a S magnetic monopole on the ends of a rotating stick of length 2d. A small sensing coil is placed outside the radius of rotation. By varying d (while keeping the dipole moment fixed) one easily reproduces the cos for small d and then flatter then double hump for d approaching the coil. There are no surprises in the model.
Sorry for the tease but I will slog through the LaTeX in the next few days. Too much sh*t on my fan right now!
EDIT: I meant that literally I need to fix my fridge!

We define the dipole using two monopoles on a rotating stick of length 2d with positions $$\mathbf r_N=\binom {d\cos (\omega t )}{d\sin(\omega t)},~~~~~~~\mathbf r_S=-\mathbf r_N $$ Pole "magnetic charges" are assigned as to keep a constant dipole moment $q_M=p/2d$ and the magnetic field from a pole M is then $$\mathbf B(\mathbf r)_M = \frac {\mu_0 q_M } {4\pi}\frac { (\mathbf r-\mathbf {r_M})} { |\mathbf r-\mathbf {r_M}|^3}$$ Our small sensing coil is at position $$\mathbf r_{sense}=\binom c 0 $$where c>d and the coil surface normal is $\hat {\mathbf x}$. The flux through the small coil will then be proportional to the x component of B $$\mathbf B(\mathbf r)_M = \frac {\mu_0 q_M } {4\pi}\frac { (\mathbf r-\mathbf {r_M})} { |\mathbf r-\mathbf {r_M}|^3}$$

Sorry I didn't get this finished. The result comports with @alan123hk exactly and I attach the EXCEL graphs for several values of d/c (note c is not the speed of light...=distance to coil) for a fixed dipole moment. If anyone wants more I will provide same...no sense duplicating effort.

 

[Charles Link]

@hutchphd For your first graph, I believe you have a ratio of $.1$ instead of $.0001$.

Edit: My mistake: I didn't read your explanation that you maintain a constant dipole moment by adjusting the magnetic charge.

 

[hutchphd]

@hutchphd For your first graph, I believe you have a ratio of $.1$ instead of $.0001$.

Edit: My mistake: I didn't read your explanation that you maintain a constant dipole moment by adjusting the magnetic charge.

This is pretty rough I realize...the arduous part is making it pretty. Happy to supply whatever is helpful. Looks like the same result. Incidently you only need define the field for one pole because the second pole is just $\omega t=\pi$ offset with opposite "charge"...makes computation easier.

 

[Charles Link]

The calculation of the magnetic field from a uniformly magnetized cylindrical magnet that we are doing here is one that I think every physics and EE major should be able to do. (See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/).

I have seen a question involving this calculation appear in the Physics Forums homework section only on rare occasion=perhaps it, along with E&M in general, needs to see more emphasis in the curriculum. In any case, this is an opportunity for the students to try the calculation, and see if they can come up with the same results that we did.

 

[vanhees71]

We define the dipole using two monopoles on a rotating stick of length 2d with positions $$\mathbf r_N=\binom {d\cos (\omega t )}{d\sin(\omega t)},~~~~~~~\mathbf r_S=-\mathbf r_N $$ Pole "magnetic charges" are assigned as to keep a constant dipole moment $q_M=p/2d$ and the magnetic field from a pole M is then $$\mathbf B(\mathbf r)_M = \frac {\mu_0 q_M } {4\pi}\frac { (\mathbf r-\mathbf {r_M})} { |\mathbf r-\mathbf {r_M}|^3}$$ Our small sensing coil is at position $$\mathbf r_{sense}=\binom c 0 $$where c>d and the coil surface normal is $\hat {\mathbf x}$. The flux through the small coil will then be proportional to the x component of B $$\mathbf B(\mathbf r)_M = \frac {\mu_0 q_M } {4\pi}\frac { (\mathbf r-\mathbf {r_M})} { |\mathbf r-\mathbf {r_M}|^3}$$

Sorry I didn't get this finished. The result comports with @alan123hk exactly and I attach the EXCEL graphs for several values of d/c (note c is not the speed of light...=distance to coil) for a fixed dipole moment. If anyone wants more I will provide same...no sense duplicating effort.

Note that this is the quasistatic approximation though!

 

[hutchphd]

How fast is the bar magnet going to be rotating? I guess it should have been said explicitly.

 

[Charles Link]

How fast is the bar magnet going to be rotating? I guess it should have been said explicitly.

I mentioned that the calculation is a static type in the last paragraph of post 205. That makes it so that it really should be fairly routine, but, I have to wonder whether the majority of the physics and EE students would know how to do the calculation of computing the magnetic field from a uniformly magnetized cylindrical magnet. See also post 217.

 

[Charles Link]

Just an additional comment or two on the above for the calculation of the magnetic field for a cylindrical magnet: Recent conversations with a physics professor who teaches E&M said that these days they are teaching the magnetic surface current method with Biot-Savart to the undergraduate students, (e.g. Griffith's textbook has a problem with a permanent magnet using magnetic surface currents), and they normally don't teach the magnetic pole method until graduate school. For this problem, the magnetic pole method of calculation is the much simpler one, but both methods will get the exact same answer for the magnetic field $B$. If anyone is interested, both methods are outlined in the "link" in post 217.

 

[hutchphd]

As I recall teaching from the first edition of Griffiths in the mid 80's it was done using surface currents. Actually I don't know that I was taught pole strength explicitly in the 70's, but I recalled Dirac's invocation of a monopole as end of a long solenoid.

 

[Charles Link]

My classmates and I were taught the pole method during the late 70's, and we were only shown the surface currents very briefly as an alternative theory. I don't know of any textbook that covers both methods and shows them to be equivalent. The pole method relies on the formula $B=\mu_o H +M$, which is often presented without proof.

See posts 14-16 of https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/. It may be of interest that the pole method formula $B=\mu_o H+M$ can be proven for a system with arbitrary magnetization $M$ starting with Biot-Savart and magnetic current density $J_m=\nabla \times M/\mu_o$, and magnetic surface current per unit length $K_m=M \times \hat{n}/\mu_o$. IMO, this is a much better justification for this formula then simply saying it is analogous to the electrostatic $D=\epsilon_o E+P$, as some textbooks do.

The above proof is done for a system without any currents in conductors. The currents in conductors are then introduced with $H$ being re-defined (besides the magnetic pole contribution), to include a Biot-Savart type contribution to $H$ for currents in conductors. This way, $B=\mu_o H +M$, which is first shown to hold for a system without any currents in conductors, will then still hold when currents in conductors are introduced. This additional $H$ from the conductors, in the form of $\mu_o H$, simply gets added to both sides of the formula $B=\mu_o H+M$, because currents in conductors are always sources for $B$, computed from Biot-Savart.