I Calculating Magnetic field strength of a magnet

[Einstein44]

@Einstein44 expresses familiarity with Biot-Savart so I used it to illustrate my point that all points on the surface, or in this case ring, contribute to the field at each point in space. Are you now expecting @Einstein44 to derive the equations from scratch using the method you propose because I thought we already discussed/decided to implement the equations from post #25? Does your method avoid elliptical integrals?

Indeed I think the method from post #25 is good, anything else would be clearly out of my capabilities I must admit. In this particular instance I am struggling to understand how to express $R$ and $\phi$ properly in the equation.

 

[Charles Link]

@Einstein44 expresses familiarity with Biot-Savart so I used it to illustrate my point that all points on the surface, or in this case ring, contribute to the field at each point in space. Are you now expecting @Einstein44 to derive the equations from scratch using the method you propose because I thought we already discussed/decided to implement the equations from post #25?

The integral for $B_z(\rho, z)$ is actually the more difficult one than $\phi(z)$, because it is a double integral. In any case, the OP @Einstein44 might try reading post 43 again on this topic as well. The idea is that whether we use polar coordinates or Cartesian coordinates, we need to integrate (sum) the contributions of the sources over the surface that all contribute to $B_z(\rho, z)$ at the location $(\rho, z)$.

We've got $\Phi$ on one axis and $R$ on the other, and we can divide the square region that goes from $0$ to $2 \pi$ and from $0$ to $a$ into a bunch of small squares. Overlaid on that is the weighting function $B_z(R, \Phi, \rho,z)$ from the integrand of the link of post 25. We need to sum all of these contributions up to get $B_z(\rho, z)$. @Einstein44 Try writing a program that will do this. :) (Suggestion is to use about one thousand increments on each axis, making for one million small squares).

 

[Einstein44]

The integral for $B_z(\rho, z)$ is actually the more difficult one than $\phi(z)$, because it is a double integral. In any case, the OP @Einstein44 might try reading post 43 again on this topic as well. The idea is that whether we use polar coordinates or Cartesian coordinates, we need to integrate (sum) the contributions of the sources over the surface that all contribute to $B_z(\rho, z)$ at the location $(\rho, z)$.

We've got $\Phi$ on one axis and $R$ on the other, and we can divide the square region that goes from $0$ to $2 \pi$ and from $0$ to $a$ into a bunch of small squares. Overlaid on that is the weighting function $B_z(R, \Phi, \rho,z)$ from the integrand of the link of post 25. We need to sum all of these contributions up to get $B_z(\rho, z)$. @Einstein44 Try writing a program that will do this. :) (Suggestion is to use about one thousand increments on each axis, making for one million small squares).

Alright I will see what I can do and get back to you when I am done. This will take a while though.

 

[hutchphd]

OK so here is my less-than-professorial attack on this problem. Please correct any flubs. An N32 magnet is typically near saturation and the surface flux density is roughly 1-2 Tesla. A Tesla is a Weber/m^2, so for a 1 cm cube magnet the flux from one pole face will be ~ $10^{-4}$ Weber. Rotated appropriately at the center of a loop at $\omega=10s^{-1}$ will produce an EMF of ~1 mV that is periodic (roughly sinusoidal as discussed in prior post previously alluded to)
Thought it was time for a sanity check for the OP (did we pass?)

 

[bob012345]

OK so here is my less-than-professorial attack on this problem. Please correct any flubs. An N32 magnet is typically near saturation and the surface flux density is roughly 1-2 Tesla. A Tesla is a Weber/m^2, so for a 1 cm cube magnet the flux from one pole face will be ~ $10^{-4}$ Weber. Rotated appropriately at the center of a loop at $\omega=10s^{-1}$ will produce an EMF of ~1 mV that is periodic (roughly sinusoidal as discussed in prior post previously alluded to)
Thought it was time for a sanity check for the OP (did we pass?)

It's a bar magnet falling through a loop.

 

[hutchphd]

Is it long and thin or short and fat? If long and thin it looks like two monopoles I recommend that as a good start. This is not easy.

 

[bob012345]

Is it long and thin or short and fat? If long and thin it looks like two monopoles I recommend that as a good start. This is not easy.

See the image in post #40. It looks like six N42 magnets and the loop is 2 or three times as big.

https://www.physicsforums.com/attachments/tempimagetzj6x4-png.287405/

No, not easy, it's a graduate level problem (at least to me).

 

[bob012345]

Well that is the question. I am unsure of the correctness of this method, but I though it would be possible to approximate where the field lines first reach the coil, and then use this distance from + to - as it moves through the coil to take the time it takes to do so. Since then the component t is present in Faradays Law, I assumed this would solve the problem of the changing flux?

This refers to the question of how to get the change in flux over time which I think also has a few issues in this problem.

So, we have the magnet of know dimensions falling from a certain height above the loop and at rest. I see a couple of issues.

First, the $B_{\rho}(z.\rho)$ component of the field at the loop will cause a force and thus a force will be on the magnet too trying to resist it's motion. If the field is weak at the edge of the loop it can be ignored.

Then, if there is no analytic expression for $B_z(z, \rho)$, there will not be an exact analytical expression of the flux change.

Since one would need $\large \frac{dB_z(z,\rho)}{dt}$ which would be $\large \frac{dB_z(z,\rho)}{dz} \frac{dz}{dt}$ which is $\large \frac{dB_z(z,\rho)}{dz} v(t)$
but there is no analytic expression so I think you have to evaluate the flux at set times
and compute the flux change as $\large \frac{ΔΦ}{Δt}$.

 

[Charles Link]

Post 104 by @hutchphd makes me wonder if we are trying to do too much with it. The flux at the surface can be estimated to be $\phi=\mu_o M A$, and that will occur in approximately a $\Delta z=.05$ m distance. If the height it is dropped from is $h=1.0$m, its speed at the coil will be about $v= 4.5$ m/sec. This makes for a $\Delta t \approx .01$ seconds. With $\mu_o M=1.3$ Tesla, and $A=.0001$ m^2, (post 26 says $d=15$ mm, so we are within a factor of 2 here), that makes for $\mathcal{E} \approx +.01$ volts, and then a corresponding pulse in the opposite direction as it passes out of the coil. If $N=10$, that would make the voltages $\pm .1$ volts.

 

[bob012345]

Post 104 by @hutchphd makes me wonder if we are trying to do too much with it. The flux at the surface can be estimated to be $\phi=\mu_o M A$, and that will occur in approximately a $\Delta z=.05$ m distance. If the height it is dropped from is $h=1.0$m, its speed will be about $v= 4.5$ m/sec. This makes for a $\Delta t \approx .01$ seconds. With $\mu_o M=1.3$Tesla, and $A=.0001$ m^2, that makes for $\mathcal{E} \approx +.01$ volts, and then a corresponding pulse in the opposite direction as it passes out of the coil. If $N=10$, that would make the voltages $\pm .1$ volts.

https://www.researchgate.net/publication/239035074_ELECTROMOTIVE_FORCE_Faraday's_law_of_induction_gets_free-falling_magnet_treatment

 

[Charles Link]

and a follow-on: The voltage is proportional to the speed, but if it is dropped from a different height, $\int V(t) \, dt$ for the positive or negative part of the pulse should remain the same. Perhaps the OP @Einstein44 should look at our latest estimates before he spends a lot of time on a lengthy computer program.

 

[bob012345]

and a follow-on: The voltage is proportional to the speed, but if it is dropped from a different height, $\int V(t) \, dt$ for the positive or negative part of the pulse should remain the same. Perhaps the OP @Einstein44 should look at our latest estimates before he spends a lot of time on a lengthy computer program.

I understand @Einstein44 has some data. It would be interesting to see that?

 

[bob012345]

Scratch that=I see I goofed. Sorry.
Edit: One other problem I see now is we are computing the field outside the magnet, using their formulas, but we also need to treat the case where the material of the magnet crosses over the plane of the coil=we need to include the extra term. It's not real difficult to include this part, but it complicates the problem.

It seems to me that the flux inside the magnet is constant so the flux change contribution from the inside of the magnet is zero as the magnet cuts through the plane.

 

[Charles Link]

It seems to me that the flux inside the magnet is constant so the flux change contribution from the inside of the magnet is zero as the magnet cuts through the plane.

The magnetization $M$ is assumed to be constant, but there is still an $H$ from the poles that points opposite the $M$, and this $H$ is not constant, but drops off in the center so that $B$ is maximized there, so that $\dot{B}=0$ at the center. I believe the peak in the EMF is likely to occur shortly before the magnet crosses the plane of the coil, but we really could use some computer simulation results to verify our conjectures.

 

[Einstein44]

Post 104 by @hutchphd makes me wonder if we are trying to do too much with it. The flux at the surface can be estimated to be $\phi=\mu_o M A$, and that will occur in approximately a $\Delta z=.05$ m distance. If the height it is dropped from is $h=1.0$m, its speed at the coil will be about $v= 4.5$ m/sec. This makes for a $\Delta t \approx .01$ seconds. With $\mu_o M=1.3$ Tesla, and $A=.0001$ m^2, (post 26 says $d=15$ mm, so we are within a factor of 2 here), that makes for $\mathcal{E} \approx +.01$ volts, and then a corresponding pulse in the opposite direction as it passes out of the coil. If $N=10$, that would make the voltages $\pm .1$ volts.

So would I have to create a special function for the induced emf depending on height? Does the time factor in the equation not solve this problem? I would rather like to solve this the proper mathematical way instead of making assumptions, since that would be too easy and not very accurate.

 

[bob012345]

So would I have to create a special function for the induced emf depending on height? Does the time factor in the equation not solve this problem? I would rather like to solve this the proper mathematical way instead of making assumptions, since that would be too easy and not very accurate.

I started to address this in post #108 but it probably needs more attention from others. As for making assumptions, that is a major part in solving difficult physics problems but if done thoughtfully it doesn't have to limit accuracy too much.

 

[hutchphd]

I would rather like to solve this the proper mathematical way instead of making assumptions

Good luck with that. Seriously you need to be able to do both. Every really good physicist I've known was facile with both ends. Life is short.

 

[bob012345]

I would rather like to solve this the proper mathematical way instead of making assumptions, since that would be too easy and not very accurate.

Why is this problem important to you? Are you just curious or is it a project for something? Also, I think you mentioned earlier you already have some experimental data. Can we see that?

 

[Charles Link]

The "link" in post 110 by @bob012345 has a graph of the experimental waveform for a similar experiment. My post 109 assigns an approximate height and width to the pulses. It is going to take some numerical computation with a computer program to do much more. I do think a computer program is likely to show reasonably good agreement with post 109 and post 110.

 

[hutchphd]

A slightly better result is to treat it as two monopoles. For a single monopole of strength $q_m$ at position z(t) the flux is just the solid angle subtended by the loop of radius R
$$flux=q_m(1-\frac z {\sqrt {z^2+R^2}})$$
The strength of each pole is as I obtained in #104 . Take some derivatives and put in z(t) or do it numerically. This will give a good physical result.

Edit: for S.I. probably need a $1/4\pi$ in that

 

[bob012345]

A slightly better result is to treat it as two monopoles. For a single monopole of strength $q_m$ at position z(t) the flux is just the solid angle subtended by the loop of radius R
$$flux=q_m(1-\frac z {\sqrt {z^2+R^2}})$$
The strength of each pole is as I obtained in #104 . Take some derivatives and put in z(t) or do it numerically. This will give a good physical result.

Edit: for S.I. probably need a $1/4\pi$ in that

Not yet considering units I got a factor of $2\pi$ in front of the charge from the integration differential element $2\pi \rho d\rho$. Also, if the origin is at zero, flux would get bigger as the charge moved through to the other side of the loop plane and away (from +$z$ to -$z$). That can be handled by making it go as;

$$flux=q_m(1-\frac {\sqrt{z^2}} {\sqrt {z^2+R^2}})$$

Edit: The issue goes away when the derivative is taken...

 

[hutchphd]

Yes you need to flip the sense of the solid angle when going through the origin. And there should be a 2pi/4pi out front so for z>0 it should be I think $$flux=\frac {q_m} 2 (1-\frac z {\sqrt {z^2+R^2}})$$ with sign change for z<0. Thanks.

 

[bob012345]

Yes you need to flip the sense of the solid angle when going through the origin. And there should be a 2pi/4pi out front so for z>0 it should be I think $$flux=\frac {q_m} 2 (1-\frac z {\sqrt {z^2+R^2}})$$ with sign change for z<0. Thanks.

In my coordinate system with $z$ positive up and the origin in the center of the coil, the positive charge is at $z$ and the negative charge is at $z'$, the flux for both charges with $z'$ = $z + L$; is;
$$flux =\frac {q_m} 2 (1-\frac z {\sqrt {z^2+R^2}} - 1 +\frac {z'} {\sqrt {z'^2+R^2}})$$

It's easier to use the chain rule;

$$ \frac{dflux}{dt} = \frac {q_m} 2 (-\frac {R^2} {\sqrt {z^2+R^2}^{3/2}} +\frac {R^2} {\sqrt {z'^2+R^2}^{3/2}}) \frac{dz}{dt}$$ since the only difference between $z'$ and $z$ is a constant.

So since the $emf = - N \Large \frac{dflux}{dt}$ and $\Large \frac{dz}{dt} = v(t) = -gt$

$$emf = \frac{q_mN R^2 g}{2} (\frac {-1} {\sqrt {(z_0 -\large \frac{g t^2}{2})^2+R^2}^{3/2}} +\frac {1} {\sqrt {(z_0 + L -\large \frac{g t^2}{2})^2+R^2}^{3/2}}) t $$

I don't have all the numbers to get the correct scale yet but using $z_0 = 25cm, R = 3cm, L = 3cm$ this is the shape of the curve I get. The $x$ axis is time in seconds and the $z$ axis is vertical. The model proposed by @hutchphd seems to capture the essential physics nicely. Notice the peak changes shape as it moves faster.

 

[Einstein44]

I think the remenance $B_r$ could be an important value here. With $B=\mu_o H +M$, (if I'm not mistaken), it gives the value of the approximate $M$ for the magnet. For N42 I see one data sheet that gives it as from 1.28 to 1.32 T. (See also post 2 of this thread by @berkeman ).
(Note: Sometimes the units are given with the formula $B=\mu_o H+\mu_o M$, so that care must be taken in using published formulas that involve $M$, to keep the units straight).

How exactly did you determine this equation? I couldn't find this anywhere so I assume you derived this one yourself ?

 

[Einstein44]

Why is this problem important to you? Are you just curious or is it a project for something? Also, I think you mentioned earlier you already have some experimental data. Can we see that?

The experimental data that I collected was pretty inconsistent, since I think the magnets are moving through the coil too fast, leaving not enough time for the voltmeter to read it properly or something. I am still trying to find ways to improve this methodology, so unit I will have done so, I will collect all the data again and make it available to you.
This is basically meant for a longer project that I am doing. I honestly don't even have to do this, this is rather the way I would prefer to do It because I am curious and I like solving these kinds of problems mathematically, in order to learn more about the maths behind the physics, if you know what I mean.

Edit: I am open to suggestions for improving the experiment

 

[cmb]

So I am going to start from the beginning now:
My aim is to find the induced emf as a cylindrical N42 magnet falls through a coil of N loops.
To calculate this I use Faradays Law.
Now for that, I need to find the magnetic flux in the first place using the equation:
$$\phi =\oint BdAcos\theta$$
This is why I am now trying to find B for the magnet, in order to work out this problem.
I am attaching a picture below that might perhaps help with visualisation.
So yes, indeed involves a moving magnet.

I think the answer to the EMF will therefore depend on the external load?

If the loop was a shorted single turn of super conductor the EMF would be zero and the magnet would slow down as it passes through the loop. (or another demonstration, the magnet passing down a copper tube).



I think trying to deduce the field strength is misleading as the question is one of total magnetic energy.

The '42' in N42 indicates the bulk magnetic energy (in MGOe). One might perform a calculation to confine the total magnetic energy of a given magnetic bulk to a given volume within a solenoid, this will then indicate the maximum conversion of the magnet's kinetic energy to electrical energy.

Clearly, if the impedance of the load on the coil is zero or near zero, as the video above shows, the conversion rate is extremely high for a very low/zero EMF. If the coil is open ended (or high impedance), the magnet will just fall through and generate a very high EMF.

The reason I think the EMF has to be difference according to the coil's external impedance is because the load/coil/magnet are a closed system whose reactance cannot be calculated independent of each other.

I have built various magnetic yokes using permanent magnets to generate uniform magnetic fields. Using the magnetic energy of the magnets used delivers a reasonably accurate indication of magnetic field (by calculating according to E=(1/2)B^2/uo). Note this is the energy per unit volume (so get your units correct on the RHS) and bear in mind fringing which will pretty much, at a first approximation, mean the total volume the field occupies is double what you might expect. For a very long copper cylinder, like the video, I'd tend to expect there would be much less fringing, but just take 'a half' as an experimental approximation for the total field the permanent magnet's field is contained within. Also bear in mind the total volume includes the magnet itself, which has an effective uo of ~1 (as it is effectively 'saturated').

As mentioned above, this is not a problem that is tractable by analytic equations and you should revert to a good 3D magnetics modelling package, as used for this sort of thing (designing solenoids and motors and such). In your case, I think just experimentation is the best means for you to deduce what you need to deduce.

 

[Einstein44]

As mentioned above, this is not a problem that is tractable by analytic equations and you should revert to a good 3D magnetics modelling package, as used for this sort of thing (designing solenoids and motors and such). In your case, I think just experimentation is the best means for you to deduce what you need to deduce.

Ok, thank you for your contribution. I will start off by doing the experiment and perhaps mention the theoretical way that this could be calculated (analytically).
Since you're saying experimental is best and you seem to have done similar stuff, do you have any suggestions as to how I could make the experiment perhaps more accurate? Currently I am dropping the magnets through the coil of N loops using a tube to keep them at the correct angle, however I think that because they move so fast the voltmeter has some trouble with accurate readings.

 

[cmb]

Ok, thank you for your contribution. I will start off by doing the experiment and perhaps mention the theoretical way that this could be calculated (analytically).
Since you're saying experimental is best and you seem to have done similar stuff, do you have any suggestions as to how I could make the experiment perhaps more accurate? Currently I am dropping the magnets through the coil of N loops using a tube to keep them at the correct angle, however I think that because they move so fast the voltmeter has some trouble with accurate readings.

Personally speaking, if this was just as a physics investigation for interest (... but it is your experiment and not sure on your desired end objectives ...) I would use an oscilloscope measuring across the open coil, then gradually add resistors.

The shorter the coil I'd expect the higher the EMF impulse for a shorter time. The longer the coil, the lower the impulse but for longer. So coil length as well as turns is a factor.

If you wanted a longer pulse, try longer magnets (put a few of the same cylindrical types end to end). Does it make a difference? If you have 3 or 4 magnets, do you get more work out of them by spacing them out (with non magnetic in between)?

I would also use a plastic tube to drop them through, makes sense. I'd use PEX heating tubing because of its particular properties, and wind the coil straight on to that. (or nested tubes, if the magnet is much smaller than the coil a smaller tube to carry the magnet within a larger tube carrying the coil, will help align the magnet and keep its orientation less of a variable).

Let's hear back from you on some tabulated results! Happy experimenting!

 

[Einstein44]

Personally speaking, if this was just as a physics investigation for interest (... but it is your experiment and not sure on your desired end objectives ...) I would use an oscilloscope measuring across the open coil, then gradually add resistors.

The shorter the coil I'd expect the higher the EMF impulse for a shorter time. The longer the coil, the lower the impulse but for longer. So coil length as well as turns is a factor.

If you wanted a longer pulse, try longer magnets (put a few of the same cylindrical types end to end). Does it make a difference? If you have 3 or 4 magnets, do you get more work out of them by spacing them out (with non magnetic in between)?

I would also use a plastic tube to drop them through, makes sense. I'd use PEX heating tubing because of its particular properties, and wind the coil straight on to that. (or nested tubes, if the magnet is much smaller than the coil a smaller tube to carry the magnet within a larger tube carrying the coil, will help align the magnet and keep its orientation less of a variable).

Let's hear back from you on some tabulated results! Happy experimenting!

Yes, the method with the tubes is something I already did to make the results more reliable, as of course the angle would vary the flux. I used paper instead, since it is hard to make plastic tubes for the right dimensions.
I will take your advice and get back with some results.
Note: I am investigating induced emf vs number of magnets

 

[Charles Link]

How exactly did you determine this equation? I couldn't find this anywhere so I assume you derived this one yourself ?

The $B=\mu_o H +M$ or $B=\mu_o H+\mu_o M$ is fairly standard, (comes from the pole model of magnetism), but perhaps it is advanced E&M (electricity and magnetism). See J.D. Jackson Classical Electrodynamics. The remanence $B_r$ as being the value of $\mu_o M$ is something that comes out of deriving the magnetic field strength using the formulas. It is somewhat difficult to find good and simple write-ups of this stuff.

 

[Charles Link]

The experimental data that I collected was pretty inconsistent, since I think the magnets are moving through the coil too fast, leaving not enough time for the voltmeter to read it properly or something. I am still trying to find ways to improve this methodology, so unit I will have done so, I will collect all the data again and make it available to you.

You really need an oscilloscope for this, rather than a voltmeter. With just a voltmeter, you could get some good readings if you built a precision rectifier circuit, followed by an integrator circuit. Otherwise, with just a voltmeter, I think your experiment is very lacking, and your results can't be compared with any accuracy to the above theoretical calculations.

 

[bob012345]

The experimental data that I collected was pretty inconsistent, since I think the magnets are moving through the coil too fast, leaving not enough time for the voltmeter to read it properly or something. I am still trying to find ways to improve this methodology, so unit I will have done so, I will collect all the data again and make it available to you.
This is basically meant for a longer project that I am doing. I honestly don't even have to do this, this is rather the way I would prefer to do It because I am curious and I like solving these kinds of problems mathematically, in order to learn more about the maths behind the physics, if you know what I mean.

Edit: I am open to suggestions for improving the experiment

I have a suggestion for that. You can simulate doing the experiment on Mars or even the Moon!

There is an easy way to effectively lowering $g$ and that is set up a simple pulley and string holding the magnet with a counterweight something like this;.

By adjusting the ratio of the masses you can control the acceleration. For example, if $m_1$ is the magnet, letting $m_2=\large \frac{m_1}{2}$ gives an acceleration of $\large \frac{g}{3}$ like Mars and letting $m_2 =\large \frac{5m_1}{7}$ gives an acceleration of $\large \frac{g}{6}$ like the Moon.

Also, I recommend using fine wire for the loop since using fat wires makes it harder to know what the effective size $R$ of the loop actually is.

 

[Einstein44]

You really need an oscilloscope for this, rather than a voltmeter. With just a voltmeter, you could get some good readings if you built a precision rectifier circuit, followed by an integrator circuit. Otherwise, with just a voltmeter, I think your experiment is very lacking, and your results can't be compared with any accuracy to the above theoretical calculations.

Yes, that actually makes much more sense. I am going to see if I have one in the lab, or else I will simply get one. That way it would be much more easy to see what the actual voltage induced is at any point in time, e.g. in the middle of the coil. (which was one of the issues I had with the voltmeter)

 

[Einstein44]

I have a suggestion for that. You can simulate doing the experiment on Mars or even the Moon!

There is an easy way to effectively lowering $g$ and that is set up a simple pulley and string holding the magnet with a counterweight something like this;.

View attachment 287537

By adjusting the ratio of the masses you can control the acceleration. For example, if $m_1$ is the magnet, letting $m_2=\large \frac{m_1}{2}$ gives an acceleration of $\large \frac{g}{3}$ like Mars and letting $m_2 =\large \frac{5m_1}{7}$ gives an acceleration of $\large \frac{g}{6}$ like the Moon.

Also, I recommend using fine wire for the loop since using fat wires makes it harder to know what the effective size $R$ of the loop actually is.

That makes sense. That wouldn't be too bad to be honest. I am going to see how much effort that will require, but I might actually do this. Also in combination with using an oscilloscope perhaps.

 

[Charles Link]

That makes sense. That wouldn't be too bad to be honest. I am going to see how much effort that will require, but I might actually do this. Also in combination with using an oscilloscope perhaps.

Yes, it could slow things down in a convenient way. One comment is the calculations are for a frictionless and massless pulley. I think you are likely to find in practice that it slows things down somewhat more than by the calculated value. Meanwhile, an oscilloscope should make for much better data.

 

[bob012345]

Yes, it could slow things down in a convenient way. One comment is the calculations are for a frictionless and massless pulley. I think you are likely to find in practice that it slows things down somewhat more than by the calculated value. Meanwhile, an oscilloscope should make for much better data.

By all means, if an oscilloscope is available that would be much better.

 

[Charles Link]

One additional comment: I think @bob012345 's post 123 is a very good one. That is almost exactly what I think you should expect to see with an oscilloscope trace. (The polarity might be reversed depending on which way you connect the leads to the oscilloscope).

 

[hutchphd]

One additional comment: I think @bob012345 's post 123 is a very good one. That is almost exactly what I think you should expect to see with an oscilloscope trace. (The polarity might be reversed depending on which way you connect the leads to the oscilloscope).

For completeness I should also point out that my outlined approximation should really include the "Dirac String" to connect the two "Dirac" monopoles. This would correspond to the body of the (thin) solenoid between the poles. I think it will look something like
$$flux _{solenoid string}=q_m\delta (x)\delta (y)[\theta(-z_2)-\theta(-z_1)]$$
where these are delta and step functions
It does not materially show up in the flux derivative I guess and it will not affect the approach or recession profile of the magnet flux.
I do rather like this cheap and dirty method...hope it works out well.

Edit: there should be a $2\pi$ in front of this.

 

[Charles Link]

For completeness I should also point out that my outlined approximation should really include the "Dirac String" to connect the two "Dirac" monopoles. This would correspond to the body of the (thin) solenoid between the poles. I think it will look something like
$$flux _{solenoid string}=q_m\delta (x)\delta (y)[\theta(-z_2)-\theta(-z_1)]$$
where these are delta and step functions
It does not materially show up in the flux derivative I guess and it will not affect the approach or recession profile of the magnet flux.
I do rather like this cheap and dirty method...hope it works out well.

It was particularly for this part of the magnet, when it crosses the plane of the coil, that I abandoned my attempt to simplify things in post 54. This approach that you @hutchphd introduced and @bob012345 applied in post 123 looks to me to be almost free of approximation. There is only two points $z=0$ and $z=-L$ where the solenoid portion (i.e the extra $\mu_o M$ term in $B=\mu_o H +\mu_o M$) will have a non-zero derivative. That seems to be erased in the mathematics, because, as has been pointed out, the flux from the poles, (i.e. the $\mu_o H$ flux), is discontinuous at these points as well, but the derivatives are steady, and it looks like they do give the correct exact answer.
The one approximation you do make is to condense each pole to a point on-axis, but otherwise I think the method may be free of approximation.

To summarize: Post 123 seems to be a very ideal way of solving this problem, and the one possible hurdle mentioned in post 138 above is, in fact, not a hurdle at all.

 

[Einstein44]

A slightly better result is to treat it as two monopoles. For a single monopole of strength $q_m$ at position z(t) the flux is just the solid angle subtended by the loop of radius R
$$flux=q_m(1-\frac z {\sqrt {z^2+R^2}})$$
The strength of each pole is as I obtained in #104 . Take some derivatives and put in z(t) or do it numerically. This will give a good physical result.

Edit: for S.I. probably need a $1/4\pi$ in that

What exactly did you do to get this equation? No need to show your working, I am just wondering what you did. Perhaps I can derive this myself for fun.

$q_{m}=B_{z}(z)$ right?

 

[Charles Link]

The formula is correct in posts 122 and 123. $H =q_m/(4 \pi \mu_o r^2)$ and flux $\phi= \mu_o H r^2 \Omega$ where $\Omega=2 \pi (1-\frac{z}{\sqrt{z^2+R^2}} )$ is the solid angle. Here a section of a sphere (with boundary being the coil, and center at $q_m$) is used to compute the flux, rather than the flat surface of the plane of the coil.
To compute the solid angle, $\Omega=2 \pi \int\limits_{0}^{\theta_o} \sin{\theta} \, d \theta$, where $\cos{\theta_o}=\frac{z}{\sqrt{z^2+R^2}}$.

 

[Einstein44]

The formula is correct in posts 122 and 123. $H =q_m/(4 \pi \mu_o r^2)$ and flux $\phi= \mu_o H r^2 \Omega$ where $\Omega=2 \pi (1-\frac{z}{\sqrt{z^2+R^2}} )$ is the solid angle. Here a section of a sphere (with boundary being the coil, and center at $q_m$) is used to compute the flux, rather than the flat surface of the plane of the coil.
To compute the solid angle, $\Omega=2 \pi \int\limits_{0}^{\theta_o} \sin{\theta} \, d \theta$, where $\cos{\theta_o}=\frac{z}{\sqrt{z^2+R^2}}$.

So does the derivation come from substituting $B_{z}(z)$ into $\phi = \int \vec{B}\cdot \vec{dA}$

 

[Charles Link]

So does the derivation come from substituting $B_{z}(z)$ into $\phi = \int \vec{B}\cdot \vec{dA}$

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

 

[Einstein44]

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

What exactly do you mean by over a spherical section? what exactly was manipulated?

 

[Charles Link]

What exactly do you mean by over a spherical section? what exactly was manipulated?

The coil is the surface boundary in both cases. The flux through the plane of the coil just uses $B_z$ because it is dotted with a surface whose unit normal is $\hat{z}$. The flux is the same for any surface with the coil as the boundary. For ease of computation, we choose to use a spherical section with the coil as the boundary and center of the sphere is the magnetic pole.
The reason the flux is the same has to do with $\nabla \cdot B=0$ and Gauss' law.

 

[Einstein44]

The coil is the surface boundary in both cases. The flux through the plane of the coil just uses $B_z$ because it is dotted with a surface whose unit normal is $\hat{z}$. The flux is the same for any surface with the coil as the boundary. For ease of computation, we choose to use a spherical section with the coil as the boundary and center of the sphere is the magnetic pole.
The reason the flux is the same has to do with $\nabla \cdot B=0$ and Gauss' law.

Perfect, that answers my question. And you did use the equation for B from post #25 am I correct

 

[Charles Link]

We are using the formula for $B$ in posts 122 and 123, or as described in more detail in post 141, with $B=\mu_o H$. The description in posts 141 and 122 is for a single point pole, and post 123 is for two poles, a plus one, and a minus one. Post 25 spreads the magnetic charge distribution evenly across the endface, and we are simplifying that slightly here. It is a reasonably good approximation to condense the magnetic charge on each endface to a point at its center, and upon doing that, the calculations can be done in closed form for an exact answer for the flux $\phi$. That's what we are doing by the method of post 123.

 

[bob012345]

Instead of using $\phi=\int B_z \cdot dA$ across the plane of the coil, the derivation uses $\phi=\int B \cdot dA$ over a spherical section. It gets the identical result, but it is much easier to compute it this way.

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

 

[Charles Link]

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

For extra detail, the $B=\frac{q_m}{4 \pi r^2}$, but yes, the integral will give the exact same thing. :)

 

[Einstein44]

I just assumed the field went as $B = \large \frac{q_m}{r^2}$ and integrated the flux element across the loop area directly and got the exact same thing.

The flux element was $B_z dA = B cos(\theta) 2\pi \rho d\rho$ where $cos(\theta) = \large \frac{z}{\sqrt{(z^2 + \rho^2)}}$.

Why are you not using an integral?
And where does the $2\pi$ come from since the area of the circle is $\pi r^{2}$