I Calculating Magnetic field strength of a magnet

[Einstein44]

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These equations are for a magnet of length $L$ and radius $a$. The coordinates in polar or cylindrical coordinates are $z$, $\rho$ and $\phi$. But at each point in space $z, \rho$ and $\phi$, there is an integration over the face of the magnet which use variables $R$ and $\phi$. $R$ is not measured, it is integrated over. The issue of complexity comes in because these integrals over $R$ and $\phi$ are not easily solved in closed form otherwise that integration would be done already and there would be no integral signs in the formula's for $B_z$ and $B_{\rho}$.

Thus there are two separate integrations, one to get the fields at a point $(z, \rho, \phi)$ and another to integrate the field at every point in the plane of the loop.

That was explained very well, but now my question is how can I possibly integrate $R$ and $\phi$ if there is no value for them? Like my question now is how can I proceed to solving this integral with these two unknowns? I also understand it is not solvable by hand and requires a special programme? What kind of programme?

 

[Charles Link]

That was explained very well, but now my question is how can I possibly integrate $R$ and $\phi$ if there is no value for them? Like my question now is how can I proceed to solving this integral with these two unknowns? I also understand it is not solvable by hand and requires a special programme? What kind of programme?

Have you had a course in integral calculus yet? If not, it is something you can always learn, but that is really a necessity for taking on a project such as this.

 

[Einstein44]

Have you had a course in integral calculus yet? If not, it is something you can always learn, but that is really a necessity for taking on a project such as this.

Yes, I know quite a bit about integral calculus. I have taken a course althoughI have taught myself a lot of it already since I did some math projects on integral calculus and used it quite a bit for several physics problems.
I mean this is not a necessity, but I thought it would be a nice addition to this project, but if this really isn't doable then I guess ill have to go without it... I don't know if there is another way of approximating this, maybe with the website I used (although I question its validity). But I definitely want to have a go a this, especially after having spent so much time on it.

 

[bob012345]

That was explained very well, but now my question is how can I possibly integrate $R$ and $\phi$ if there is no value for them? Like my question now is how can I proceed to solving this integral with these two unknowns? I also understand it is not solvable by hand and requires a special programme? What kind of programme?

$R$ and $\phi$ are variables. $R$ goes from zero → $a$ and $\phi$ from 0 → $2\pi$.

The idea is that each point on both end faces of the magnet contributes to the total field at a point in space at $(z, \rho, \phi)$.

It's just like the Biot- Savart Law being used to calculate the magnetic field at one point from an line of current. All points on the current line contribute to each point in space for the field.

 

[Einstein44]

$R$ and $\phi$ are variables. $R$ goes from zero → $a$ and $\phi$ from 0 → $2\pi$.

The idea is that each point on both end faces of the magnet contributes to the total field at a point in space at $(z, \rho, \phi)$.

So if I understand this correctly $R$ is going from the centre of the enface to $a$, which is the radius of the magnet... Wait no that doesn't make sense. By this logic it would simply be the radius. Is perhaps point 0 in the very middle of the magnet?
For $\phi$ I understand you have to integrate because we need it for every point along the circumference of the circle $2\pi$?

 

[bob012345]

So if I understand this correctly $R$ is going from the centre of the enface to $a$, which is the radius of the magnet... Wait no that doesn't make sense. By this logic it would simply be the radius. Is perhaps point 0 in the very middle of the magnet?
For $\phi$ I understand you have to integrate because we need it for every point along the circumference of the circle $2\pi$?

See my post above. It is just the radius of the magnet.

I propose you try the problem of using the Biot-Savart Law to set up the magnetic field at some point off-axis for a simple current loop. It's easy on the axis of symmetry but difficult to solve off-axis. I'm not asking you to solve the equations because they will look complicated like in post #25. Just set it up and you will quickly see what is happening with $R$ above.

 

[Charles Link]

See my post above. It is just the radius of the magnet.

I propose you try the problem of using the Biot-Savart Law to set up the magnetic field at some point off-axis for a simple current loop. It's easy on the axis of symmetry but difficult to solve off-axis. I'm not asking you to solve the equations because they will look complicated like in post #25. Just set it up and you will quickly see what is happening with $R$ above.View attachment 287459

It's actually simpler than Biot-Savart. The pole model of magnetism is what we are doing here, and the solution for $H$ works just like electrical charges for $E$. Post 25 uses some difficult mathematics to get their result (a magnetic potential), but the easier approach is to treat $H$ like the electric field $E$. See also post 35.

There is also a Biot-Savart way to work this problem, but that is much more difficult. See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ The second part of the first post summarizes this approach. The first part of the first post summarizes how we are solving this with the pole model.

 

[Charles Link]

I mean this is not a necessity, but I thought it would be a nice addition to this project, but if this really isn't doable then I guess ill have to go without it... I don't know if there is another way of approximating this, maybe with the website I used (although I question its validity). But I definitely want to have a go a this, especially after having spent so much time on it.

I think it might take a little work to learn the process of the double integral, and doing it numerically. Once you understand the concepts, it really is not too difficult to program. I presently don't have that capability on my Chromebook, but I think we've got a couple people on here who could program the $B(\rho, z)$ and $\phi(z)$ results very routinely in about 30 minutes or less. @hutchphd Might you lend a hand? (see post 65).

 

[bob012345]

It's actually simpler than Biot-Savart. The pole model of magnetism is what we are doing here, and the solution for $H$ works just like electrical charges for $E$. Post 25 uses some difficult mathematics to get their result (a magnetic potential), but the easier approach is to treat $H$ like the electric field $E$. See also post 35.

There is also a Biot-Savart way to work this problem, but that is much more difficult. See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ The second part of the first post summarizes this approach. The first part of the first post summarizes how we are solving this with the pole model.

@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?

 

[Einstein44]

I think it might take a little work to learn the process of the double integral, and doing it numerically. Once you understand the concepts, it really is not too difficult to program. I presently don't have that capability on my Chromebook, but I think we've got a couple people on here who could program the $B(\rho, z)$ and $\phi(z)$ results very routinely in about 30 minutes or less. @hutchphd Might you lend a hand? (see post 65).

Solving the double integral itself should not be a problem, my issue is still that I don't know how exactly to express all of this into the formula in the correct manner.
This is something I mentioned in post #79... perhaps someone can help a bit with that...
I have the same problem for the long Integral of B, where I don't know in what way I am supposed to express $R$ and $\phi$

 

[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