I Calculating Magnetic field strength of a magnet

[Einstein44]

You are getting help on multiple levels at once here. I hope it's not too confusing!

The calculation of the magnetic field everywhere in 3D space is needed in order to know what it is in the loop as the magnet moves through the loop. It is a very difficult calculation. What is your EM background level?

The components $\rho$ , $z$ and $\theta$ are the variables of the cylindrical or polar coordinate system. I believe $R$ was used above just to represent integration over the radius of the magnet.

The problem will be ill defined until you establish a fixed coordinate system to work in. That means placing the magnet in that coordinate system say, with the exact center of the magnet at the origin.

My EM background is not that advanced to be honest, I mean I've covered Biot Savart, Lenzs/ Faradays Law (more familiar with EM induction) than with this kind of stuff to be honest, but I am always willing to learn.

So from post #49 I understand that z is the distance from the center point of the magnet (now the question is TO where), so I assume what is means is the distance to the coil from that point? Please tell me if I am correct, because looking at the Cylindrical coordinated, the z component is vertical and not horizontal ($\rho$).

 

[Charles Link]

You got the origin correct, (for z), and note that the z-axis is also the axis of the cylindrical magnet. The magnetization of the magnet $\vec{M}$ points along z as well. The z is also used to measure the distance to the plane of the coil, starting from the origin. The origin, just to make it clear, is right in the middle of the magnet. The two endfaces are located at $z=-L/2$ and $+L/2$.

In the above integrals, $R$ is the radial distance in the plane of the endface, (either one), and $\rho$ is the radial distance in the plane of the coil. In cylindrical coordinates, these are all specifying the same coordinate, which is the radial distance from the z-axis=known as $\rho$ in a standard cylindrical coordinate system. For clarity, perhaps the letters $\rho'$ and $\rho''$ would make more sense, but the one publication used $R$ for their integral,(in the plane of the endface), and I used simply $\rho$ for the other integral, (in the plane of the coil).

 

[bob012345]

My EM background is not that advances to be honest, I mean I've covered Biot Savart, Lenzs/ Faradays Law (more familiar with EM induction) than with this kind of stuff to be honest, but I am always willing to learn.

So from post #49 I understand that z is the distance from the center point of the magnet (now the question is TO where), so I assume what is means is the distance to the coil from that point? Please tell me if I am correct, because looking at the Cylindrical coordinated, the z component is vertical and not horizontal ($\rho$).

Ok, thanks. That is useful.

Delving further into this it appears to me that this problem is not easily solvable even by experts and the solution involves messy elliptical integrals. I could not find this problem actually solved in my graduate level texts. There are papers written with exact solutions but they are very likely beyond what you want to deal with. There are simpler formula's but they are only for limited solutions such as along the $z$ axis only.

So before struggling through complex calculations I think it would be instructive for you to think about what it should look like when the magnet goes through the loop. What would a sketch of the $emf$ look like? Then we can discuss ways to simplify the calculation.

 

[Charles Link]

and a follow-on: It might be asking a lot for you to follow some of the advanced physics and complicated integrals above, when there is a shortcut that should get you reasonably accurate results:
(Note: I see I did make one error above: $B_z(\rho)$ should be $B_z(\rho, z)$).
Here we are going to have an approximation for $B_z$ that is independent of $\rho$, and we will call it $B_z(z)$:
$B_z(z)=\mu_o M (\frac{ a^2}{4})(\frac{1}{(z-L/2)^2}-\frac{1}{(z+L/2)^2})$
and flux $\phi=B_z(z) \pi b^2$, where $b$ is the radius of the coil. (If you have $N$ loops in the coil, you need to multiply by $N$).
Here $\mu_o M=1.28$ T.
The above formulas involve approximations, Edit: and I see I goofed. They don't work as good as I thought they might. They only work well if $z-L/2>> a$ and it helps to have $z-L/2>> b$ as well.

 

[bob012345]

and a follow-on: It might be asking a lot for you to follow some of the advanced physics and complicated integrals above, when there is a shortcut that should get you reasonably accurate results:
(Note: I see I did make one error above: $B_z(\rho)$ should be $B_z(\rho, z)$).
Here we are going to have an approximation for $B_z$ that is independent of $\rho$, and we will call it $B_z(z)$:
$B_z(z)=\mu_o M (\frac{ a^2}{4})(\frac{1}{(z-L/2)^2}-\frac{1}{(z+L/2)^2})$
and flux $\phi=B_z(z) \pi b^2$, where $b$ is the radius of the coil. (If you have $N$ loops in the coil, you need to multiply by $N$).
Here $\mu_o M=1.28$ T.
The above formulas involve approximations, but they should give you a very good approximate result.

Does this assume most of the field outside the magnet is small compared to the magnet itself with respect to the size of the loop? Is this essentially assuming the loop is just barely bigger than the magnet?

 

[Charles Link]

Does this assume most of the field outside the magnet is small compared to the magnet itself with respect to the size of the loop? Is this essentially assuming the loop is just barely bigger than the magnet?

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.

 

[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.

Who's formula's and what does that part mean? Crosses over means something different than cuts across or the same?

 

[Charles Link]

Who's formula's and what does that part mean? Crosses over means something different than cuts across or the same?

My formula for post 54 would work fairly well when the magnet is far from the coil. We've been using the OP's "link" of post 25 to compute the field, but I don't think it works for $-L/2<z<+L/2$. In fact, I think that region needs to be treated with additional details. I think the additional calculations will be fairly routine in any case. (Edit note: See post 60 for the necessary corrections to the formula of post 25).

 

[bob012345]

Aren't those elliptical integrals? Have you integrated them?

 

[Charles Link]

The $B_z(\rho, z)$ field outside the magnet ($\rho>a$) should be ok for all z. If I'm not mistaken, (I will need to double-check this [Edit: I'm pretty sure I have it right]), for $\rho<a$ with $-L/2<z<+L/2$, all that may be necessary is to add a term $+\mu_o M$ to $B_z(\rho, z)$. For $z$ outside this interval, everything is ok as is.

This means the program needs to check each computation: If $-L/2<z<+L/2$, and $\rho<a$, then an extra $+\mu_o M$ needs to be added to the formula for $B_z(\rho,z)$ of post 25. (The reason for this is the observation point is inside the magnet, and $B=\mu_o H+\mu_o M$, instead of just $B=\mu_o H$). This is really a very simple correction, and it should be easy to implement.

Meanwhile, to answer post 59, the integrals are being done numerically.

 

[bob012345]

The $B_z(\rho, z)$ field outside the magnet ($\rho>a$) should be ok. If I'm not mistaken, (I will need to double-check this), for $\rho<a$ with $-L/2<z<+L/2$, all that may be necessary is to add a term $+\mu_o M$ to $B_z(\rho, z)$.
Meanwhile, to answer post 59, the integrals are being done numerically.

Is the OP doing them?

 

[bob012345]

I found this free Python package that computes magnetic fields for various shapes;

https://www.sciencedirect.com/science/article/pii/S2352711020300170?via=ihub

 

[Charles Link]

Is the OP doing them?

See posts 45 and 46 and 49. That's the direction that I'm hoping it goes. It appears even numerical integration may be new to the OP.

 

[bob012345]

See posts 45 and 46 and 49. That's the direction that I'm hoping it goes. It appears even numerical integration may be new to the OP.

I think it would be instructive to use your simplified model first so the OP can get the mechanics of the problem down by direct integration. Once the essential features of the physics are captured, the numerical integration can be attempted but it is up to the OP.

 

[Charles Link]

I think it would be instructive to use your simplified model first so the OP can get the mechanics of the problem down by direct integration. Once the essential features of the physics are captured, the numerical integration can be attempted but it is up to the OP.

Looking it over some more, I think my result of post 60 is correct, and that makes things easier than I originally anticipated. The physics including the pole model could be explained in more detail, but the numerical integrals should be very manageable, and I look forward to seeing how the OP @Einstein44 does with them. :) Posts 25, 45, and 60 contain most of what the OP needs. (Note: I've added some detail to posts 45 and 60 to make them easier to follow).

Note in post 25 that the $R$ in the numerators is actually from being part of $dA=R \, dR \, d \Phi$, but that doesn't change the computation that you perform.

Item of interest is we recently did something similar on Physics Forums with this thread:
See https://www.physicsforums.com/threa...agnetic-induction.1003690/page-5#post-6502079
@alan123hk You might find our latest exercise here of interest, where we are computing the waveform as the magnet is moved through the coil.

 

[Einstein44]

and a follow-on: It might be asking a lot for you to follow some of the advanced physics and complicated integrals above, when there is a shortcut that should get you reasonably accurate results:
(Note: I see I did make one error above: $B_z(\rho)$ should be $B_z(\rho, z)$).
Here we are going to have an approximation for $B_z$ that is independent of $\rho$, and we will call it $B_z(z)$:
$B_z(z)=\mu_o M (\frac{ a^2}{4})(\frac{1}{(z-L/2)^2}-\frac{1}{(z+L/2)^2})$
and flux $\phi=B_z(z) \pi b^2$, where $b$ is the radius of the coil. (If you have $N$ loops in the coil, you need to multiply by $N$).
Here $\mu_o M=1.28$ T.
The above formulas involve approximations, Edit: and I see I goofed. They don't work as good as I thought they might. They only work well if $z-L/2>> a$ and it helps to have $z-L/2>> b$ as well.

For the part where you say that the flux would have to be multiplied by N loops, don't you simply have to multiply this using Faradays Law for N loops at then end? Or are you saying you also have to multiply the flux before hand

 

[bob012345]

For the part where you say that the flux would have to be multiplied by N loops, don't you simply have to multiply this using Faradays Law for N loops at then end? Or are you saying you also have to multiply the flux before hand

Faraday’s experiments showed that the EMF induced by a change in magnetic flux depends on only a few factors. First, EMF is directly proportional to the change in flux Δ. Second, EMF is greatest when the change in time Δt is smallest—that is, EMF is inversely proportional to Δt. Finally, if a coil has N turns, an EMF will be produced that is N times greater than for a single coil, so that EMF is directly proportional to N. The equation for the EMF induced by a change in magnetic flux is;

$$EMF=−N\frac{ΔΦ}{Δt}$$ or in the limit;

$$EMF =−N\frac{dΦ}{dt}$$

https://courses.lumenlearning.com/b...ter/magnetic-flux-induction-and-faradays-law/

 

[Einstein44]

Faraday’s experiments showed that the EMF induced by a change in magnetic flux depends on only a few factors. First, EMF is directly proportional to the change in flux Δ. Second, EMF is greatest when the change in time Δt is smallest—that is, EMF is inversely proportional to Δt. Finally, if a coil has N turns, an EMF will be produced that is N times greater than for a single coil, so that EMF is directly proportional to N. The equation for the EMF induced by a change in magnetic flux is;

$$EMF=−N\frac{ΔΦ}{Δt}$$ or in the limit;

$$EMF =−N\frac{dΦ}{dt}$$

https://courses.lumenlearning.com/b...ter/magnetic-flux-induction-and-faradays-law/

Exactly what I meant, so since the N turns are already mentioned in Faradays Law for N loops, there is no need to multiply the flux by the number of loops am I correct? Just making sure because that's what @Charles Link mentioned.

 

[Einstein44]

I found this free Python package that computes magnetic fields for various shapes;

https://www.sciencedirect.com/science/article/pii/S2352711020300170?via=ihub

I tried downloading this, had a few issues so I tried something different (I don't know its validity)...
Just used this website that calculated the B field at a given position for a cylindrical magnet (N42) and this is what I got.
I selected a position in x from the center of the coil where the magnet is positioned to the coil, since this is where the magnetic field is going to interact with the coil.
(to at least have a value temporarily although I would have liked to work this out)
Is this the same thing as what you said I should do using the python code ? Or is there any important bits missing

Edit: I think the field lines are wrong...

 

[Einstein44]

and a follow-on: It might be asking a lot for you to follow some of the advanced physics and complicated integrals above, when there is a shortcut that should get you reasonably accurate results:
(Note: I see I did make one error above: $B_z(\rho)$ should be $B_z(\rho, z)$).
Here we are going to have an approximation for $B_z$ that is independent of $\rho$, and we will call it $B_z(z)$:
$B_z(z)=\mu_o M (\frac{ a^2}{4})(\frac{1}{(z-L/2)^2}-\frac{1}{(z+L/2)^2})$
and flux $\phi=B_z(z) \pi b^2$, where $b$ is the radius of the coil. (If you have $N$ loops in the coil, you need to multiply by $N$).
Here $\mu_o M=1.28$ T.
The above formulas involve approximations, Edit: and I see I goofed. They don't work as good as I thought they might. They only work well if $z-L/2>> a$ and it helps to have $z-L/2>> b$ as well.

Where does that first expression even come from?
And why are you representing the flux in the form of $\phi=B_z(z) \pi b^2$? I thought in this case it would rather be $\phi =\oint BdAcos\theta$ --> $\phi =\oint B_{z}(z)d\pi r^{2}$

You said they don't work as well as you thought. So I assume I should not look at these, but I was still wondering.

 

[Einstein44]

Looking it over some more, I think my result of post 60 is correct, and that makes things easier than I originally anticipated. The physics including the pole model could be explained in more detail, but the numerical integrals should be very manageable, and I look forward to seeing how the OP @Einstein44 does with them. :) Posts 25, 45, and 60 contain most of what the OP needs. (Note: I've added some detail to posts 45 and 60 to make them easier to follow).

Note in post 25 that the $R$ in the numerators is actually from being part of $dA=R \, dR \, d \Phi$, but that doesn't change the computation that you perform.

Item of interest is we recently did something similar on Physics Forums with this thread:
See https://www.physicsforums.com/threa...agnetic-induction.1003690/page-5#post-6502079
@alan123hk You might find our latest exercise here of interest, where we are computing the waveform as the magnet is moved through the coil.

I will work this out now.

 

[Einstein44]

The $B_z(\rho, z)$ field outside the magnet ($\rho>a$) should be ok for all z. If I'm not mistaken, (I will need to double-check this [Edit: I'm pretty sure I have it right]), for $\rho<a$ with $-L/2<z<+L/2$, all that may be necessary is to add a term $+\mu_o M$ to $B_z(\rho, z)$. For $z$ outside this interval, everything is ok as is.

This means the program needs to check each computation: If $-L/2<z<+L/2$, and $\rho<a$, then an extra $+\mu_o M$ needs to be added to the formula for $B_z(\rho,z)$ of post 25. (The reason for this is the observation point is inside the magnet, and $B=\mu_o H+\mu_o M$, instead of just $B=\mu_o H$). This is really a very simple correction, and it should be easy to implement.

Meanwhile, to answer post 59, the integrals are being done numerically.

what does "a" represent as you say p>a?
and looking at the cylindrical coordinate system, $\rho$ is going to be the radius of the cylinder, however you have already used a different letter b for this, so I am getting confused with the symbols. It would be helpful to have a clear description of what you have used for what in order to have some clarity.

 

[Charles Link]

$a$ is the radius of the cylindrical magnet, and $b$ is the radius of the coil. $\rho$ is an arbitrary radius in the cylindrical coordinate system.

If you look at the integrals of the "link" of post 25, they introduce $a$ as the limit of the $R \, dR \, d \Phi$ integration over the endface of the magnet. Meanwhile, I introduced the letter $b$ for the radius of the coil in post 45.

 

[bob012345]

Where does that first expression even come from?
And why are you representing the flux in the form of $\phi=B_z(z) \pi b^2$? I thought in this case it would rather be $\phi =\oint BdAcos\theta$ --> $\phi =\oint B_{z}(z)d\pi r^{2}$

You said they don't work as well as you thought. So I assume I should not look at these, but I was still wondering.

I believe because in this case, the field is a function of $z$ only while the loop is 90° to the $z$ axis so $cos\theta = 1$. Therefore the flux it any $z$ on the axis is simply $\phi = B_z(z) \oint dA = B_z(z)\pi b^2$

 

[bob012345]

I tried downloading this, had a few issues so I tried something different (I don't know its validity)...
Just used this website that calculated the B field at a given position for a cylindrical magnet (N42) and this is what I got.
I selected a position in x from the center of the coil where the magnet is positioned to the coil, since this is where the magnetic field is going to interact with the coil.
(to at least have a value temporarily although I would have liked to work this out)
Is this the same thing as what you said I should do using the python code ? Or is there any important bits missing

Edit: I think the field lines are wrong...

Why do you think the field lines are wrong?

The Python program was just one way of calculating the field. You still need to integrate it. I thought it could be used as a subroutine. What I am getting at is if you have a math program on your computer that can evaluate those integrals from post #25?

 

[Charles Link]

Where does that first expression even come from?
And why are you representing the flux in the form of $\phi=B_z(z) \pi b^2$? I thought in this case it would rather be $\phi =\oint BdAcos\theta$ --> $\phi =\oint B_{z}(z)d\pi r^{2}$

You said they don't work as well as you thought. So I assume I should not look at these, but I was still wondering.

In the integrals of post 25, if $a$ is small, the denominators can be simplified and come out of the integral after partial cancellation of the numerator. I think it is important here to treat the case where the magnet passes near and into the loop, and for that case the approximation doesn't work, so post 54 can be ignored.

 

[Charles Link]

I want to take a couple minutes to describe the numerical integration process mentioned in post 45. I'm hoping you @Einstein44 have enough calculus to know that an integral is the area under the curve of the graph of the function. In doing the integration numerically, you don't get the exact answer, but you can get very close to it, if you use enough rectangles in approximating the area under the curve. Let me see if I can find you a "link" on the topic... See https://en.wikipedia.org/wiki/Numerical_integration

 

[Einstein44]

I believe because in this case, the field is a function of $z$ only while the loop is 90° to the $z$ axis so $cos\theta = 1$. Therefore the flux it any $z$ on the axis is simply $\phi = B_z(z) \oint dA = B_z(z)\pi b^2$

why does the integral go away? The part with cos I got.

 

[Einstein44]

I want to take a couple minutes to describe the numerical integration process mentioned in post 45. I'm hoping you @Einstein44 have enough calculus to know that an integral is the area under the curve of the graph of the function. In doing the integration numerically, you don't get the exact answer, but you can get very close to it, if you use enough rectangles in approximating the area under the curve. Let me see if I can find you a "link" on the topic... See https://en.wikipedia.org/wiki/Numerical_integration

I know enough calculus to solve all these integrals, which is not really my concern. It is more with all these components / length that you have added to these equations which make it hard to follow along. (Now I know what you mean by the dimensions.) Although here is what I don't get for instance:
what doe sit mean when you have $B_{z}(\rho )\rho d\rho$
Like what is the calculation you're supposed to do when you calculate B for a certain component like this, substitute its value into the bracket? (This is just me being confused as to what to do with it, once I figure that out I should be able to do to the calculations

Now I have to calculate this long integral for B?

 

[bob012345]

why does the integral go away? The part with cos I got.

The integral doesn't go away, but when the plane of the loop is 90° to the axis, the integration is just the area of the loop which is $\pi b^2$.

 

[Einstein44]

The integral doesn't go away, but when the plane of the loop is 90° to the axis, the integration is just the area of the loop which is $\pi b^2$.

Yes, ok I get it.

 

[bob012345]

Yes, ok I get it.

And only because in that example, $B_z$ was constant over the loop.

 

[Einstein44]

And only because in that example, $B_z$ was constant over the loop.

By the way, so @Charles Link said that R was the radial distance in the plane of the enface. The distance to where exactly? How do I measure it?

 

[bob012345]

By the way, so @Charles Link said that R was the radial distance in the plane of the enface. The distance to where exactly? How do I measure it?

.

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.

 

[vanhees71]

I guess this is just derivable from the formula, I've given above for the magnetostatic scalar potential. You just have to take the gradient wrt. $\vec{r}$ to get these integrals for the field components. It's just written in cylinder coordinates, which is natural treating a homogeneously magnetized cylinder.

 

[Charles Link]

I guess this is just derivable from the formula, I've given above for the magnetostatic scalar potential. You just have to take the gradient wrt. $\vec{r}$ to get these integrals for the field components. It's just written in cylinder coordinates, which is natural treating a homogeneously magnetized cylinder.

That's how the "link" of post 25 does it also. I prefer to simply compute $H$ from $\sigma_m=M \cdot \hat{n}$, and the inverse square law. Then $B$ is obtained using $B=\mu_o H+ M$, which will work both outside and inside the magnet. See also post 60 for the extra detail.

 

[vanhees71]

That's of course also equivalent to what I wrote. The reason is that for homogeneously magnetized materials
$$\rho_m=-\vec{\nabla} \cdot \vec{M}$$
is in fact an effective magnetic "surface charge", because
$$\vec{M}=\vec{M} \chi_V(\vec{x}),$$
where
$$\chi_V(\vec{x})=\begin{cases} 1 & \text{for} \quad \vec{x} \in V \\
0 & \text{for} \quad \vec{x} \notin V. \end{cases}$$
It's clear that $\rho_{m}$ is $\delta$-function like along the boundary $\partial V$.
The direct calculation of $\vec{\nabla} \cdot \vec{M}$ is not that simple as in the case of the sphere, where simply $\chi_V(\vec{x})=\Theta(a-r)$, which I've used above as a shortcut.

The calculation can however be done using the integral formula for the surface divergence. Take a point on the surface and take an infinitesimal cylinder $Z$ around the point with two of it's surfaces parallel to $\partial V$. Then you define
$$-\sigma_m\mathrm{Div} \vec{M}=\lim_{Z \rightarrow \{\vec{x} \}} \frac{1}{S_Z} \int_Z \mathrm{d}^2 \vec{f} \cdot \vec{M}(\vec{x}) = -\vec{n} \cdot \vec{M}(\vec{x}).$$
Then you find indeed
$$\Phi_m(\vec{x})=\int_{\partial V} \mathrm{d}^2 \vec{f}' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
For a cylinder polarized parallel to its axis you only need to consider the bottom and the top surfaces, because the $\mathrm{d}^2 \vec{f}$ along the mantle is $\perp \vec{M}$.

 

[Charles Link]

View attachment 287447

View attachment 287448.

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.

This will take a little work to write a program to do these integrations. The 2-D integral can be done with a two dimensional array and nested "Do" loops. Suggest a third Do loop to get the result for different z's. It shouldn't be too difficult to program, but if it's the first time for @Einstein44 , it might be a challenge. Perhaps wolfram has a routine to do this sort of thing?

 

[bob012345]

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.

So now that the issue of calculating the field is being addressed it is time to think about the setup of the problem as a whole. You might set it up something like this diagram where the center of the magnet is some distance, say $nL$ above the plane of the loop which is fixed. We have discussed getting the flux at anyone time above but how do you think you can calculate the change in flux as time becomes involved?

 

[Einstein44]

So now that the issue of calculating the field is being addressed it is time to think about the setup of the problem as a whole. You might set it up something like this diagram where the center of the magnet is some distance, say $nL$ above the plane of the loop which is fixed. We have discussed getting the flux at anyone time above but how do you think you can calculate the change in flux as time becomes involved?View attachment 287455

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?

 

[Einstein44]

View attachment 287447

View attachment 287448.

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 $\phi$