Calculating Magnetic Field Strength and Force of Eddy Currents in a Solenoid

[Delta2]

Ok, however I am still a bit confused about what this force really represents. The BIL force is the magnetic force of the magnet itself? And on what exactly does it act upon?
And what exactly does the "d" in that equation represent? Because the normal equation is: F= BIL sinθ

Ok I guess you know about the BIL force. Here because we don't have a straight conductor but multiple turns of coil, we model each turn as a conductor at the shape of the circle. We split this circle into infinitesimal segments of length $dl$, each segment tangential to the circle at that point. The infinitesimal force from the magnetic field $B$ to this infinitesimal length $dl$ will $BIdl\sin\theta$. The total force will be the sum or integral of these infinitesimal forces (and no it will not be equal to $BI2\pi L$ where L the radius of one turn, I can expand on this if you wish).

Anyway this total force is the force from the magnet to one turn. But due to Newton's 3rd Law, the turn will apply an equal and opposite force to the magnet which will be essentially the breaking force to the magnet from this one turn. If the coil has many closely spaced turns, the magnet will be affected by more than one turns , so we have to multiply the braking force of post #23 by some number, but for simplicity we put this number equal to 1.

 

[Delta2]

A related thread on how a bar magnet will produce a force on one turn is this
https://www.physicsforums.com/threa...-ring-with-a-bar-magnet.1005442/#post-6519455

 

[Einstein44]

Ok I guess you know about the BIL force. Here because we don't have a straight conductor but multiple turns of coil, we model each turn as a conductor at the shape of the circle. We split this circle into infinitesimal segments of length $dl$, each segment tangential to the circle at that point. The infinitesimal force from the magnetic field $B$ to this infinitesimal length $dl$ will $BIdl\sin\theta$. The total force will be the sum or integral of these infinitesimal forces (and no it will not be equal to $BI2\pi L$ where L the radius of one turn, I can expand on this if you wish).

Anyway this total force is the force from the magnet to one turn. But due to Newton's 3rd Law, the turn will apply an equal and opposite force to the magnet which will be essentially the breaking force to the magnet from this one turn. If the coil has many closely spaced turns, the magnet will be affected by more than one turns , so we have to multiply the braking force of post #23 by some number, but for simplicity we put this number equal to 1.

Thank you, but I fail to see the connection between post #5 by @kuruman and what you have showed me in post #23...
I understand you used some assumptions as mentioned, but the method seems to be different?

 

[Einstein44]

Ok I guess you know about the BIL force. Here because we don't have a straight conductor but multiple turns of coil, we model each turn as a conductor at the shape of the circle. We split this circle into infinitesimal segments of length $dl$, each segment tangential to the circle at that point. The infinitesimal force from the magnetic field $B$ to this infinitesimal length $dl$ will $BIdl\sin\theta$. The total force will be the sum or integral of these infinitesimal forces (and no it will not be equal to $BI2\pi L$ where L the radius of one turn, I can expand on this if you wish).

Anyway this total force is the force from the magnet to one turn. But due to Newton's 3rd Law, the turn will apply an equal and opposite force to the magnet which will be essentially the breaking force to the magnet from this one turn. If the coil has many closely spaced turns, the magnet will be affected by more than one turns , so we have to multiply the braking force of post #23 by some number, but for simplicity we put this number equal to 1

Lets call it our basic assumption that the braking force on the magnet will be of the form $$F=-kv$$ where $v$ the velocity of the magnet and k a constant which encloses all the parameters of the setup (like ohmic resistance, magnet field strength, radius of coil).

This assumption is inspired by the aerodynamic drag case, and in this electromagnetic case is at least partially justified by the comments 3. and 4. of post #5 by @kuruman (the velocity of the magnet is $v=\frac{dz}{dt}$ in case you wonder where it does appear there). Also by comments 1 to 4 of post #5 we can conclude that the constant $k$ will be inversely proportional to the ohmic resistance R of the coil and proportional to the radius of the coil L and the strength of the field of the magnet B. A more careful analysis using dimensional analysis (I don't know if you have heard this term before, in short it means that we have to take care the units of k such that when multiplied by the units of velocity which are m/s (meter per second) will give us unit of force, that is Newton), reveals that k must be proportional to the square of the magnet field strength and to the square of the radius L. So we can write $$k=\frac{B^2L^2}{R}$$ and the braking force will be $$F=-\frac{B^2L^2}{R}v$$

So, now that you are given a good approximation for the braking force that the magnet experiences can you proceed with the rest of your work?

I tried to understand the thing with the units, could you please show mathematically how you concluded that B and L must be squared?

 

[Einstein44]

Ok I guess you know about the BIL force. Here because we don't have a straight conductor but multiple turns of coil, we model each turn as a conductor at the shape of the circle. We split this circle into infinitesimal segments of length $dl$, each segment tangential to the circle at that point. The infinitesimal force from the magnetic field $B$ to this infinitesimal length $dl$ will $BIdl\sin\theta$. The total force will be the sum or integral of these infinitesimal forces (and no it will not be equal to $BI2\pi L$ where L the radius of one turn, I can expand on this if you wish).

Anyway this total force is the force from the magnet to one turn. But due to Newton's 3rd Law, the turn will apply an equal and opposite force to the magnet which will be essentially the breaking force to the magnet from this one turn. If the coil has many closely spaced turns, the magnet will be affected by more than one turns , so we have to multiply the braking force of post #23 by some number, but for simplicity we put this number equal to 1.

what is confusing me, is that you gave me one expression for the force on post #23, but now we have this expression for the BIL force, which I don't know how it is related to the expression or any of this? Especially because current is not even mentioned in the other one?

If you have spare time, I would appreciate it if you could put the important elements into one post, because I lost the overview of all this and I am really trying hard to put all the pieces together from all the posts.

 

[Delta2]

Ok, understandable, you might be feeling lost and confused because the knowledge wasn't presented to you 100% properly, I ll try my next post to be a summary of the most important keys of knowledge presented in the most understandable way.

But you might have to wait till tomorrow, not sure when I can find the time to post (its going to be somewhat big post).

 

[Einstein44]

Ok, understandable, you might be feeling lost and confused because the knowledge wasn't presented to you 100% properly, I ll try my next post to be a summary of the most important keys of knowledge presented in the most understandable way.

But you might have to wait till tomorrow, not sure when I can find the time to post (its going to be somewhat big post).

yes that's absolutely fine, I really appreciate your help.

 

[Delta2]

We mainly used two approaches for this problem:

1. To use the laws of electromagnetism to find the braking force. This approach is complex and cumbersome and it requires many proper approximations to be made in order to derive simple and useful results.
2. To make a basic assumption (that it might seem that it drops down from heavens) about the braking force that it is similar to a linear drag . Then to use dimensional analysis to approximate the constant that appears in the expression of the drag. This approach mostly bypasses the laws of electromagnetism (though we will still refer to them to some extent) and it is much faster and easier to understand

Lets start from the second approach which is faster and easier:

So we make the assumption that the braking force is $\vec{F}=-k\vec{v}$. We know that the units of Force $\vec{F}$, are Newtons (N) and the units of velocity $\vec{v}$ are $m/s$. So the units of the constant $k$ must be $\frac{N}{\frac{m}{s}}$.
We also know that k must some how encapsulate the magnetic field $B$ of the magnet (since we expect the stronger the magnet is the stronger the braking force to it) the radius $L$ of the coil (the biggest the radius L, the more EMF generated by each turn hence more current generated, hence the interaction between the current and the magnetic field of the magnet will be stronger) and the ohmic resistance R of the coil (more ohmic resistance means less current).

From all the above we can conclude that the units of k will contain Tesla $T$ (the unit of magnetic field), meters $m$ (the unit of the radius L) and ohms $\Omega$ (the units of the ohmic resistance R). If we know some basic relations about the units in S.I we will find that one suitable selection for the units of k that satisfies all the above conditions is $$\frac{T^2\cdot m^2}{\Omega}$$ because it is $$\frac{T^2\cdot m^2}{\Omega}\cdot \frac{m}{s}=Newton$$
so we can infer that the constant $k$ will contain the term $\frac{B^2L^2}{R}$ and it might contain some other constants as well , for example it might have $\pi$ because the EMF per turn is proportional to the flux per turn which is proportional to the area per turn which area is $\pi L^2$. But $\pi$ is dimensionless, that is it is a pure number so its participation in the constant $k$ cannot be derived by dimensional analysis.

Now let's see what we can get from the first approach, which seems to be more valid since it will use directly the laws of electromagnetism as they apply to this specific problem.

(Sadly I am abit tired now I will continue this post tomorrow...)

 

[Delta2]

Ok let me continue with the first approach now.
I kind of get my self lost and confused too trying to apply the EM laws for this problem. I ll try to explain my approach the best way I can.

First we 'll do the simplifying assumption that the coil consists only of 1 turn. If we don't this , then the expression for the total flux through the coil will be a sum of N different terms (N the number of turns of coil) and things might get quite complex.

Second simplifying assumption is that we shall ignore the self inductance of the coil. If we don't neglect it then we 'll have to solve a system of two coupled differential equations with two unknown functions the current $I(t)$ and the velocity of the bar magnet $v(t)$ and things will also get more complex too.

if z(t) is the distance of the dipole magnet from the center of the coil then a simplifying expression for the magnetic field of the z-component (which is perpendicular to the cross section area of the coil) will be $$B_z(z(t))=\frac{1}{z(t)^2+C}B_0$$ where $\frac{B_0}{C}$ is the magnetic field at the surface of the bar magnet.

Then the magnetic flux through the coil will be $$\Phi=\pi L^2B_z(z(t))$$ and hence the EMF produced will be $$\mathcal {E}=\frac{d\Phi}{dt}=\frac{d\Phi}{dz}\frac{dz}{dt}=\pi L^2 \frac{dB_z(z)}{dz}v(t)$$ where $$v(t)=\frac{dz}{dt}$$ the speed of the magnet. Here we ll do another "fat" approximation and we ll set $$\frac{dB_z(z)}{dz}=C'$$ that is to set this expression to a constant C', though in reality it will not be a constant it will depend on time t. A good choice for the constant C' will be the average value of $\frac{dB_z(z)}{dz}$ over time t.

The current through the coil , since we neglect the self inductance of the coil, will be $$I=\frac{\mathcal{E}}{R}=\pi L^2C' \frac{v(t)}{R}$$ where $R$ the ohmic resistance of the coil.

Then the force from the magnet to the coil will be $$F_B=\int Id\vec{l}\times \vec{B}$$ where $\vec{B}$ is the total magnetic field of the dipole magnet at the points of the turn of the coil. This $\vec{B}=B_x\hat x+B_y\hat y+B_z\hat z$ will have x,y and of course the z component which is already mentioned. Now if we do this integral we will find that the total force due to the z component $B_z(z)$ is zero (at all times), so it remains only to calculate the integral for the x and y components of the $B$. We further do the simplifying assumption that those x and y components are such that their sum $\vec{B_N}=B_x\hat x+B_y\hat y$ is normal to the $d\vec{l}$. Then it will be $$F_B=\int B_NIdl=2\pi LB_NI =2\pi^2\frac{ L^3C'B_N}{R}v(t)$$. The braking force on the magnet, is due to Newton's 3rd law $-F_B$.

So at the end of the day we have find a different expression for the force than the other approach. The approach with the linear drag and dimensional analysis gives the force as $$-\frac{B^2L^2}{R}v(t)$$, while the approach from applying the EM laws(including a hell load of simplifying assumptions) gives the force as $$-2\pi^2\frac{L^3C'B_N}{R}v(t)$$.

I am not sure at all in what I did when applying the laws of EM but that's the best I could do, if someone else can take over, please be my guest.

 

[Einstein44]

Ok I guess you know about the BIL force. Here because we don't have a straight conductor but multiple turns of coil, we model each turn as a conductor at the shape of the circle. We split this circle into infinitesimal segments of length $dl$, each segment tangential to the circle at that point. The infinitesimal force from the magnetic field $B$ to this infinitesimal length $dl$ will $BIdl\sin\theta$. The total force will be the sum or integral of these infinitesimal forces (and no it will not be equal to $BI2\pi L$ where L the radius of one turn, I can expand on this if you wish).

Anyway this total force is the force from the magnet to one turn. But due to Newton's 3rd Law, the turn will apply an equal and opposite force to the magnet which will be essentially the breaking force to the magnet from this one turn. If the coil has many closely spaced turns, the magnet will be affected by more than one turns , so we have to multiply the braking force of post #23 by some number, but for simplicity we put this number equal to 1

Ok let me continue with the first approach now.
I kind of get my self lost and confused too trying to apply the EM laws for this problem. I ll try to explain my approach the best way I can.

First we 'll do the simplifying assumption that the coil consists only of 1 turn. If we don't this , then the expression for the total flux through the coil will be a sum of N different terms (N the number of turns of coil) and things might get quite complex.

Second simplifying assumption is that we shall ignore the self inductance of the coil. If we don't neglect it then we 'll have to solve a system of two coupled differential equations with two unknown functions the current $I(t)$ and the velocity of the bar magnet $v(t)$ and things will also get more complex too.

if z(t) is the distance of the dipole magnet from the center of the coil then a simplifying expression for the magnetic field of the z-component (which is perpendicular to the cross section area of the coil) will be $$B_z(z(t))=\frac{1}{z(t)^2+C}B_0$$ where $\frac{B_0}{C}$ is the magnetic field at the surface of the bar magnet.

Then the magnetic flux through the coil will be $$\Phi=\pi L^2B_z(z(t))$$ and hence the EMF produced will be $$\mathcal {E}=\frac{d\Phi}{dt}=\frac{d\Phi}{dz}\frac{dz}{dt}=\pi L^2 \frac{dB_z(z)}{dz}v(t)$$ where $$v(t)=\frac{dz}{dt}$$ the speed of the magnet. Here we ll do another "fat" approximation and we ll set $$\frac{dB_z(z)}{dz}=C'$$ that is to set this expression to a constant C', though in reality it will not be a constant it will depend on time t. A good choice for the constant C' will be the average value of $\frac{dB_z(z)}{dz}$ over time t.

The current through the coil , since we neglect the self inductance of the coil, will be $$I=\frac{\mathcal{E}}{R}=\pi L^2C' \frac{v(t)}{R}$$ where $R$ the ohmic resistance of the coil.

Then the force from the magnet to the coil will be $$F_B=\int Id\vec{l}\times \vec{B}$$ where $\vec{B}$ is the total magnetic field of the dipole magnet at the points of the turn of the coil. This $\vec{B}=B_x\hat x+B_y\hat y+B_z\hat z$ will have x,y and of course the z component which is already mentioned. Now if we do this integral we will find that the total force due to the z component $B_z(z)$ is zero (at all times), so it remains only to calculate the integral for the x and y components of the $B$. We further do the simplifying assumption that those x and y components are such that their sum $\vec{B_N}=B_x\hat x+B_y\hat y$ is normal to the $d\vec{l}$. Then it will be $$F_B=\int B_NIdl=2\pi LB_NI =2\pi^2\frac{ L^3C'B_N}{R}v(t)$$. The braking force on the magnet, is due to Newton's 3rd law $-F_B$.

So at the end of the day we have find a different expression for the force than the other approach. The approach with the linear drag and dimensional analysis gives the force as $$-\frac{B^2L^2}{R}v(t)$$, while the approach from applying the EM laws(including a hell load of simplifying assumptions) gives the force as $$-2\pi^2\frac{L^3C'B_N}{R}v(t)$$.

I am not sure at all in what I did when applying the laws of EM but that's the best I could do, if someone else can take over, please be my guest.

Thank you so much for this, Ill have a closer look at it soon when I have time and will start doing the calculations. Ill come back to you once this is done.

I really appreciate the help and time you have invested!

 

[kuruman]

if z(t) is the distance of the dipole magnet from the center of the coil then a simplifying expression for the magnetic field of the z-component (which is perpendicular to the cross section area of the coil) will be $$B_z(z(t))=\frac{1}{z(t)^2+C}B_0$$ where $\frac{B_0}{C}$ is the magnetic field at the surface of the bar magnet.

Not that it matters much, but I think the dependence goes as the inverse cube power. The magnetic field due to a magnetic dipole $\vec m$ is $$\vec B=\frac{\mu_0}{4\pi}\frac{(3\vec m\cdot\hat r)\hat r-\vec m}{r^3}.$$I spent some time exploring the magnet falling through a conducting tube alternative. The theoretical description is just as involved as the falling magnet through a ring.

At this point I think @Einstein44 needs to give us an outline of the project. After all that has been said and written, the actual project is still unfocused and we are groping in the dark.

 

[Delta2]

Thanks @kuruman for the correction, but my main concern was to come up with something that it doesn't give an infinity for z=0 (z=0 when the magnet is at the coil's turn plane). How the inverse cube equation can be modified in order to reflect the magnetic field of a real dipole (should we say bar) magnet.

 

[kuruman]

Thanks @kuruman for the correction, but my main concern was to come up with something that it doesn't give an infinity for z=0 (z=0 when the magnet is at the coil's turn plane). How the inverse cube equation can be modified in order to reflect the magnetic field of a real dipole (should we say bar) magnet.

I fully understand your concern. I made my suggestion because your expression does not asymptotically reduce to a $z^{-3}$ dependence when $z>>C.$ If you wrote $B_z(z(t))=\dfrac{B_0}{z(t)^3+C},$ I would not have said anything. I think that just adding a constant in the denominator is a fine way to get around the infinity at $z=0$ and it's not the only adjustment or approximation made here.

 

[Einstein44]

Not that it matters much, but I think the dependence goes as the inverse cube power. The magnetic field due to a magnetic dipole $\vec m$ is $$\vec B=\frac{\mu_0}{4\pi}\frac{(3\vec m\cdot\hat r)\hat r-\vec m}{r^3}.$$I spent some time exploring the magnet falling through a conducting tube alternative. The theoretical description is just as involved as the falling magnet through a ring.

At this point I think @Einstein44 needs to give us an outline of the project. After all that has been said and written, the actual project is still unfocused and we are groping in the dark.

What the actual project was about, was to calculate the force that eddy currents apply on a magnet falling through a copper coil. This is where we made some corrections:
I realized that this was a more complex task than I thought, so the goal was to create a mathematical model for the Magnetic Force acting on the magnet, as a function of velocity.
I still don't know if this force is even noticeable in a coil, so I am not sure if this model can also be used for a tube let's say?
I would calculate the Forces for many different numbers of magnet that I let fall, by putting them into a small object where I can keep the weight constant as @kuruman suggested in a previous post. I would probably aim to test this mathematical model doing some sort of experiment. This would then be a theoretical project as opposed to an experimental one.
Please tell me if this is what you need to know, or if there is anything I should elaborate on. I am also open to suggestions for improvement.

 

[kuruman]

What the actual project was about, was to calculate the force that eddy currents apply on a magnet falling through a copper coil. This is where we made some corrections:
I realized that this was a more complex task than I thought, so the goal was to create a mathematical model for the Magnetic Force acting on the magnet, as a function of velocity.
I still don't know if this force is even noticeable in a coil, so I am not sure if this model can also be used for a tube let's say?
I would calculate the Forces for many different numbers of magnet that I let fall, by putting them into a small object where I can keep the weight constant as @kuruman suggested in a previous post. I would probably aim to test this mathematical model doing some sort of experiment. This would then be a theoretical project as opposed to an experimental one.
Please tell me if this is what you need to know, or if there is anything I should elaborate on. I am also open to suggestions for improvement.

It seems that you want to do a theoretical calculation with no experimental component. That's fine. However, if that is the case, then
1. You don't need to be concerned whether the force is noticeable because you are not going to measure it. The model cannot be used for a tube because the geometry is different.
2. You don't need separate calculations for magnets of different strengths or mass. These would be represented by symbols in your theoretical expression.

I don't want to push you in any specific direction, but is it fair to say that you will proceed along the lines suggested by @Delta2 in post #39 with my suggested correction in post #43?

 

[Einstein44]

It seems that you want to do a theoretical calculation with no experimental component. That's fine. However, if that is the case, then
1. You don't need to be concerned whether the force is noticeable because you are not going to measure it. The model cannot be used for a tube because the geometry is different.
2. You don't need separate calculations for magnets of different strengths or mass. These would be represented by symbols in your theoretical expression.

I don't want to push you in any specific direction, but is it fair to say that you will proceed along the lines suggested by @Delta2 in post #39 with my suggested correction in post #43?

Yes, I will do exactly that. I will do all the calculations and then send you what I did on here again. I will probably have to do this this weekend when I have time.
I assume it would be good if I do the calculation using this model one time?

 

[Einstein44]

It seems that you want to do a theoretical calculation with no experimental component. That's fine. However, if that is the case, then
1. You don't need to be concerned whether the force is noticeable because you are not going to measure it. The model cannot be used for a tube because the geometry is different.
2. You don't need separate calculations for magnets of different strengths or mass. These would be represented by symbols in your theoretical expression.

I don't want to push you in any specific direction, but is it fair to say that you will proceed along the lines suggested by @Delta2 in post #39 with my suggested correction in post #43?

Would this model work in some way for a loop like showing in the picture? I know this model is made for one loop of a coil, but would that also work for this ??
I have connected it to an ammeter and have found that by dropping the magnet through it there was a noticeable current induced.
I think I could perhaps make an experimental part to this and then do the theoretical part using the model @Delta2 suggested.

 

[Einstein44]

if z(t) is the distance of the dipole magnet from the center of the coil then a simplifying expression for the magnetic field of the z-component (which is perpendicular to the cross section area of the coil) will be $$B_z(z(t))=\frac{1}{z(t)^2+C}B_0$$ where $\frac{B_0}{C}$ is the magnetic field at the surface of the bar magnet.

How exactly do you calculate B_o ? Is it just simply the formula for magnetic field strength B=μ_oNI / L ?

 

[Delta2]

How exactly do you calculate B_o ? Is it just simply the formula for magnetic field strength B=μ_oNI / L ?

$B_0$ is the magnetic field strength at the surface of the magnet, it is not the magnetic field of the coil due to the induced current.

 

[Iyeron]

A related thread on how a bar magnet will produce a force on one turn is this
https://www.physicsforums.com/threads/swinging-a-hanging-copper-worktime-ring-with-a-bar-magnet.1005442/#post-6519455

Thanks for this!