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

[Einstein44]

Homework Statement

I want to find out the force that eddy currents apply on a magnet that is moving though a copper coil. I have not been able to find out how this works, although I know hoe to find the magnitude of the induced current occurring from the change in magnetic flux using Faraday's Law.

Relevant Equations

Emf = -N dΦB/dt
This is Faradays Law I used to find the induced EMF.

How can I find out its effect on the magnet itself?

I have jus thought about how maybe I could sue the magnetic field strength equation for a solenoid (that I used to find the Emf in the first place) again, to find the magnetic field strength arising from the induced current:

B= μNI / L
where μ is the permativity of free space
N= Number of turns in the coil
L = length of the coil

Here I could just plug in the newly found Current using the calculations from before when I found the EMF and then plugged it into the equation I=V/R. Would this work? And if so, how does this even tell me what the force acting on the magnet is? Is there any way to calculate this?

 

[kuruman]

You cannot use the solenoid formula because the induced currents are (a) not uniform (they depend on the position of the magnet) and (b) do not flow in the same direction. See
https://www.physicsforums.com/threa...e-process-occur-visually.999296/#post-6452919
for a qualitative explanation of what's going on. If you use a copper tube (continuous coil) that is long enough for a strong enough magnet to reach terminal velocity inside the tube, then a good estimate for the size of the magnetic force is the weight of the magnet (neglecting air resistance, that is).

Is this actually a homework problem? If yes, please post the exact statement of the problem as given to you. If no, this thread should be moved to Classical Physics.

 

[Einstein44]

You cannot use the solenoid formula because the induced currents are (a) not uniform (they depend on the position of the magnet) and (b) do not flow in the same direction. See
https://www.physicsforums.com/threa...e-process-occur-visually.999296/#post-6452919
for a qualitative explanation of what's going on. If you use a copper tube (continuous coil) that is long enough for a strong enough magnet to reach terminal velocity inside the tube, then a good estimate for the size of the magnetic force is the weight of the magnet (neglecting air resistance, that is).

Is this actually a homework problem? If yes, please post the exact statement of the problem as given to you. If no, this thread should be moved to Classical Physics.

Thank you for your reply. Although I need to do some further research to grasp this better, as I always encounter different equations for everything, making this fairly complicated.
This is actually for a physics project I am currently working on (for school), so there is no homework statement to this and yes maybe this should be moved to classical physics (my bad), I am new on here and did not know where this corresponds.

What I am exactly doing, is I am trying to find the effect of eddy currents on a neodymium magnet moving (or falling) through a coiled up copper wire. I am trying to find out exactly how much braking force due to eddy currents acts on this magnet by doing the calculations. (which has turned out more complicated than I thought, with all the different forms of Faradays Laws and the Maxwell equations to find the induced emf, and then find out how to calculate what the actual braking force these eddy currents create on the magnet(s).

 

[Steve4Physics]

A few random thoughts in random order...

Is the coil closed - i.e. are the 2 ends of the coil joined together? If not, you will probably not be able to detect any effects at all.

Is the wire insulated? (If not are the coils touching? - if they are touching and uninsulated, the wire is not acting as a coil.)

What are you actually going to measure?

The effects you are trying to detect are small and the experiment would usually be conducted using a thick hollow copper tube to make the effect larger as noted by @kuruman. So you are making life very difficult for yourself by using a coil.

An ‘eddy current’ is a term used for an induced current loop in conductor such as a block of metal. The term eddy current is generally not used for the induced current around a coil.

Calculating the magnitude of the force would be a hugely complex task – the force will be a complicated function of many variables such as the speed of magnet, the shape and orientation of the magnet, the strength of magnet, the geometrical arrangement of the individual turns of wire in the coil, the length and diameter of the wire and the resistivity of copper Calculating the force as a function of these parameters is simply not feasible.

You haven’t stated the specific objective of the investigation. Do you have one? You might want to limit the scope (if you are allowed) to make things more manageable – e.g. investigate the effect of number of turns of wire on time for magnet to fall. Or, maybe, investigate the effect of different sizes of copper tube on time for magnet to fall.

 

[kuruman]

I assume this is a theoretical (as opposed to experimental) project. You will have to make assumptions and approximations and even then it may not be an easy problem. For example, you can approximate the magnet as a magnetic dipole which will give you a (relatively) simple expression for the magnetic field.
Then you drop this dipole along the axis of the coil (z-axis). At this point, you will have to
1. Find the z-component of the magnetic field at an arbitrary point on the face of the coil, $B_z(\rho,z)$ in cylindrical coordinates.
2. Integrate to find the magnetic flux through the coil, $\Phi_M (z)=\int B_z(\rho,z)~2\pi \rho d\rho.$
3. Calculate the induced current using $$I(z)=\frac{1}{R}\frac{d\Phi_M}{dt}=\frac{1}{R}\frac{d\Phi_M}{dz}\frac{dz}{dt}.$$Now comes the fun part. You should probably consider a very thin loop of a single turn. Then you will have to
4. Calculate the force on the dipole using $\vec F_M(z)=- \int I(z)d\vec l\times \vec B$, where $\vec B$ is the total field, not just the z-component.
5. Solve Newton's second law equation $$m\frac{d^2z}{dt^2}=mg-F_{M,z}.$$This last equation assumes that "down" is positive and that there are no lateral forces by symmetry.
6. Once you have the solution $z(t)$ you can easily figure out the magnetic force acting on the magnet.

The whole calculation is a tall order, to say the least. It's the way I would try doing it, but only if I had to, e.g. if my dog's life depended on it.

On edit: Placed a negative sign to indicate that the magnetic force on the falling dipole is the 3rd law counterpart of the magnetic force exerted on current element $I(z) d\vec l$.

 

[Einstein44]

A few random thoughts in random order...

Is the coil closed - i.e. are the 2 ends of the coil joined together? If not, you will probably not be able to detect any effects at all.

Is the wire insulated? (If not are the coils touching? - if they are touching and uninsulated, the wire is not acting as a coil.)

What are you actually going to measure?

The effects you are trying to detect are small and the experiment would usually be conducted using a thick hollow copper tube to make the effect larger as noted by @kuruman. So you are making life very difficult for yourself by using a coil.

An ‘eddy current’ is a term used for an induced current loop in conductor such as a block of metal. The term eddy current is generally not used for the induced current around a coil.

Calculating the magnitude of the force would be a hugely complex task – the force will be a complicated function of many variables such as the speed of magnet, the shape and orientation of the magnet, the strength of magnet, the geometrical arrangement of the individual turns of wire in the coil, the length and diameter of the wire and the resistivity of copper Calculating the force as a function of these parameters is simply not feasible.

You haven’t stated the specific objective of the investigation. Do you have one? You might want to limit the scope (if you are allowed) to make things more manageable – e.g. investigate the effect of number of turns of wire on time for magnet to fall. Or, maybe, investigate the effect of different sizes of copper tube on time for magnet to fall.

Yes, I think that changing the direction of the topic would be a good idea.
As you just mentioned, maybe I could investigate the time taken for magnets to fall through a copper tube (instead of a coil). However: What would be calculations I would have to do here?

I was thinking of investigating how the number of magnets (magnetic field strength) will affect the time taken for it to fall through the tube: However here I would have two variables: The magnetic Field strength and the weight of the magnets, so I assume this wouldn't work?

Please let me know if you have any suggestions.
And thank you for your response!

 

[Einstein44]

I assume this is a theoretical (as opposed to experimental) project. You will have to make assumptions and approximations and even then it may not be an easy problem. For example, you can approximate the magnet as a magnetic dipole which will give you a (relatively) simple expression for the magnetic field.
Then you drop this dipole along the axis of the coil (z-axis). At this point, you will have to
1. Find the z-component of the magnetic field at an arbitrary point on the face of the coil, $B_z(\rho,z)$ in cylindrical coordinates.
2. Integrate to find the magnetic flux through the coil, $\Phi_M (z)=\int B_z(\rho,z)~2\pi \rho d\rho.$
3. Calculate the induced current using $$I(z)=\frac{1}{R}\frac{d\Phi_M}{dt}=\frac{1}{R}\frac{d\Phi_M}{dz}\frac{dz}{dt}.$$Now comes the fun part. You should probably consider a very thin loop of a single turn. Then you will have to
4. Calculate the force on the dipole using $\vec F_M(z)= \int I(z)d\vec l\times \vec B$, where $\vec B$ is the total field, not just the z-component.
5. Solve Newton's second law equation $$m\frac{d^2z}{dt^2}=mg-F_{M,z}.$$This last equation assumes that "down" is positive and that there are no lateral forces by symmetry.
6. Once you have the solution $z(t)$ you can easily figure out the magnetic force acting on the magnet.

The whole calculation is a tall order, to say the least. It's the way I would try doing it, but only if I had to, e.g. if my dog's life depended on it.

Thank you very much for taking the time,
Indeed I realize that this might be way too complicated to actually do, which is also what a university professor just suggested me when U told him about the idea.
I was actually trying to make this an experimental project and I wanted to do the needed calculations to investigate the effect of eddy currents on a magnet falling through a copper coil...

I just thought I might be able to do the following:
I was thinking of investigating how the number of magnets (magnetic field strength) will affect the time taken for it to fall through a copper tube: However here I would have two variables: The magnetic Field strength and the weight of the magnets, so I assume this wouldn't work?
I also don't know which calculations I could do here, so if you have any ideas or suggestions I would really appreciate that.

 

[kuruman]

You have not specified whether this project is theoretical, experimental, or both. You keep talking about calculations. Are you going to take measurements and see what you can make of them? If you are planning to take measurements, I think stacking up magnets is a good idea. You can take a pillbox of appropriate diameter, put your magnet in, drop it through the tube and measure the time. Then you can add something non-magnetic in the pillbox, like sand, to increase the mass continuously but not the magnetic field. Then you can plot "Drop time" (dependent variable) vs. "Pillbox Mass" (independent variable) and see if you can model mathematically the law that governs the acceleration due to magnetic braking. You can repeat with a second magnet in the pillbox.

I think that would make a reasonably fun project.

 

[Einstein44]

You have not specified whether this project is theoretical, experimental, or both. You keep talking about calculations. Are you going to take measurements and see what you can make of them? If you are planning to take measurements, I think stacking up magnets is a good idea. You can take a pillbox of appropriate diameter, put your magnet in, drop it through the tube and measure the time. Then you can add something non-magnetic in the pillbox like sand to add mass continuously but not magnetic field. Then you can plot "Drop time" (dependent variable) vs. "Pillbox Mass" (independent variable) and see if you can model mathematically the law that governs the acceleration due to magnetic braking. You can repeat with a second magnet in the pillbox.

I think that would make a reasonably fun project.

That is absolutely brilliant, I think that is completely doable. I think I will do that then. And yes, the goal was to make this experimental, however since this is supposed to be more advanced than the curriculum I need to have some calculations in order to find something out mathematically.

What exactly do you mean by "model the law that governs the acceleration due to magnetic braking"? I am not sure I understood that.

Thank you again!

 

[kuruman]

That is absolutely brilliant, I think that is completely doable. I think I will do that then. And yes, the goal was to make this experimental, however since this is supposed to be more advanced than the curriculum I need to have some calculations in order to find something out mathematically.

What exactly do you mean by "model the law that governs the acceleration due to magnetic braking"? I am not sure I understood that.

Thank you again!

I am glad you find my suggestion doable. By "model" I mean find a mathematical expression for the magnetic force $F_M$ that I discussed earlier. You know that it is not constant like the force of gravity. What could it depend on other than the strength and geometry of the magnet? Hint: When is the magnetic braking force zero and when is it non-zero?

 

[Einstein44]

I am glad you find my suggestion doable. By "model" I mean find a mathematical expression for the magnetic force $F_M$ that I discussed earlier. You know that it is not constant like the force of gravity. What could it depend on other than the strength and geometry of the magnet? Hint: When is the magnetic braking force zero and when is it non-zero?

Magnetic Force depends on magnetic flux density, current and length. (From the formula F= BIl), if this is even what you meat?

For the Hint --> I assume the magnetic braking force is zero when the magnets is stationary and non-zero when moving. So velocity would be a factor that affects the magnetic force, since I know that the faster the movement, the faster the braking force due to eddy current is. This is because the bigger the change in magnetic flux, the bigger the emf induced and therefore the bigger the eddy current creating the braking effect.

Please tell me if this is what you meant?

 

[kuruman]

Magnetic Force depends on magnetic flux density, current and length. (From the formula F= BIl), if this is even what you meat?

For the Hint --> I assume the magnetic braking force is zero when the magnets is stationary and non-zero when moving. So velocity would be a factor that affects the magnetic force, since I know that the faster the movement, the faster the braking force due to eddy current is.

Please tell me if this is what you meant?

Yes, this is what I meant. Very good. The other variables that you mentioned, determine the overall strength of the force. The important variable here is the velocity. At this point, you need to write an expression for the dependence of the magnetic force on the velocity. You can lump all the other variables under a common constant called $k$. How would you write an expression for the magnetic braking force that includes the velocity and why this particular form?

 

[Einstein44]

Yes, this is what I meant. Very good. The other variables that you mentioned, determine the overall strength of the force. The important variable here is the velocity. At this point, you need to write an expression for the dependence of the magnetic force on the velocity. You can lump all the other variables under a common constant called $k$. How would you write an expression for the magnetic braking force that includes the velocity and why this particular form?

What are you referring to by the constant of k?

And here I am again with the same problem:
It is incredibly complicated to find an expression for the magnetic braking force of those eddy currents. I tried finding expressions all over the internet and all I found was: "its very complex". This is because there are so many factors that affect it, such as:

1. strength of magnetic field within the metal and the magnitude of the change in field strength
2. the position of the magnets, which related to field strength
3. the shape, thickness and geometry of the metal
4. and of course the SPEED of the magnet, which is what we were discussing

I have tried to find clues on the internet, which was not a success. Can you help me understand a bit better what you had in mind?

And wouldn't the equations you stated above work for this? (when solving for Fm)

 

[kuruman]

Yes, it is incredibly complicated, but you picked this problem. If you cannot solve it, then you need to "model" it, i.e. make simplifying assumptions that would make the problem solvable. It's either that or find another problem. I will give you some hints, but you have to do the thinking.

If you drop a magnet through a metal tube you have eddy currents above and below the magnet, both opposing the fall. This means that as the magnet moves closer to the bottom, eddy currents are removed from below the magnet and added to above the magnet. This might be enough justification to make the simplifying assumption that the magnetic force is independent of the position. That addresses item 2 above.

The dependence of the quantities you mentioned in items 1 and 3 is, as we have seen, complex. They contribute to the magnetic braking force (MBF) in some unknowable manner. However, the most important piece of information is that the MBF is non-zero only when the velocity is non-zero. This means that you can write the MBF as $F_M=f(v)$ where $f(v)$ is a function of velocity such that $f(0)=0$. What form. do you guess, $f(v)$ should have? Don't worry that your guess might be incorrect at this point. You will make measurements to verify it. As the saying goes, "if we knew what we are doing, it wouldn't be called research."

 

[Einstein44]

Yes, it is incredibly complicated, but you picked this problem. If you cannot solve it, then you need to "model" it, i.e. make simplifying assumptions that would make the problem solvable. It's either that or find another problem. I will give you some hints, but you have to do the thinking.

If you drop a magnet through a metal tube you have eddy currents above and below the magnet, both opposing the fall. This means that as the magnet moves closer to the bottom, eddy currents are removed from below the magnet and added to above the magnet. This might be enough justification to make the simplifying assumption that the magnetic force is independent of the position. That addresses item 2 above.

The dependence of the quantities you mentioned in items 1 and 3 is, as we have seen, complex. They contribute to the magnetic braking force (MBF) in some unknowable manner. However, the most important piece of information is that the MBF is non-zero only when the velocity is non-zero. This means that you can write the MBF as $F_M=f(v)$ where $f(v)$ is a function of velocity such that $f(0)=0$. What form. do you guess, $f(v)$ should have? Don't worry that your guess might be incorrect at this point. You will make measurements to verify it. As the saying goes, "if we knew what we are doing, it wouldn't be called research."

I have tried to come up with something: no success.

Here was my thought process and yes I know this will be dead wrong, but since you told me to try something...

Using Faradays Law: EMF= dφ/dt = d/dt ∫B⋅dA
equating this to the velocity of the magnet --> converting the differential area to a known height times a differential with (dA=Ldx), and relating the differential width to a differential time using velocity cancels out the integral:
EMF= BLv
L=Length of conductor
v=speed of magnet
to find B, I can simply use the equation B=μI /2πr
and to find the velocity: simply do v=s/t

After calculating this EMF, I can find the current induced by plugging it into the formula I= V/R , while finding R with an ohmmeter.

Now plugging everything into a formula for "Retarding Force associated with the braking (don't know if this is the correct one, but this is all I have for now:
F=ILBas I said, I know this is absolute nonsense, but here I am...

 

[kuruman]

as I said, I know this is absolute nonsense, but here I am...

You shouldn't be there. It's more than absolute nonsense, it's a mish-mash of nonsense, hastily assembled with total disregard to the applicability of the equations that you are quoting.
EMF = B L v is the equation for motional EMF when a conductor of length L moves with velocity v in a magnetic field B such that the velocity, the length of the rod and the B field are mutually perpendicular. Here you have a magnet falling inside a metal tube.

B=μI /2πr is the equation giving the B-field at distance r from a very long wire carrying current I. The metal tube carries no current except for the eddy currents which are not along the length of the wire but circulate about the axis. Even if there were a current established from one end of the tube to the other, the B-field inside the tube would be zero. This makes the measurement of the resistance of the tube with an ohmmeter irrelevant.

v = s/t is an equation that is applicable only if the acceleration is zero. In fact, you want to find a measure of the magnetic braking acceleration. You know that the acceleration of the magnet is not zero because when you let it go from rest it starts moving. Thus, using this equation can only lead to an incorrect result.

I am not going to do this project for you. Read post #14 and think. If nothing comes to mind, then someone else might be able to help you but not me.

 

[Einstein44]

You shouldn't be there. It's more than absolute nonsense, it's a mish-mash of nonsense, hastily assembled with total disregard to the applicability of the equations that you are quoting.
EMF = B L v is the equation for motional EMF when a conductor of length L moves with velocity v in a magnetic field B such that the velocity, the length of the rod and the B field are mutually perpendicular. Here you have a magnet falling inside a metal tube.

B=μI /2πr is the equation giving the B-field at distance r from a very long wire carrying current I. The metal tube carries no current except for the eddy currents which are not along the length of the wire but circulate about the axis. Even if there were a current established from one end of the tube to the other, the B-field inside the tube would be zero. This makes the measurement of the resistance of the tube with an ohmmeter irrelevant.

v = s/t is an equation that is applicable only if the acceleration is zero. In fact, you want to find a measure of the magnetic braking acceleration. You know that the acceleration of the magnet is not zero because when you let it go from rest it starts moving. Thus, using this equation can only lead to an incorrect result.

I am not going to do this project for you. Read post #14 and think. If nothing comes to mind, then someone else might be able to help you but not me.

I think this topic is too hard for me, so I am thinking of altering the investigation:

One idea I had in mind was: investigating the effect of magnetic field strength (by adding magnets with sand to keep the weight constant like you suggested) on the voltage induced in the copper cylinder. For this I would only have to use Faradays/ Lenz's Law, which I think is doable:

Emf= - ∂Φ/∂t
to find the change in magnetic flux I would just have to use the following formula:
Φb = ABcosθ

I could also talk about the applications of generating a voltage using a magnet moving through a conducting materials, as if I am not mistaken this is how wind turbines work, as well as trains using this as a braking mechanism (eddy currents). For the braking effect I could potentially talk about how the emf induced creates the eddy currents in detail without actually doing the calculations?

Could I please have your opinion on this idea and if you think this would work in your opinion?
(if there is scope to write an essay about this)

I'd like to thank you for the help. I appreciate that a lot.

 

[kuruman]

Where on the copper cylinder will you attach the voltmeter? How do you visualize the eddy currents in the cylinder? Your best bet is to wind a pick-up coil around (use insulated wire or better yet magnet wire) and attach the ends of the coil to the voltmeter. An oscilloscope would be a better way to detect the induced voltage but you probably don't have access to one.

Is an experimental component necessary to this project? You can write an essay if you feel more comfortable with it but you need to clear it with the person who assigned this project to you. I am not qualified to say that it's OK or not OK.

 

[Delta2]

This problem is indeed too complex, and if someone is going to look for an exact solution, he will have to write a Master thesis . However, as @kuruman suggested with the proper simplifying assumptions you can find a relatively simple form for the braking force as a function of the velocity/speed of magnet.

Hint: do you know the expression for the aerodynamic drag that an object experiences due to its velocity within an environment of air? Aerodynamic drag is also a kind of braking force.

 

[Einstein44]

This problem is indeed too complex, and if someone is going to look for an exact solution, he will have to write a Master thesis . However, as @kuruman suggested with the proper simplifying assumptions you can find a relatively simple form for the braking force as a function of the velocity/speed of magnet.

Hint: do you know the expression for the aerodynamic drag that an object experiences due to its velocity within an environment of air? Aerodynamic drag is also a kind of braking force.

do you mean this?

F_D= 1/2 ρv²C_DA

And how is that related to the problem?

 

[Delta2]

Yes but three things:

1. Instead of aerodynamic drag we have electromagnetic drag here , if I can call it that way
2. Instead of the constants $\rho,C_D,A$ you will have some other constants, like strength of magnetic field B of the magnet, radius L of the coil, ohmic resistance R of the coil, and maybe some others.
3. Instead of $v^2$ I think you will have just $v$, I think this (electromagnetic) drag is proportional to the speed not to the square of speed of the magnet.

 

[Einstein44]

Yes but three things:

1. Instead of aerodynamic drag we have electromagnetic drag here , if I can call it that way
2. Instead of the constants $\rho,C_D,A$ you will have some other constants, like strength of magnetic field B of the magnet, radius L of the coil, ohmic resistance R of the coil, and maybe some others.
3. Instead of $v^2$ I think you will have just $v$, I think this (electromagnetic) drag is proportional to the speed not to the square of speed of the magnet.

and how exactly did you get to this point? How do I know that I can just change these variables?

 

[Delta2]

and how exactly did you get to this point? How do I know that I can just change these variables?

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?

 

[Einstein44]

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?

Yes, this is great! Thank you so much, now I understand what you did! I appreciate it a los, I have been trying to figure things out for so long...

One other question: I have heard that there is a Law called the "Biot-Savart Law" for finding the magnetic field due to a current, which is:

B= μI /4π ∫ds x r^{^} /r²

I am not exactly certain in which case exactly you use this formula, instead of:
The regular formula for magnetic field strength of a solenoid?
B=μNI /l

 

[Delta2]

In our approximation I totally ignore the magnetic field generated by the coil due to self induction. The B present in the formulas in post #23 is something like the average magnetic field around the magnet..

 

[Delta2]

The braking force experienced by the magnet is not due to the magnetic field generated by the coil current but due to the so called BIL force (Laplace force) that the magnet exerts on the current (and hence by Newton's 3rd Law the current exerts on the magnet an equal and opposite force).

 

[Einstein44]

The braking force experienced by the magnet is not due to the magnetic field generated by the coil current but due to the so called BIL force (Laplace force) that the magnet exerts on the current (and hence by Newton's 3rd Law the current exerts on the magnet an equal and opposite force).

I am not exactly sure what you mean. How does the BIL Force apply a force on the current? I only know that the induced current creates a magnetic field that opposes the polarity of the magnet and therefore creates the braking force.

 

[Delta2]

I am not exactly sure what you mean. How does the BIL Force apply a force on the current? I only know that the induced current creates a magnetic field that opposes the polarity of the magnet and therefore creates the braking force.

Hold on, me and @kuruman might be wrong that the braking force is due to the BIL force, I need to think this a bit more...

 

[Delta2]

Ok, I think I have thought about this considerably in the last 24 hours and my final resolve is as follows:

• If we model the magnet as a linear dipole magnet, the force from the magnetic field generated by the coil to the magnet will be approximately zero and that is because the magnetic field generated by the coil will be approximately uniform, and it is a well known result that a uniform magnetic field applies zero force to a dipole magnet.
• Thus the only force on the dipole magnet comes from being the 3rd Newton law pair to the force that is the sum of the $BIdl$ forces that the magnet applies to elements $Idl$ of the turn of the coil that is closest to the magnet. A good approximation to this force I believe is indeed as presented in post #23

 

[Einstein44]

Ok, I think I have thought about this considerably in the last 24 hours and my final resolve is as follows:

• If we model the magnet as a linear dipole magnet, the force from the magnetic field generated by the coil to the magnet will be approximately zero and that is because the magnetic field generated by the coil will be approximately uniform, and it is a well known result that a uniform magnetic field applies zero force to a dipole magnet.
• Thus the only force on the dipole magnet comes from being the 3rd Newton law pair to the force that is the sum of the $BIdl$ forces that the magnet applies to elements $Idl$ of the turn of the coil that is closest to the magnet. A good approximation to this force I believe is indeed as presented in post #23

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

 

[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.