# How to find the Braking Torque of an Eddy Current Braking Setup?

#### IndianGeek234

Homework Statement
Finding the braking torque applied on a moving disk as a result of the Eddy Currents
Homework Equations
$$F = \sigma v V B^2$$
Homework Statement: Finding the braking torque applied on a moving disk as a result of the Eddy Currents
Homework Equations: $$F = \sigma v V B^2$$

So right,

Basically my problem is how to find the torque exerted by an eddy current braking setup.

My setup consists of a disk rotating on an axle and five magnets on a frame next to the disk. When the magnets are directly over the disk, the eddy current force produced retards and stops the disk. I'm trying to quantify this force based off of the magnetic field strength and velocity or frequency of the disk.
Where I started off was this PF thread where they concluded the Eddy Current Braking force as $$J\times B$$ Which when $J$ is substituted as $\sigma(v \times B)$ simplifies to $$F = \sigma v V B^2$$ However, the problems I had with this, is that firstly it didn't match my data, when I empirically calculated torque through rate of change of angular momentum. Also, on the last few messages on that thread, there is a mention of integration over r, which I'm not sure what that means.

I have seen this paper by Smythe on several sites, as well as others by Wiederick and others, trying to make an analytical model of eddy-current braking. However, unfortunately the maths used looks quite confusing for me. So I'd appreciate it if someone over here could help me out with this problem..

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#### scottdave

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The math is not going to be simple as you had hoped.
Let's just look at what you mentioned integrating. What can you say about the velocity of a portion of the disc in relation to the location (distance from the center)?
Then how does torque relate to distance from center?

#### IndianGeek234

The math is not going to be simple as you had hoped.
Let's just look at what you mentioned integrating. What can you say about the velocity of a portion of the disc in relation to the location (distance from the center)?
Then how does torque relate to distance from center?
Hi,

As far as I know, the velocity is defined as $$\omega r$$ and the torque would be defined as $$Fd$$ I'm sorry I'm not able to get at what you're suggesting..

#### haruspex

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Hi,

As far as I know, the velocity is defined as $$\omega r$$ and the torque would be defined as $$Fd$$ I'm sorry I'm not able to get at what you're suggesting..
What is the relationship between the r and the d?

#### scottdave

Homework Helper
Does the magnetic field only act at a single distance from the center, or is there a range of distances? If it is not a single distance, how can you take this into account?

#### IndianGeek234

What is the relationship between the r and the d?
In my case, both would be the same.

Does the magnetic field only act at a single distance from the center, or is there a range of distances? If it is not a single distance, how can you take this into account?
It would act at a range of distances, however, not the whole disk as the magnetic field strength would drop off, after a distance. I feel like the answer would be to integrate. However, I'm not sure what to integrate, and over what range

#### haruspex

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In my case, both would be the same.

It would act at a range of distances, however, not the whole disk as the magnetic field strength would drop off, after a distance. I feel like the answer would be to integrate. However, I'm not sure what to integrate, and over what range
So consider a small element of the field, area dA, at distance r from the axis. What is the tangential speed of this element? What torque does it exert?

#### IndianGeek234

So consider a small element of the field, area dA, at distance r from the axis. What is the tangential speed of this element? What torque does it exert?
The torque and the tangential speed would be irrespective of which element and depend only on the distance from the centre. Thus, respectively they would be $T=Fd$ and $v=\omega r$

#### haruspex

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The torque and the tangential speed would be irrespective of which element and depend only on the distance from the centre. Thus, respectively they would be $T=Fd$ and $v=\omega r$
We already established that d=r, so no need to introduce d.
Substitute your equation for F, except ... what is V? You have vV in there. Isn't the V redundant?

#### IndianGeek234

We already established that d=r, so no need to introduce d.
Substitute your equation for F, except ... what is V? You have vV in there. Isn't the V redundant?
Small v is velocity. Capital V is volume. This is because $J \times B$ only gives force per unit volume. So, in order to get the total force, it has to be multiplied by the volume

#### haruspex

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Small v is velocity. Capital V is volume. This is because $J \times B$ only gives force per unit volume. So, in order to get the total force, it has to be multiplied by the volume
Ok, that's good, because I was puzzling over how to get dA into the equation. Express V in terms of that and the thickness.

#### IndianGeek234

Ok, that's good, because I was puzzling over how to get dA into the equation. Express V in terms of that and the thickness.
Err.. Then it would become $$F= \sigma \cdot \omega r \cdot (dA \cdot \tau) B^2$$
Where $\tau$ represents the thickness of the disk.

If I'm right you now want me to integrate this whole equation with respect to $dA$ ?

#### haruspex

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If I'm right you now want me to integrate this whole equation with respect to dAdAdA ?
Not quite. It's the torque that you need to integrate. What is the torque exerted by this element?

#### IndianGeek234

Not quite. It's the torque that you need to integrate. What is the torque exerted by this element?
The torque would be $$T= F \cdot r$$ which is equivalent to $$\sigma \cdot \omega r \cdot (dA \cdot \tau) B^2 \cdot r$$ $$\sigma \cdot \omega r^2 \cdot (dA \cdot \tau) B^2$$ So the total torque is $$\sigma \cdot \omega r^2 \cdot \tau \cdot B^2 \int dA$$ Or am I missing something?

#### haruspex

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The torque would be $$T= F \cdot r$$ which is equivalent to $$\sigma \cdot \omega r \cdot (dA \cdot \tau) B^2 \cdot r$$ $$\sigma \cdot \omega r^2 \cdot (dA \cdot \tau) B^2$$ So the total torque is $$\sigma \cdot \omega r^2 \cdot \tau \cdot B^2 \int dA$$ Or am I missing something?
r varies, so belongs inside the integral.
τdA is an element of volume. If you think of it as an element of mass you have an integral like ∫r2dm. Remind you of anything?

#### IndianGeek234

r varies, so belongs inside the integral.
τdA is an element of volume. If you think of it as an element of mass you have an integral like ∫r2dm. Remind you of anything?
Wait, I'm a bit confused. I don't understand exactly what you mean by an element of volume....

#### haruspex

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Wait, I'm a bit confused. I don't understand exactly what you mean by an element of volume....
We have dA as an element of area, i.e. the area of a small patch on the surface in one of the magnetic fields. Since the thickness is τ, its volume is τdA. You have the expression for the torque exerted there (the element of torque), and we wish to add all these elements up. That's what integration is.
So, does an integral like ∫r2dm remind you of anything?

#### IndianGeek234

We have dA as an element of area, i.e. the area of a small patch on the surface in one of the magnetic fields. Since the thickness is τ, its volume is τdA. You have the expression for the torque exerted there (the element of torque), and we wish to add all these elements up. That's what integration is.
So, does an integral like ∫r2dm remind you of anything?
Oh.Thats the moment of inertia isn't it?

#### haruspex

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Oh.Thats the moment of inertia isn't it?
Right. So you can use the usual methods to find the total 'MoI' of your magnetic fields about the central axis. I'm guessing each is a disc, roughly.

#### IndianGeek234

Right. So you can use the usual methods to find the total 'MoI' of your magnetic fields about the central axis. I'm guessing each is a disc, roughly.
Right,
So, What I did is equate the torque to $\sigma \cdot \omega \cdot B^2 \cdot I_c$

So, what I did for finding $I_c$, was firstly approximate the total area of effect of the magnets, which is approximately in the shape of a ring around the disk, but only covering half of the disk, as the magnets are basically mounted on a wooden semi-circular frame over the disk I then found the $I_c$ of this area, which I then implemented in the end-formula.

However, the problem I'm facing now, is that the data predicted by this is nowhere near my experimental torque results. Its not a small error, it's in the range of a few orders of magnitude. So I'm not sure what to do.

#### haruspex

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Right,
So, What I did is equate the torque to $\sigma \cdot \omega \cdot B^2 \cdot I_c$

So, what I did for finding $I_c$, was firstly approximate the total area of effect of the magnets, which is approximately in the shape of a ring around the disk, but only covering half of the disk, as the magnets are basically mounted on a wooden semi-circular frame over the disk I then found the $I_c$ of this area, which I then implemented in the end-formula.

However, the problem I'm facing now, is that the data predicted by this is nowhere near my experimental torque results. Its not a small error, it's in the range of a few orders of magnitude. So I'm not sure what to do.
Please show the details of your Ic calculation. It sounds like that is the second moment about the "mass centre" of magnet area. As I wrote, you want it about the central axis. Use the parallel axis theorem.

#### IndianGeek234

Please show the details of your Ic calculation. It sounds like that is the second moment about the "mass centre" of magnet area. As I wrote, you want it about the central axis. Use the parallel axis theorem.
Ok, right,

So firstly, to find the moment of inertia of this semicircular region, I first find the total moment of inertia and then the moment of inertia of the inner ring, and then subtract to get the moment of inertia of the ring itself. This is then divided by two, to give me the moment for the semicircular region.

If it helps, then imagine a wooden semicircle sitting atop the disk. This area of the wooden frame is the same as the area of the disk whose moment I'm trying to find.

So, the total moment of inertia is $$1/2 MR^2 = 4.21 \cdot 10^{-3} kg m^2$$ Now for the inner disk, the radius would be 0.06984 m, the thickness is 3.76 mm, and I'm using the density as 2810 kg/m3. So, the expression for the moment of inertia of the inner disk, using the volume of a cylinder times the density to get the mass, becomes: $$1/2 \cdot \pi (0.06984)^2 \cdot 3.76 \cdot 10^{-3} \cdot 2810 \cdot (0.06984)^2 = 3.948\cdot 10^{-4}$$ Subtracting these and dividing by 2, gives me the final moment value.

#### haruspex

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I'm confused by your description. You describe it as semicircular, then mention an "inner ring". Is it a semicircular annulus? If so, the MoI about the central axis is $\frac 12m(r_1^2+r_2^2)$, where r1 and r2 are the inner and outer radii.
You might be surprised by the + sign. If we express it in terms of density rather than mass, $m=\rho \pi(r_2^2-r_1^2)$, so $I=\frac 12\rho\pi(r_2^4-r_1^4)$.

#### IndianGeek234

I'm confused by your description. You describe it as semicircular, then mention an "inner ring". Is it a semicircular annulus? If so, the MoI about the central axis is $\frac 12m(r_1^2+r_2^2)$, where r1 and r2 are the inner and outer radii.
You might be surprised by the + sign. If we express it in terms of density rather than mass, $m=\rho \pi(r_2^2-r_1^2)$, so $I=\frac 12\rho\pi(r_2^4-r_1^4)$.
Hi,
Sorry for the delay,
So, right after looking at what is an annulus, I decided that it is in fact an annulus. My values for inner and outer radii are: $$r_1 =0.6984 m ; r_2 =0.11984 m$$. Using these values and the density of aluminium as 2810 kg/m3, I get the value of the moment of inertia as 0.805. Plugging these values back into the original equation, gives me exorbitantly high values of torque, which don't seem to correspond with the experimentally derived torque. So, I don't know what I'm doing wrong.

#### haruspex

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the density of aluminium
Density?!
Thinking of it as mass was only an analogy so that you could know to apply the moment of annulus formula, or parallel axis theorem, or whatever.

In post #14 you had $\sigma\omega\tau r^2B^2\int dA$.
But the r should have been inside the integral: $\sigma\omega\tau B^2\int r^2dA$.
That integral is the second moment of area (like moment of inertia, but without the density factor), and in post #23 I gave you the solution for this when it is an annulus:
had $\sigma\omega\tau B^2\frac 12\pi(r_2^4-r_1^4)$.
But your shape is only half an annulus, so $\sigma\omega\tau B^2\frac 14\pi(r_2^4-r_1^4)$.

"How to find the Braking Torque of an Eddy Current Braking Setup?"

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