Why does the current have no ##\phi## component in a toroidal coil?

In summary: There is no current in the phi direction because the problem defines it that way. Look at the drawing: the wires go up in z, in in r, down in z, and out in r. They do not go in phi at all. Since the...In summary, these images show an image from the book Introduction to Electrodynamics by David J. Griffiths. The problem is that the author is unable to understand how the current has zero phi component. However, in spite of this, the current will always have a component from the toroid. If n is made very large, then the toroid current can be minimized. Finally, the author is ready to move on to the problem at hand.
  • #1
Adesh
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These are images from the book Introduction to Electrodynamics by David J. Griffiths .

Electro 1.png
.
Electro 2.png
.

My problem is that I'm unable to understand how the current has zero ##\phi## component (I have underlined it in the first image)? I do understand cylindrical coordinates, I know cylindrical coordinates involve three components ##(r,ϕ,z)##and ##\hat{r}##r^ points radially outwards, ##\hat{\phi}## points perpendicular to ##\hat{r}## and even to ##z## axis.

I fully understand this image (credit: https://physics.stackexchange.com/users/110781/frobenius)
HnqbE.png
.

But in spite of all these I can't seem to understand how the current have no ##\phi## component. Well, in that toroid we have current going on in circles and really I don't anything more than this.

Please help me!
 
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  • #2
There will always be a ## \phi ## component of ## I ## from the toroid, but if ##n=N/L ## is made very large, then the toroid current ## I ## can be minimized. This step is really necessary, because it really simplifies things.
 
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  • #3
It is just an approximation/simplification
 
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  • #4
Dale said:
It is just an approximation/simplification
Can you please make a figure and show me what really happens when we increase number of turns per unit length too much.
 
  • #5
Theoretically, we can also make the wire as fine as possible. The result is a current per unit length exactly how it is modeled, without any ## \phi ## component, with ##n=N/L \rightarrow + \infty ##, and ## I \rightarrow 0 ##, and finite current per unit length ##K=n I ##.
 
  • #6
Adesh said:
Can you please make a figure and show me what really happens when we increase number of turns per unit length too much.
I cannot, I am not that good of an artist.
 
  • #7
Charles Link said:
Theoretically, we can also make the wire as fine as possible. The result is a current per unit length exactly how it is modeled, without any ## \phi ## component, with ##n=N/L \rightarrow + \infty ##, and ## I \rightarrow 0 ##, and finite current per unit length ##K=n I ##.
I request you to please explain me how ##n \to \infty## implies ##I \to \infty##.
 
  • #8
In my view, trying to model the toroid without the triple integral (1 integral for each dimension), means that you're probably trying to model it as a disc -- that's possibly a mistake -- I'm presently too lazy to do the proof . . .
 
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  • #9
sysprog said:
means that you're probably trying to model it as a disc -- that's possibly a mistake --
How to picture it?

Well, I’m ready to do everything. Just tell me, just give me directions :-)
 
  • #10
Adesh said:
How to picture it?

Well, I’m ready to do everything. Just tell me, just give me directions :-)
Nice eager attitude, but are you ready for triple integrals? If you haven't yet become comfortable with single variable calculus, you probably aren't yet ready . . .
 
  • #11
sysprog said:
Nice eager attitude, but are you ready for triple integrals? If you haven't yet become comfortable with single variable calculus, you probably aren't yet ready . . .
Well, I know Multivariable Calculus. Once I got a problem about finding the volume between the intersection of three orthogonal cylinders which took me 2 weeks, finally I took help from @Charles Link for that.

So, in short I’m ready for the drill !
 
  • #12
This thread is a mess.

I refer to the first drawing. The other two seem unrelated to the problem.

There is no calculus required. There is no current in the phi direction because the problem defines it that way. Look at the drawing: the wires go up in z, in in r, down in z, and out in r. They do not go in phi at all. Since the wires don't go in phi, neither does the current.

If you want to argue that this is an approximation of a physical toroid, we can go down that route, but that is not the problem as asked.
 
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  • #13
Adesh said:
How to picture it?

Well, I’m ready to do everything. Just tell me, just give me directions :-)

Let's consider a single electron. Let's say it takes ##1s## to go round one loop of the wire. If there are ##n## loops of wire, then it takes ##n## seconds to travel all the way round the torus. The current in the ##r-z## circumferential direction is, therefore, ##n## times the current in the ##\phi## direction.

If ##n## is large enough, therefore, the ##\phi## current is negligible.

Now, time to move on to the problem at hand.
 
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  • #14
Vanadium 50 said:
This thread is a mess.

I refer to the first drawing. The other two seem unrelated to the problem.

There is no calculus required. There is no current in the phi direction because the problem defines it that way. Look at the drawing: the wires go up in z, in in r, down in z, and out in r. They do not go in phi at all. Since the wires don't go in phi, neither does the current.

If you want to argue that this is an approximation of a physical toroid, we can go down that route, but that is not the problem as asked.
But in the second figure (where the cross-sectional area is weird) I’m having trouble in seeing how it doesn’t have a phi component.

You have made it very clear to me about how phi component is zero in the first figure.
 
  • #15
I got your PM. I do not owe you an explanation when you demand it, and certainly not in less than 24 hours. I will not be participating in this thread any more.
 
  • #16
I intended to show how to determine the volume of each loop of the wire, along with the volume of the toroid, (and then take it from there to Maxwell's for the induction) -- that's where I was going with the triple integrals and other things -- but I was redirected by someone (@Vanadium 50, who knows his stuff and thousands of times has graciously shared his knowledge, thinking, and understanding) letting me know that all that wasn't necessary, and on reflection, I think that he was definitely right.
 
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  • #17
sysprog said:
I intended to show how to determine the volume of each loop of the wire, along with the volume of the toroid, (and then take it from there to Maxwell's for the induction) -- that's where I was going with the triple integrals and other things -- but I was redirected by someone (@Vanadium 50, who knows his stuff and thousands of times has graciously shared his knowledge, thinking, and understanding) letting me know that all that wasn't necessary, and on reflection, I think that he was definitely right.
Since he has sworn not to come back here, can we follow your route or any other body’s ?
 
  • #18
Adesh said:
Since he has sworn not to come back here, can we follow your route or any other body’s ?
@Vanadium 50 didn't swear to do or not do anything here; he just said he was done in this thread; he probably won't change his mind about that, but I think that he's at liberty to do so; as for the question, he already schooled me about that, so, no, I won't be playing math boy in this thread, but that's not a sworn statement.
 
  • #19
sysprog said:
@Vanadium 50 didn't swear to do or not do anything here; he just said he was done in this thread; he probably won't change his mind about that, but I think that he's at liberty to do so; as for the question, he already schooled me about that, so, no, I won't be playing math boy in this thread, but that's not a sworn statement.
All right, no problem!
 
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  • #20
Adesh said:
Since he has sworn not to come back here, can we follow your route or any other body’s ?
Is there another question about the problem? The matter of (approx) zero current in the ##\phi## direction must be settled now, surely?
 
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  • #21
PeroK said:
Is there another question about the problem? The matter of (approx) zero current in the ##\phi## direction must be settled now, surely?
In my opinion, yes, the answers of @Vanadium 50 settled that -- I wanted to hold forth about centroids and such, and he gently settled my hash . . .
 
  • #22
Is the issue that the windings have a slight (nonzero) pitch? Jeez, just wind two layers, one "around the donut" and one "back". Zero pitch.
 
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  • #23
PeroK said:
Is there another question about the problem? The matter of (approx) zero current in the ##\phi## direction must be settled now, surely?
Thanks for asking me :-)

Actually, if you look at that example, Griffiths is explaining everything with the help of that 2nd figure (that weird cross sectional area in the 1st post) so it was quite obvious for me to take his words “the ##\phi## component of the current is zero” for that 2nd figure.

I want to know is the ##\phi## current zero in that 2nd figure (of my first post)?
 
  • #24
hutchphd said:
Is the issue that the windings have a slight (nonzero) pitch? Jeez, just wind two layers, one "around the donut" and one "back". Zero pitch.
I want to know is the ##\phi## component of current is zero even in that 2nd figure?
 
  • #25
People sometimes use 'zero' when they mean 'negligibly different from zero'.
 
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  • #26
I suppose there is a reason for the weird cross-section shape ...maybe this is a fusion " tokamak' coil and you want some lensing? Some one must know (not me) . But it still clearly is toroidally wound.
 
  • #27
hutchphd said:
I suppose there is a reason for the weird cross-section shape ...maybe this is a fusion " tokamak' coil and you want some lensing? Some one must know (not me) . But it still clearly is toroidally wound.
When the cross section is rectangular I can see that ##\phi## current is zero.

But now let’s imagine, we got a slinky and we join it end to end to form a toroid, now since the cross-sectional area is circular then definitely we have a ##\phi## component. Even if the turns are so close to each other then also we have a circular cross-sectional area, and a circle’s line element does involve a ##\phi## component.
 
  • #28
Yes it likely will have a very very small phi component but it could be wound in such a way that it was actually zero in the unlikely event it really mattered. Nothing is ever perfectly exact!
 
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  • #29
hutchphd said:
Yes it likely will have a very very small phi component but it could be wound in such a way that it was actually zero in the unlikely event it really mattered. Nothing is ever perfectly exact!
Can you please draw a figure (or at least guide me) to show how it got a negligible ##\phi## component? I’m really unable to see it.
 
  • #30
Here are two figures that I tried to draw
F40E39CB-C547-40F8-8470-7CBB5898C0D4.jpeg


8D6B1A85-46F4-4B3F-A4A0-9B9102F4590E.jpeg
 
  • #31
Adesh said:
But in spite of all these I can't seem to understand how the current have no ##\phi## component. Well, in that toroid we have current going on in circles and really I don't anything more than this.

Please help me!
You can make the ##\phi## component of current = 0 by the way you wind the wire.

If you use just one winding then obviously there must be a ##\phi## component of current since each turn of wire is connected to the next by a short run of ##\phi## section to get around the toroid's periphery.

But if you do 2 windings (layers) the second winding's sections' current can cancel out the first if properly wound.
 
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  • #32
rude man said:
You can make the ##\phi## component of current = 0 by the way you wind the wire.

If you use just one winding then obviously there must be a ##\phi## component of current since each turn of wire is connected to the next by a short run of ##\phi## section to get around the toroid's periphery.

But if you do 2 windings (layers) the second winding's sections' current can cancel out the first if properly wound.
Yes, I got it thanks. You know my main problem was that I was unable to decompose a circle (that is the loop of a wire wound around the toroid) into just ##\hat{r}## and ##\hat{z}## components, because you know when we draw a circle in ##xy## plane we decompose it’s line elements into ##\hat{r}## and ##\hat{\phi}## directions (if we use polar coordinates) . So, it was hard for me to imagine such a situation where we had no ##\phi## component.
 
  • #33
Adesh said:
Yes, I got it thanks. You know my main problem was that I was unable to decompose a circle (that is the loop of a wire wound around the toroid) into just ##\hat{r}## and ##\hat{z}## components, because you know when we draw a circle in ##xy## plane we decompose it’s line elements into ##\hat{r}## and ##\hat{\phi}## directions (if we use polar coordinates) . So, it was hard for me to imagine such a situation where we had no ##\phi## component.
Right. The trick is two windings with the second nulling the ##\phi ## component from the first.
 
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FAQ: Why does the current have no ##\phi## component in a toroidal coil?

Why does the current have no ##\phi## component in a toroidal coil?

The current in a toroidal coil is induced by a changing magnetic field, which is perpendicular to the plane of the coil. Since the coil is circular in shape, the induced current flows in a circular path, parallel to the plane of the coil. This means that there is no component of the current in the ##\phi## direction, which is perpendicular to the plane of the coil.

How does the shape of a toroidal coil affect the direction of the induced current?

The shape of a toroidal coil is specifically designed to create a magnetic field that is perpendicular to the plane of the coil. This results in an induced current that flows in a circular path, parallel to the plane of the coil. Therefore, the shape of the coil plays a crucial role in determining the direction of the induced current.

Can a toroidal coil have a component of current in the ##\phi## direction?

No, a toroidal coil is designed to have a magnetic field that is perpendicular to the plane of the coil. This means that the induced current will always flow in a circular path, parallel to the plane of the coil, and there will be no component of current in the ##\phi## direction.

Why is it important for a toroidal coil to have no component of current in the ##\phi## direction?

The circular path of the induced current in a toroidal coil allows for a more efficient transfer of energy. This is because the magnetic field created by the current is confined to the inside of the coil, resulting in a stronger and more concentrated field. If there was a component of current in the ##\phi## direction, the magnetic field would not be as strong and the energy transfer would be less efficient.

Are there any real-life applications of toroidal coils with no ##\phi## component of current?

Yes, toroidal coils are commonly used in transformers, which are essential components in many electrical devices. The circular path of the induced current in a toroidal coil allows for a more efficient transfer of energy, making it a popular choice for use in transformers. Additionally, toroidal coils are also used in particle accelerators to create strong magnetic fields that are necessary for accelerating particles.

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