Solenoid Problem: B Field Constant on Plane of Loop

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In summary, the B field is not constant at all points within the loop, but it is uniform for any point located on the circumference of the loop.
  • #1
Cyrus
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Lets say there is a single loop of wire. Is the B field at exactly of the plane of the loop going to all be perpendicular and of constant magnitude, at exactly the plane of the loop. I know this is true for the center. Doesnt it also hold for points off center of the loops axis, as long as we stay on the plane of the loop and don't move away from it. Also, won't the magnitude of B be the same at all points on the plane inside the loop as well?

I thought of this as follows. If we do the right hand rule at all points on the plane, then the B field is going to be perpendicular to the loop, becuase there's nothing that will make it angle from being perpendicular. The current I is constant so we can factor that out. So any point off center will be the sum of all the R's times the I around the loop. But that always works out to the area of the circle times I. So the B field should be uniform, and constant, when looking at any point contained in the plane of the loop.
 
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  • #2
No, the magnitude of B will not be the same everywhere in the plane. Obviously, the direction of the field inside and outside the loop will be opposite and, as you go to infinity, the magnetic field will go to zero so the field cannot be constant in magnitude. Try Biot-Savart yourself and see.
 
  • #3
Only a solenoid of essentially infinite length and large number of turns compared with its radius would have a constant B field inside. The B field will be perpendicular to the plane of the loop (but either positive or negative direction depending on whether you're looking at the field inside or outside) though if I understand the setup correctly but it is definitely not constant.
 
  • #4
Tide said:
No, the magnitude of B will not be the same everywhere in the plane. Obviously, the direction of the field inside and outside the loop will be opposite and, as you go to infinity, the magnetic field will go to zero so the field cannot be constant in magnitude. Try Biot-Savart yourself and see.

Ok, let me restate what I said to be clearer. Let's say you have a loop with a current in it. What I mean is any point within the loop. So if the loop is on the x-y plane, i mean all points in the x-y plane that are within the loop. (No z-component to it. )
 
  • #5
cyrusabdollahi said:
Ok, let me restate what I said to be clearer. Let's say you have a loop with a current in it. What I mean is any point within the loop. So if the loop is on the x-y plane, i mean all points in the x-y plane that are within the loop. (No z-component to it. )

The answer is the same - Biot-Savart will show you the magnitude of the field is not constant. The only way to get it uniform, as Vsage already suggested, is if you have an infinitely long solenoid. In fact, for a single loop you will end up with VERY large fields in the vicinity of the loop.
 
  • #6
You lost me there tide, I am talking about inside the loop. If I am inside the loop, then it says db= mu I dl x r / 4 pi r^2.

the integral of dl will be the circumference of the circle, no matter what point i am inside the loop. Similarly, the integral of the the 1/r will be the same if I am at the center of the current loop or off center of the current loop.
 
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  • #7
The magnetic field will most definitely have [itex] r [/itex] dependence, where [itex] r [/itex] is the distance between the source point (on the circumference) and the field point (somewhere within or outside the loop, but constrained to lie in the plane of the loop, as you said). So, let's set up a cylindrical coordinate system [itex] (s, \phi, z) [/itex], with s being the radial coordinate. From Biot-Savart:

[tex] \vec{B}(\vec{r}) = \frac{\mu_0 I}{4 \pi} \int{\frac{d\vec{l'} \times \hat{r}}{r^2}} [/tex]

In fact, since all field points at which we are evaluating the magnetic field are constrained to lie in the plane of the loop, B is at most a function of two of the position coordinates:

[tex] \vec{B}(s, \phi) = \frac{\mu_0 I}{4 \pi} \int{\frac{d\vec{l'} \times \hat{r}}{r^2}} [/tex]

Furthermore, we can see that B has no phi dependence, since any two points within the loop the same distance from the centre (which we'll call the origin) are also the same distance (in the radial direction) from the loop, and in the above expression, B depends only on r. What the hell is r? Well, r is the separation between source point and field point...so since I'm using primed coordinates to designate source coordinates (like dl'), we can express this separation as follows: a position vector of a source point on the circumerence is [itex] \vec{s'} [/itex]. A position vector of a field point is given by [itex] \vec{s} [/itex]. For points within the loop, s < s', so the separation between source point and field point is given by:

[tex] \vec{s} - \vec{s'} = \vec{r} [/tex]

which points inward (i.e in the [itex] -\hat{s} [/itex] direction.) Biot-Savart becomes:

[tex] \vec{B}(s) = \frac{\mu_0 I}{4 \pi} \int{\frac{d\vec{l'} \times -\hat{s}}{|\vec{s} - \vec{s'}|^2}} [/tex]

A small bit of the circumference:
[tex] d\vec{l'} = (dl') \hat{\phi} = (s'd\phi ') \hat{\phi} [/tex]

[tex] \vec{B}(s) = \frac{\mu_0 I}{4 \pi} \int{\frac{(s'd\phi ') \hat{\phi} \times -\hat{s}}{|\vec{s} - \vec{s'}|^2}} [/tex]

note: [itex] \hat{\phi} \times -\hat{s} = \hat{z} [/tex]

[tex] \vec{B}(s) = \frac{\mu_0 I}{4 \pi} (\hat{z}) \int{\frac{(s'd\phi ') }{|\vec{s} - \vec{s'}|^2}} [/tex]

[tex] = \frac{\mu_0 Is'}{4 \pi (s - s')^2} \hat{z} \int_0^{2\pi}{d\phi ' } [/tex]

[tex] = \frac{\mu_0 I(2\pi s')}{4 \pi (s - s')^2} \hat{z} [/tex]

[tex] = \frac{\mu_0 Is'}{2(s - s')^2} \hat{z} [/tex]

I hope that's right :redface:
 
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  • #8
Cyrus,

1/r is not constant throughout the region and Cepheid did a nice job of providing you with the details.
 
  • #9
thanks tide and cepheid. Lately my brain has been mush. I can't wait for school to be over.
 
  • #10
I hear you man!...but there's still final exams... :(

Yeah, I went through all that stuff because I'm taking E&M so I figured I ought to be able to do it in excruciating detail.
 
  • #11
Cepheid, when looking at points that lie on the plane inside the loop, is the B field perpendicualr to the plane of the loop at all points? Its just that the magnitude is not constant, or is both the mag and direc change?
 
  • #12
cepheid -- Unfortunately you neglected to take the vector aspect of the s--s' distance.

|s - s'|**2 = s**2 + s'**2 -2*s*s'*cos(phi)

It turns out it is easier to evaluate the integral for the vector potential, which boils down to


INTEGRAL (0 TO 2PI) OVER PHI{ COS(PHI)/( s**2 + s'**2 -2*s*s'*cos(phi))**1/2



This turns out to be an Elliptic Integral. Elliptic functions are hard. Many people, in the physics world, don't much like Elliptic Integrals.

It's only at the center that the contributions from the points in the wire totally cancel each other. Move into the interior and that circular symmetry is broken. And, the sizes of the point contributions vary. If you draw a picture or two, you'll see that the B field is not uniform inside the loop.

Regards,
Reilly Atkinson
 
  • #13
I forgot to mention that this problem is in Jackson, Sec. 5.5, p141. My edition is somewhat aged, so newer editions may cover this material in a different plae. RA
 
  • #14
cyrusabdollahi said:
..when looking at points that lie on the plane inside the loop, is the B field perpendicualr to the plane of the loop at all points? Its just that the magnitude is not constant, or is both the mag and direc change?
Yes: in the plane of the loop, the magnetic field is perpendicular to the plane of the loop. I believe you made the correct symmetry argument for this in your first post. (BTW, making symmetry arguments like this is a very important skill for a physicist to have.)

An easy way to see that the magnitude varies (without doing any integrals) is to imagine being very very close to the wire of the loop. When this close, the wire looks like a straight line and the field is [itex]\mu_0I/(2\pi s)[/itex] where s is the distance to the wire, and s<<R where R is the loop radius. But at the centre, the field is [itex]\mu_0I/(2R)[/itex]. These 2 are definitely not the same, in fact, for an infinitely thin wire, the former field goes to infinity as s -> 0.
 
  • #15
reilly said:
If you draw a picture or two, you'll see that the B field is not uniform inside the loop.

Regards,
Reilly Atkinson

Hmmm...but I did draw a picture or two...and I already knew that the B field inside the loop was non-uniform...that's what we were all trying to explain to cyrus! I stated so in my post...and then attempted to prove it. Well, I thought I had succeeded, because I got B as a function of s.

It wouldn't surprise me if I had made some mistake somewhere...but I don't understand what you mean by your correction:

|s - s'|**2 = s**2 + s'**2 -2*s*s'*cos(phi)

I guess by ** you mean 'to the power of' (in this context, I don't see what else it could be). Where does that term involving 2ss'cos(phi) come in? s and s' for me lay along the same line...radial.
 
  • #16
I have a follow up question krab. If the field is always perpendicular at the plane, then that means that one coil is good enough to be a solenoid, (except for the fact that the field is not constant.). The book says that you need many many coils to be a good solenoid. But it seems the just one single coil is good enough to give you a uniform B field, in terms of direction, not in magnitude.

So how come it becomes uniform in both magnitude AND direction as you have many of these coils? Let's say the center of the axis of the loop is where the fields strongest, hypothetically. The more coils I add, the stronger and stronger the B field should be along the axis, no? It would seem that the strongest location just gets proportionally stronger realtive to the weaker areas. How does it manage to be uniform everywhere in magnitude as well as this thing gets longer and longer?

(oh, and I am not studying physics, although I wish I were, I am studying mechanical engineering :frown:, its not nearly as neat, but I don't want to start a new major and have to toss out all the work I already did.)
 
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  • #17
cepheid: Reilly is right. You started off solving the problem correctly, but then it went off the rails. Your final answer appears to depend on s'. How could it? Primed variables refer to the location of dl, and are therefore integrated out.
 
  • #18
cyrusabdollahi said:
I have a follow up question krab. If the field is always perpendicular at the plane, then that means that one coil is good enough to be a solenoid, (except for the fact that the field is not constant.). The book says that you need many many coils to be a good solenoid. But it seems the just one single coil is good enough to give you a uniform B field, in terms of direction, not in magnitude.
"Uniform" usually refers to magnitude rather than direction.

So how come it becomes uniform in both magnitude AND direction as you have many of these coils? Let's say the center of the axis of the loop is where the fields strongest, hypothetically.
as I pointed out, the field is weakest at the centre.
The more coils I add, the stronger and stronger the B field should be along the axis, no? It would seem that the strongest location just gets proportionally stronger realtive to the weaker areas. How does it manage to be uniform everywhere in magnitude as well as this thing gets longer and longer?
The field gets more and more uniform as you add loops. Of course if each loop is an infinitesimal filament, the field is still very large right next to this filament. However, the limiting case of a solenoid is not in fact an infinite number of loops each with current I (that would give an infinite field everywhere inside), but it is actually a current sheet with a given current per unit length along its length.

Below is a link to solved problems of this nature. Look especially at Chapter 3.

http://www.sfu.ca/physics/people/faculty/cochran/outline.html [Broken]
 
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  • #19
krab said:
cepheid: Reilly is right. You started off solving the problem correctly, but then it went off the rails. Your final answer appears to depend on s'. How could it? Primed variables refer to the location of dl, and are therefore integrated out.

Cepheid's s' stands for the radius of the loop (i.e. it doesn't integrate out) and s stands for the distance of a point from the center. However, I am perplexed that he properly evaluated the integral but seems not to have used the [itex]\cos \phi[/itex] term in the denominator (in which case he would have gotten a different and incorrect result!). Cepheid? 'Splain yourself! :-)
 
  • #20
Could you explain what you mean by more and more uniform please.

If the axis of the loop is the weakest location of the B field, then all the other points have a stronger B field. Adding n loops has the same effect as increasing the amount of B fields proportionally. So if one loop has a value of 'a' at the axis, and '3a' at some point off center, when i add n loops, won't the two respective points be 'na' and '3na'. How will the 'na' term possibly catch up with the '3na' to be uniform. so that it becomes some contant B field of a value, say 'Ca', where C is a constant number at all points off center or on the center of the loop axis. If n is infinity, then I can see that the B field would equal to 'infinity*a' and 'infinity*3a' which is equivalent. But for any reasonable length, I can't see this working out as a good approximation. It seems that infinity is the only number that satisfies this condition.
 
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  • #21
Tide said:
Cepheid's s' stands for the radius of the loop (i.e. it doesn't integrate out) and s stands for the distance of a point from the center. However, I am perplexed that he properly evaluated the integral but seems not to have used the [itex]\cos \phi[/itex] term in the denominator (in which case he would have gotten a different and incorrect result!). Cepheid? 'Splain yourself! :-)

Huh? Sorry...I can't. I don't know where my error is, if any. Could YOU please explain it Tide... :confused:
 
  • #22
The correct expression involves elliptic integrals and can be found for example here:

http://www-mucool.fnal.gov/mcnotes/public/pdf/muc0281/muc0281.pdf [Broken]

As Reily said, the |s-s'|^2 in the denominator is

[tex]s^2+s'^2-2ss'\cos(\phi-\phi')[/tex] (don't you know the cosine rule?)
and you are integrating over [itex]\phi'[/itex]
 
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  • #23
Yes, I know the cosine rule :rolleyes: ...to me the question seems a little bit..insulting. I'm sorry, but I don't know how to interpret it any other way. As I mentioned before, my s' was just the radius of the loop, and there was never any angle between it and s! So I guess I set up the problem incorrectly in the first place...but it's hard to show you what I did...without a picture. Sheesh...I've never heard of elliptic integrals before, so I wasn't able to follow the info from that link very well.
 
  • #24
If [itex]\vec \rho[/itex] is a point on the loop and [itex]\vec \rho '[/itex] is a point at which you want to calculate the field then [itex]|\vec \rho - \vec \rho '|^2 = \rho^2 + {\rho '}^2 - 2 \vec \rho \cdot \vec \rho '[/itex].

[itex]\vec \rho \cdot \rho '[/itex] is just [itex]\rho \rho ' \cos (\phi - \phi ')[/itex] where [itex]\phi[/itex] and [itex]\phi '[/itex] are the respective polar angles.

Your other problem was the [itex]d \vec l \times \hat r[/itex]. The unit vector [itex]\hat r[/itex] is a unit vector from [itex]\vec \rho[/itex] to [itex]\vec \rho '[/itex]. When you put it all together you get the result Krab indicated (I had originally only checked your integration which looked okay but you hadn't set up the intergrand properly.)
 
  • #25
Looks like Krab beat me to it! Too much multitasking here! ;-)
 
  • #26
Ill post my follow up again, I think it took backseat to cepheids proof but that's ok.

Could you explain what you mean by more and more uniform please.

If the axis of the loop is the weakest location of the B field, then all the other points have a stronger B field. Adding n loops has the same effect as increasing the amount of B fields proportionally. So if one loop has a value of 'a' at the axis, and '3a' at some point off center, when i add n loops, won't the two respective points be 'na' and '3na'. How will the 'na' term possibly catch up with the '3na' to be uniform. so that it becomes some contant B field of a value, say 'Ca', where C is a constant number at all points off center or on the center of the loop axis. If n is infinity, then I can see that the B field would equal to 'infinity*a' and 'infinity*3a' which is equivalent. But for any reasonable length, I can't see this working out as a good approximation. It seems that infinity is the only number that satisfies this condition.
 
  • #27
cyrusabdollahi said:
If the axis of the loop is the weakest location of the B field, then all the other points have a stronger B field. Adding n loops has the same effect as increasing the amount of B fields proportionally. So if one loop has a value of 'a' at the axis, and '3a' at some point off center, when i add n loops, won't the two respective points be 'na' and '3na'.
No, because the added loops are not all in one plane, and the way they add contributes more to the on-axis than to the off-axis. This is because on axis, the contributions are all in one direction, while off axis, they are not, so the axial component of field falls off more quickly with distance away from the plane of the loop.
 
  • #28
cepheid said:
Yes, I know the cosine rule :rolleyes: ...to me the question seems a little bit..insulting. I'm sorry, but I don't know how to interpret it any other way. As I mentioned before, my s' was just the radius of the loop, and there was never any angle between it and s! So I guess I set up the problem incorrectly in the first place...but it's hard to show you what I did...without a picture. Sheesh...I've never heard of elliptic integrals before, so I wasn't able to follow the info from that link very well.
You could interpret it this way: I did not know what level you are at, and wanted to find out. If my words were clumsy, I apologize.

The thing in the denominator is the squared distance from the line element of the loop to the observation point. Unless you are at the centre, this distance varies as you integrate around the loop with your line element. Since it varies, you cannot take this quantity outside the integral as you did.
 
  • #29
krab and Tide...I learned something from my mistake...thanks very much for the explanations...this problem is not as simple as I thought. Is there no way to use some other method? I thought about Ampere's law...but it didn't seem useful...amperian loops in the plane don't enclose a current, and B is nonuniform anyways, so there's no simplification.

Edit: In the original integral, here is my 'reasoning' for always assuming the two vectors lay on the same radial line: (I will use Tide's notation).

So if the field point rho prime lies somewhere within the loop, and the source point we are considering rho are located by vectors [itex] \mathbf{\rho'} [/itex] and [itex] \mathbf{\rho} [/itex], and there is some angle between these vectors...wouldn't the magnetic field due to another source point on the loop whose vector was also the same angle away from [itex] \mathbf{\rho'} [/itex], but in the opposite direction just cancel out any phi component of the magnetic field at that point due to the first source point? Wouldn't this be true for any field point? So why can't you always consider the field at a point only due to the source point on the loop that lies along a radial line from the centre to rho' to rho? Am I making sense?
 
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  • #30
krab said:
No, because the added loops are not all in one plane, and the way they add contributes more to the on-axis than to the off-axis. This is because on axis, the contributions are all in one direction, while off axis, they are not, so the axial component of field falls off more quickly with distance away from the plane of the loop.

I was assuming that because the B fields are perpendicular at all points on the plane of a loop, adding many loops would just be like multiplying the loop by some number n.

I think I see what your saying now, but to be clear Ill explain my new interpretation of the solenoid, and you can fix any mistakes I make.

Lets break the solenoid down into individual loops. For the first loop, we say that the B field is not uniform, but it does have the same direction everywhere in the plane of the loop.

Now let take away this loop. And put a new loop just next to where the old one was in space. This new loop will cause a B field which are circles. Again, this loop has a perpendicular B field just at the plane of the loop that is all in the same direction. But this, all perpendicular B field exists at the plane of the loop, not at the points inside the plane of the first loop. This is why they don't just add directly. So if we move away from the second loop, we start to see the B field curve with the loops the B field makes. So when we arrive at the points inside the plane of the first loop, its these curved B fields that will contribute. The exception is the on axis B field, that one stays on axis no matter how far away you are, so its magnitude at that point is greater than all the rest, becuase the really strong B fields will be curved the most, so their component along the length of the solenoid will actually be less than the weakest B field along the axis.

I remember you said that the B field very near the wire should be huge. But at the same token, the circles radius would be the tightest, so its component along the axis would be the smallest of them all.

So far, I only have only added one other loop to my origional first loop. So it seems that at the first loop, the B field due to the second loops will cause the B field at the plane of the first loop to no longer be all perpendicular. But to remidy this, I have to place another loop on the other side. That way, the components that are nonaxial will cancel out.

But what happens to the loops that are not at the center. Say we look at a loop at the very end of the solenoid. In that case there is only loops to its one side. so there's nothing on its other side to make the field uniform in direction. So let's say this is the last loop on the left side of the solenoid. That means there are only loops to the right of this last loop. And each one will contribute a B field at the location of the last loop. It would seem that the B field should be horribly nonuniform in direction at this end loop. It just gets worse as you add more loops, because there is no component to cancel out the nonaxial B field, so the nonaxial component linger and sums up as more and more loops are added to the other end.
 

1. What is a solenoid?

A solenoid is a long, cylindrical coil of wire that produces a magnetic field when an electric current is passed through it.

2. What is the "B field constant" in relation to a solenoid problem?

The "B field constant" refers to the strength or magnitude of the magnetic field produced by the solenoid. It is a measure of the intensity of the magnetic field.

3. What is the "plane of loop" in a solenoid problem?

The "plane of loop" refers to the imaginary flat surface that is perpendicular to the axis of the solenoid and contains the wire loop. It is used to calculate the magnetic field strength at a specific point.

4. How is the B field constant affected by the plane of loop in a solenoid problem?

The B field constant is directly proportional to the number of turns in the solenoid and the current passing through it. It is also affected by the distance from the center of the solenoid to the plane of loop, with the field strength decreasing as the distance increases.

5. What are some practical applications of solenoids?

Solenoids have many practical applications, including as electromagnets in electric motors, generators, and speakers. They are also used in various industries for tasks such as controlling valves, locking mechanisms, and sorting objects.

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