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Solenoid problemby Cyrus
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#1
Nov2004, 08:09 PM

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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 dont move away from it. Also, wont 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 theres 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. 


#2
Nov2004, 08:17 PM

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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 BiotSavart yourself and see.



#3
Nov2004, 08:46 PM

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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
Nov2004, 10:48 PM

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Solenoid problem



#5
Nov2104, 02:35 AM

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#6
Nov2104, 02:01 PM

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You lost me there tide, im talking about inside the loop. If im 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. 


#7
Nov2104, 04:05 PM

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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 BiotSavart:
[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.) BiotSavart 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 


#8
Nov2104, 04:49 PM

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Cyrus,
1/r is not constant throughout the region and Cepheid did a nice job of providing you with the details. 


#9
Nov2104, 04:58 PM

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thanks tide and cepheid. Lately my brain has been mush. I cant wait for school to be over.



#10
Nov2104, 06:56 PM

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I hear ya 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
Nov2104, 07:25 PM

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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
Nov2104, 07:42 PM

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cepheid  Unfortunately you neglected to take the vector aspect of the ss' 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
Nov2104, 07:44 PM

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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
Nov2104, 11:57 PM

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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
Nov2204, 02:25 AM

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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
Nov2204, 05:52 PM

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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? Lets 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 im not studying physics, although I wish I were, im studying mechanical engineering , its not nearly as neat, but I dont want to start a new major and have to toss out all the work I already did.) 


#17
Nov2204, 10:24 PM

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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
Nov2204, 10:43 PM

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Below is a link to solved problems of this nature. Look especially at Chapter 3. http://www.sfu.ca/physics/people/fac...n/outline.html 


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