How does the magnetic field strength decay(radially) inside a solenoid coil?


by merper
Tags: coil, decayradially, field, inside, magnetic, solenoid, strength
merper
merper is offline
#1
Oct25-11, 02:07 PM
P: 5
I'm looking at redesigning a hydraulic solenoid valve with the aim of reducing the package size. The solenoid coil is by far the largest component of the device, with about 4000 coils arrayed in about 90 layers stacked on top of each other. I'm wondering if I can reduce the size by removing some of these layers of coils and pumping an increased current through the rest.

However, given the 90 layers of coils there is a difference of about 30% in the radius of the coils in the inner layer vs the coils in the outermost layer. I am having trouble understanding how (or if ) the coils effect on the moving solenoid core varies with radial distance.

My intuition is that the magnetic field generated by current passing through a coil in an outer layer is going to have less of an impact on the solenoid core than a coil in the innermost layer. In addition, the outermost coils have a larger circumference and thus higher coil resistance. However, none of the derivations i've seen of magnetic field within a solenoid seem to talk at all about the path length of the magnetic circuit induced by the coils. The closest I've come to an understanding is via this site, which indicates that I could rewrite the magnetic circuit with the reluctances of the moving core, the copper wiring of interior or exterior current loops, and a small air gap in series, but i have no idea what loop size I'm dealing with.

Can anyone help me understand what's going on qualitatively? Or better yet, mathemetically?
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Termotanque
Termotanque is offline
#2
Oct25-11, 06:21 PM
P: 34
I don't know your specific application, but as a first approximation we can suppose that the inner radius a, and the outer radius b are both much smaller than the length L of the coil.

In this case, the magnetic field due to one layer is B = u_0 i N / L, where u_0 is the vacuum permeability, i the current in one wire, N the number of turns of that layer, and L the length of the coil. B is directed axially.

Note that under these assumptions, B is essentially uniform (does not depend on the distance r from the axis).

Multiplying this by the number of layers, you get the B due to your coil. This tells you that under the assumptions above, mainly: the coil is long (and radii are short), the turns are closely wound, and carries a constant current i, you can halve the layers and double your current.

If some of these assumptions can't hold for some reason, the problem can be solved but takes a little more work.
Termotanque
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#3
Oct25-11, 06:26 PM
P: 34
Also, the current necessarily is the same for every turn. The resistance is characteristic of the coil as a whole, and that fixes the current. The layers, if you want to see it this way, are connected in series.

merper
merper is offline
#4
Oct26-11, 09:40 AM
P: 5

How does the magnetic field strength decay(radially) inside a solenoid coil?


Unfortunately, the length and diameters are the same order of magnitude. The outer layer of the coil is actually the same diameter as the length of the coil(~2in) and the inner diameter is maybe 2/3rds that.

I agree that the current is constant. I was just noticing I could increase the current in fewer coils to achieve the same magnetomotive force without increasing the power by the same current multiplier ratio - since the same current in a wider outer coil consumes more power.

Anyway, it's the whole breakdown of the L>>>r assumption that I can't figure out. If you have any ideas on how I can go about solving this, I would really appreciate it.

Thanks!
Philip Wood
Philip Wood is offline
#5
Oct26-11, 12:43 PM
P: 861
Off-axis fields in coils of circular cross-section are devils to calculate (except in the case of a solenoid much longer than its diameter). What you could treat by elementary methods is a coil of square cross-section, by adding together (vectorially) the fields due to the straight bits of which it is made. [You'd treat the sums as integrals.] This is clearly a somewhat tedious job, but it is do-able, and will give you some idea of what happens for a circular cross-section. When I've got spare hour...
Termotanque
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#6
Oct26-11, 02:48 PM
P: 34
You could calculate the magnetic field B due to one layer of a short coil at any point along its axis, but I'll write the expression for B exactly in the middle (L/2) so that it's simple to explain the formula.

At half the length of the coil from each end, on its axis, the field is B(r) = u_0*i*N / 2sqrt( (L/2)^2 + r^2). Where N is the number of turns (assumed tight) in *one* layer, i is the current, L is the length, and r is the radial distance from the axis to the layer. You can see that now, B does depend on r. You can check that if r << L, we recover the expression u_0*i*N/2sqrt(L^2 / 4) = u_0*i*N/L, which is exactly the formula for the long coil.

Since there are a finite number of layers, you cannot in general find a closed formula for the total magnetic field, and so you have to manually add the B due to each of them, using their corresponding r. Remember to use the correct N, and not the total number of turns.

Finally, this formula does in fact tell you that the outermost layers produce less field along the axis.
merper
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#7
Oct26-11, 03:22 PM
P: 5
Termo, that's exactly the sort of formula I'm looking for. Cheers! However, I am a bit curious as to where it came from. Like Philip said, the exact derivation does sound extremely tedious. On the other hand, the 1/r decay does seem appropriate since the field is axially confined.
Termotanque
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#8
Oct26-11, 03:34 PM
P: 34
It's not tedious at all. It's one of the prototypical problems arising when one studies electrodynamics. A good book to study this is Griffiths's Introduction to Electrodynamics.

Basically what you do is use Biot-Savart law applied to a ring of infinitesimal length dx, carrying an infinitesimal current di = Ni/L dx. That's why assuming that the turns are tight is crucial. You calculate the infinitesimal magnetic field dB due to that ring, and then integrate along its length using some symmetry, to get B due to one coil. Then multiply as needed.
Philip Wood
Philip Wood is offline
#9
Oct26-11, 04:07 PM
P: 861
Termotanque. I was talking about off-axis points (which I thought Merper wanted to know about). Forgive me if I've misread, but you seem to be dealing with points on the axis, which, I agree, are perfectly straightforward.
merper
merper is offline
#10
Oct26-11, 05:13 PM
P: 5
Ok, I'm not sure I understand the implications of this "tight ring" assumption. Is that to say that the cross sectional area of each ring is small compared to the length of the overall solenoid?

I guess my topic title left my overall question a bit ambiguous, but my end goal is not learning how the B field varies inside the solenoid(though I'm always up for learning a bit more about magnetics), but how the radius of the loop will affect its contribution to the overall magnetic field acting on the center where the moving core is contained. I believe B(r) = u_0*i*N / 2sqrt( (L/2)^2 + r^2) does address this?
Termotanque
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#11
Oct26-11, 08:26 PM
P: 34
It means that the pitch of the coil is small enough that the current can be regarded as flowing through a conducting sheet, instead of discrete wires.
merper
merper is offline
#12
Oct27-11, 09:51 AM
P: 5
So radius of the conducting wire <<<< axial length of solenoid ?


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