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Winding a coil for maximum magnetic field

Thread starter #1
I am trying to create as high a magnetic field as I can in a solenoid, but have read and seen in pictures alot of doubling up of the wrapping... as in having multiple layers of wire one on top of one another.....

Is this recommended to get more loops in a smaller space or is this only used for other applications.?

Also, if it is recommended to have multiple layers to get more loops in a smaller space... must you make sure the direction of the wrapping is always constant? I.e begin wrapping left to right and complete a layer then back up again all the way to the left for another layer? or Just try to get as many layers as possible in the left side and begin moving towards the right as you go?

Thanks
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#2
shrewd said:
I am trying to create as high a magnetic field as I can in a solenoid, but have read and seen in pictures alot of doubling up of the wrapping... as in having multiple layers of wire one on top of one another.....

Is this recommended to get more loops in a smaller space or is this only used for other applications.?

Also, if it is recommended to have multiple layers to get more loops in a smaller space... must you make sure the direction of the wrapping is always constant? I.e begin wrapping left to right and complete a layer then back up again all the way to the left for another layer? or Just try to get as many layers as possible in the left side and begin moving towards the right as you go?
You just wind from end to end, going back to the same end each layer would be very wasteful, and probably perform more poorly. You have to have multiple layers to get enough turns in the space available.
 
#3
i found that winding the backwards and forwards tends to counteract itself so i would always start at yhe same end with wire trailing back to the start each time
 
#4
I am trying to create as high a magnetic field as I can in a solenoid...

Does this have a ferromagnetic core or not? If it does have a core, then you have a limit as these cores saturate. Once that happens, any additional coil current serves no further purpose other than to make the wire hotter.

The formulae you need are:

|H|= (NI)/L Where:

H: Magnetic Field Intensity (Amp * Turns/meter)
N: No. of Turns
I: Current (Amps)
L: Solenoid length (meters)

B= uH Where:

B: Flux density (Wb/sq. meter)
u: u(0)u(r)
u(0)= 4pi X 10^-7 (Permeablility of free space) (Henry/m)
u(r): Relative permeability of core material (dimensionless)
H: Magnetic field intensity (Amp * Turns/m)

For silicon steel core material, B(sat)= 1.6 Wb/m^2 is a nominal value. Ferrite materials have B(sat)'s of 0.2 -- 0.6 Wb/m^2 for low frequency (< 1.0MHz) cores.
 
Thread starter #5
It will use an air core.

I just need a couple tenths of a tesla to 1 tesla.

Actually I would be satisfied and be able to do this experiment with only .1 tesla with an air core.

Just to check, since I have been only using the formulas you just gave for the idea... I have not found anywhere in any books where the diameter of the coild plays a significant role in the strength of the field... It seems like it should but i am uncertain on this. Anyway I have a 4.5 inch air core.

The problems I am running into conceptually is the wire (for amperage) sizing compared to the number of turns you can put in the space.. of course thats always gonna be the comprimise you hit unless you can refrigerate the environment... but this is a poor mans experiment so if I see the results I think I will I will invest more afterward.

So far I am looking at 1000 turns of 10 gauge magnetic wire for 1 meter. (gonna be pretty expensive) and 10 amps current... If I went with lower amperage I may be able to come up with the same for less money in wire but a ton more windings... I will need to look a this more...
 
#6
I have not found anywhere in any books where the diameter of the coild plays a significant role in the strength of the field... It seems like it should but i am uncertain on this.

That's because coil diameter plays no part in that. Coil diameter determines the X-Section area, which, in turn, determines the magnetic flux. You would use magnetic flux in a Faraday's Law calculation, or to calculate inductance (two different ways to do the same thing). Magnetic field intensity (H) is dependent on the current, length, and number of turns.

So far I am looking at 1000 turns of 10 gauge magnetic wire for 1 meter. (gonna be pretty expensive) and 10 amps current... If I went with lower amperage I may be able to come up with the same for less money in wire but a ton more windings... I will need to look a this more...

So far, here's what you have:

D= 114.3E-3 meters
l= 1.0 meters
N= 1000
I= 10.0A

H= (1000 * 10)/1.0= 10E3 Amp * Turn/meter
B= 10E3 * (4piE-7)= 12.566E-3 Wb/m^2

So far, you're coming up way short of what you want. For this coil, you'd need 79.577A to get your minimum required 0.1 Wb/m^2.

However, there is another solution. Since you specify a maximum of 1.0 Wb/m^2, why not use an iron core? If the u(r) of your core is ~80 (not too demanding) you will get the flux density you want.

What, exactly, are you trying to do here?
 
Thread starter #7
Perhaps I am miscalculating but doesn't more wrappings as in layers just increase the amount of loops per meter?

And your right looks like I lost an order of magnitude... bleh. I love how in programming languages the ^ symbol doesn't always work and Math.pow must be used instead.... bleh.

Its basically a sort fo gun like thing, but instead of using an induced magnetic field... I will be using an electronically switched uniform field... but thats far away... for now I am just trying to construct .. a field and work with the difficulties it has.
 
#8
:lol:
Increase the number of turns as much you can and current as much you can for high magnetic field. Also in case of an air core coil, the coil geometry is very important. The magnetic field can be approximed as uniform distributed on inside section of coil only if the coil lenght is much higher than coil diameter, otherwise is not uniform. But the field is stronger in coil plan if you use a short coil than a long one, but in case of a short coil not have so mach space for winding. For compute the magnetic field in a stationary case (no field variation) the Ampere theorem is appropriate, but drive at very dificult math in case of ununiform field. But the simples way is to try.
 
#9
Here's what I have so far:

B= 0.1 Wb/m^2 and #10 wire.

The current limit for #10 (AWS) is 32.5A. Call that 32.0A. Let's also increase the length to 1.14m (length= 10X diameter).

B= 0.1 Wb/m^2
H= B/u(0)= 0.1/(4piE-7)= 79.577E3 (Call this 80E3) (Amp * Turns/m)
H= (NI)/l

NI= Hl= 80E3 X 1.14= 91.2E3 (Amp * Turns)
N= 91.2E3/32= 2850 turns.

Running this through my coil calculator gives:

Wire Length: 3776.9 feet
Wire Weight: 118.7 pounds
R= 3.19 Ohms
Turns per Layer: 440
Layers: 7
Finished Diameter: 5.93 inches.
 

williB

New Member
#10
Miles Prower said:
Here's what I have so far:

B= 0.1 Wb/m^2 and #10 wire.

The current limit for #10 (AWS) is 32.5A. Call that 32.0A. Let's also increase the length to 1.14m (length= 10X diameter).

B= 0.1 Wb/m^2
H= B/u(0)= 0.1/(4piE-7)= 79.577E3 (Call this 80E3) (Amp * Turns/m)
H= (NI)/l

NI= Hl= 80E3 X 1.14= 91.2E3 (Amp * Turns)
N= 91.2E3/32= 2850 turns.

Running this through my coil calculator gives:

Wire Length: 3776.9 feet
Wire Weight: 118.7 pounds
R= 3.19 Ohms
Turns per Layer: 440
Layers: 7
Finished Diameter: 5.93 inches.
nice job..! i think.because i'm not following where you got the 80,000??
but have you considered a short pulse (one second)of maybe 100A..
because i am assuming the the 32A is a sustained current, while the wire can handle a much larger short pulse of current..without overheating the wire..
 
#11
nice job..! i think.because i'm not following where you got the 80,000??

By rounding up. Originally, he wanted 0.1 -- 1.0 Wb/m^2. So I rounded up to get a bit more than the specified minimum.

but have you considered a short pulse (one second) of maybe 100A..

So far, he's been saying that he wants it to produce the magnetic field in steady state form. As for further developments, we'll see what he needs later. Though, yeah, it could handle larger currents of shorter duration.
 
Thread starter #12
Miles Prower said:
nice job..! i think.because i'm not following where you got the 80,000??

By rounding up. Originally, he wanted 0.1 -- 1.0 Wb/m^2. So I rounded up to get a bit more than the specified minimum.

but have you considered a short pulse (one second) of maybe 100A..

So far, he's been saying that he wants it to produce the magnetic field in steady state form. As for further developments, we'll see what he needs later. Though, yeah, it could handle larger currents of shorter duration.
I have been doing finals so have not been keeping up.

I am clueless about wire conductivity and heat constraints... Is there any type of table or page online that speaks of how much current you can put through a wire for small bursts like he is pointing out here?

The few pages I have found are all conservative in their estimates generally..

ala http://www.powerstream.com/Wire_Size.htm

They show a bit higher amperage for the chassis wiring, but that seems really conservative compared to the above claim.

A pulsed solution may be a great boom to help with the costs and weight. Where as the poster above speaks of 100 amps in a 1 second burst...

The couple things specifically I would love to know ... how can you tell the approx max current you can get through a wire for a short burst.

After the burst is there a way to calculate the heat dissipation or even an approx time between bursts that would still not damage the wire... 5 seconds? 30 seconds? 5 minutes etc.. Since the field will be used for a projectile it doesn't need to be active for much longer than a second.
 
#13
After the burst is there a way to calculate the heat dissipation or even an approx time between bursts that would still not damage the wire... 5 seconds? 30 seconds? 5 minutes etc.. Since the field will be used for a projectile it doesn't need to be active for much longer than a second.

I haven't been able to come up with anything on this, so I'll "wing it". The values for the current-carrying capacity of these wires are either DC or RMS values. So I figure that one should be able to figure the current pulse on the basis of its RMS value. The lower the duty cycle, the more current you can pulse the coil with.

If the RMS current is 32.0A (selected previously) and the peak current is 100A, then:

t(on)= 1.0 sec
t(off)= 8.77 sec

This will give you an RMS current equal to 32.0A DC or sine wave RMS.

Of course, the next problem becomes forcing the necessary current through the coil's inductance.

Hope this helps.
 

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audioguru

Well-Known Member
Most Helpful Member
#14
You might need to tie-down the projectile with a strong metal latch then release it when the current has built-up enough in the inductance! Or else the projectile might slowly crawl out the end of the coil and flop onto the ground before the wires and power supply melt. :lol:
 
#15
You might need to tie-down the projectile with a strong metal latch then release it when the current has built-up enough in the inductance! Or else the projectile might slowly crawl out the end of the coil and flop onto the ground before the wires and power supply melt.

Not necessarily. Now that we know what he's doing, it's possible to wind a coil with fewer turns, use bifilar windings to keep the inductance down, and use a high voltage capacitor discharge to get that current pulse.

Of course, given the prior specifications, I wonder just what he's trying to launch here?

In high school, the physics prof got the bright idea of putting six solenoids (much smaller than we're talking here) end-to-end, asking six students to make connections by hand, to fire a bar magnet through the thing. Despite the lack of accurate timing, that magnet shot out of the end of the improvised EM cannon with enough force to actually smash through a plaster board wall. Surprised |-|311 out of him, to say the least. :shock:
 

MrAl

Well-Known Member
Most Helpful Member
#17
Hi,

NOTE: 10 year old thread here :)
 

JimB

Super Moderator
Most Helpful Member
#18
Miles was last here on ETO in May last year.
JimB
 

dr pepper

Well-Known Member
Most Helpful Member
#19
Heres the universal transformer calc, you can work out b flux density by thinking the solenoid as a transformer primary.
You can still use this for dc but you'll need to fiddle the frequency part, for that you'll need to have an idea on the sol's inductance.

 

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