Air Core Electro Magnet Construction

B

Bill Bowden

Guest
Given a volume and dimensions for an air core electro magnet, how does one
determine the wire gauge and number of turns for maximum strength and
efficiency? It seems using several wire gauges starting with a smaller gauge
near the core and then a larger gauge on the outside would be optimum since
the resistance of a single turn near the center would be minimal, while a
larger gauge wire on the outside would have less resistance for the
increased circumference. Is there some formula for this stuff?


.



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On Friday, October 30, 2015 at 11:01:21 PM UTC-4, Bill Bowden wrote:
Given a volume and dimensions for an air core electro magnet, how does one
determine the wire gauge and number of turns for maximum strength and
efficiency? It seems using several wire gauges starting with a smaller gauge
near the core and then a larger gauge on the outside would be optimum since
the resistance of a single turn near the center would be minimal, while a
larger gauge wire on the outside would have less resistance for the
increased circumference. Is there some formula for this stuff?
Click to expand...
Changing wire size would be a pain.
You need more than just the dimensions, one usually designs
for a given impedance.. the power supply you are using to run it.
You can get the same field with a few fat wires and lots of
current, or smaller with higher voltage.

George H.
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Click to expand...
 
"George Herold" <gherold@teachspin.com> wrote in message
news:19622ddb-9f45-4af3-a9b9-a244214ae76c@googlegroups.com...
On Friday, October 30, 2015 at 11:01:21 PM UTC-4, Bill Bowden wrote:
Given a volume and dimensions for an air core electro magnet, how does
one
determine the wire gauge and number of turns for maximum strength and
efficiency? It seems using several wire gauges starting with a smaller
gauge
near the core and then a larger gauge on the outside would be optimum
since
the resistance of a single turn near the center would be minimal, while a
larger gauge wire on the outside would have less resistance for the
increased circumference. Is there some formula for this stuff?


Changing wire size would be a pain.
You need more than just the dimensions, one usually designs
for a given impedance.. the power supply you are using to run it.
You can get the same field with a few fat wires and lots of
current, or smaller with higher voltage.

George H.
Click to expand...

The power supply is two 'D' cell batteries for an initial voltage of 3.2. It
has to work down to where the batteries are dead at 2 volts. How do I design
the electro-magnet for maximum strength and battery life? I think the
magenetic force is a function of only the supply voltage and volume of the
copper, but there are many combinations of turns,resistance, and amps to
get the same result. I could write a power basic program to show the results
based on turns and resistance of each turn, but I thought there was an
easier way.





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In article <n140pk$hcs$1@adenine.netfront.net>,
bperryb@bowdenshobbycircuits.info says...
"George Herold" <gherold@teachspin.com> wrote in message
news:19622ddb-9f45-4af3-a9b9-a244214ae76c@googlegroups.com...
On Friday, October 30, 2015 at 11:01:21 PM UTC-4, Bill Bowden wrote:
Given a volume and dimensions for an air core electro magnet, how does
one
determine the wire gauge and number of turns for maximum strength and
efficiency? It seems using several wire gauges starting with a smaller
gauge
near the core and then a larger gauge on the outside would be optimum
since
the resistance of a single turn near the center would be minimal, while a
larger gauge wire on the outside would have less resistance for the
increased circumference. Is there some formula for this stuff?


Changing wire size would be a pain.
You need more than just the dimensions, one usually designs
for a given impedance.. the power supply you are using to run it.
You can get the same field with a few fat wires and lots of
current, or smaller with higher voltage.

George H.

The power supply is two 'D' cell batteries for an initial voltage of 3.2. It
has to work down to where the batteries are dead at 2 volts. How do I design
the electro-magnet for maximum strength and battery life? I think the
magenetic force is a function of only the supply voltage and volume of the
copper, but there are many combinations of turns,resistance, and amps to
get the same result. I could write a power basic program to show the results
based on turns and resistance of each turn, but I thought there was an
easier way.





--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
Click to expand...

The magnetic field is a function of current and turns.

if you want it to work down to 2 volts, then calculate the
mag field strength to opereate at 2 volts and use a current
limiter.

Set the limiter to the field strength you need so basically
when the two cells hit 2 volts total the regulator will be
fully on. You must also calculate losses. What ever current you
get at 2 volts is the current you want to maintain at any voltage
above that.

Using a switcher supply for the current regulation would most likely
be the most efficient way to go. You need to look for cell voltage
switcher chips.

Not sure how you are doing your mag calculations? I suppose you
could be using Weber(Wb)/(T)Tesla/Gauss etc ?

Trying to remember the single wire basic math.. ?

W = (I * u) / Pi * R

I = current;
u = permittivity of space
R = radius

Something like that. It's been a while.

Jamie
 
On Fri, 30 Oct 2015 19:01:06 -0800, Bill Bowden wrote:

Given a volume and dimensions for an air core electro magnet, how does
one determine the wire gauge and number of turns for maximum strength
and efficiency? It seems using several wire gauges starting with a
smaller gauge near the core and then a larger gauge on the outside would
be optimum since the resistance of a single turn near the center would
be minimal, while a larger gauge wire on the outside would have less
resistance for the increased circumference. Is there some formula for
this stuff?
Click to expand...

I'm pretty sure that changing the wire size would not get you the most
field for the power input.

Google "solenoid formula" and see what pops up.

Most people end up using a finite element analysis program. FEMM is free
and seems to be pretty good.

Why air core?

--
www.wescottdesign.com
 
On Saturday, October 31, 2015 at 8:26:49 PM UTC-7, Bill Bowden wrote:

[about an air-core electromagnet]

The power supply is two 'D' cell batteries for an initial voltage of 3.2. It
has to work down to where the batteries are dead at 2 volts. How do I design
the electro-magnet for maximum strength and battery life?
Click to expand...

The internal resistance of the D cells is an ohm or so when they're at end-of-life,
so you need to plan on 2A maximum (and copper resistance means more like 1A
will be achievable).

There is no way to maximize strength that also maximizes battery life. You can
at best choose a battery technology and working time (NiCd, NiMH, alkaline,
and 'standard' D cells will all be different enough to matter). Then, you
can increase or decrease the working volume (size of the coil) and solve for
the best achievable field.

Adding iron gives a major improvement in field strength (factor of 100). Air core
magnets aren't very power-efficient, usually only used for high field when
the field desired is higher than the saturation limit for iron, or when the
windings can be superconductors.
 
On Fri, 30 Oct 2015, Bill Bowden wrote:

Given a volume and dimensions for an air core electro magnet, how does one
determine the wire gauge and number of turns for maximum strength and
efficiency? It seems using several wire gauges starting with a smaller gauge
near the core and then a larger gauge on the outside would be optimum since
the resistance of a single turn near the center would be minimal, while a
larger gauge wire on the outside would have less resistance for the
increased circumference. Is there some formula for this stuff?
Click to expand...
Is this to have some "supermagnet" that you can turn on and off? Unless
you need that ability, I'd just pull a magnet out of a hard drive, or a
good speaker.

Popular Electronics in the early sixties, I'd guess 1963 or 64 but that is
a guess, had what seemed to be a pretty good electro magnet on the cover,
details inside. Since those are now available online, one could figure
out the issue and download it.

Michael
 
Bill Bowden wrote:
It's air core because it interacts with a permanent magnet. But look at it
this way. Suppose you use a single turn of a certian volume and the
resistance was 1 ohm and the voltage is 1 volt for total amp-turns of 1
which produces 1 unit of field and the power input is 1 watt. Now if we use
2 turns of a smaller wire of twice the resistance with the same total
volume, the current is reduced 50%
Click to expand...

** No, it is reduced to 25%.

Two turns of double resistance = 4 times resistance.

and the power input is also reduced 50%
but we still get the same field (amp-turns) since amps is cut in half while
turns has doubled.
Click to expand...

** Not correct.

The mag field is now half what it was, you have to double the voltage to get the same field.

We have been down this path before.


..... Phil
 
"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:5LOdnWcsK4fN-avLnZ2dnUU7-ROdnZ2d@giganews.com...
On Fri, 30 Oct 2015 19:01:06 -0800, Bill Bowden wrote:

Given a volume and dimensions for an air core electro magnet, how does
one determine the wire gauge and number of turns for maximum strength
and efficiency? It seems using several wire gauges starting with a
smaller gauge near the core and then a larger gauge on the outside would
be optimum since the resistance of a single turn near the center would
be minimal, while a larger gauge wire on the outside would have less
resistance for the increased circumference. Is there some formula for
this stuff?

I'm pretty sure that changing the wire size would not get you the most
field for the power input.

Google "solenoid formula" and see what pops up.

Most people end up using a finite element analysis program. FEMM is free
and seems to be pretty good.

Why air core?

--
www.wescottdesign.com
Click to expand...

It's air core because it interacts with a permanent magnet. But look at it
this way. Suppose you use a single turn of a certian volume and the
resistance was 1 ohm and the voltage is 1 volt for total amp-turns of 1
which produces 1 unit of field and the power input is 1 watt. Now if we use
2 turns of a smaller wire of twice the resistance with the same total
volume, the current is reduced 50% and the power input is also reduced 50%
but we still get the same field (amp-turns) since amps is cut in half while
turns has doubled. So, it appears using more turns of a smaller wire is an
advantage in reducing power input without losing field strength for a given
voltage. But that only works for a single layer since additional layers will
have a larger circumference and more resistance per turn.


. . .





--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
 
On Sun, 01 Nov 2015 19:37:05 -0800, Bill Bowden wrote:

"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:5LOdnWcsK4fN-avLnZ2dnUU7-ROdnZ2d@giganews.com...
On Fri, 30 Oct 2015 19:01:06 -0800, Bill Bowden wrote:

Given a volume and dimensions for an air core electro magnet, how does
one determine the wire gauge and number of turns for maximum strength
and efficiency? It seems using several wire gauges starting with a
smaller gauge near the core and then a larger gauge on the outside
would be optimum since the resistance of a single turn near the center
would be minimal, while a larger gauge wire on the outside would have
less resistance for the increased circumference. Is there some formula
for this stuff?

I'm pretty sure that changing the wire size would not get you the most
field for the power input.

Google "solenoid formula" and see what pops up.

Most people end up using a finite element analysis program. FEMM is
free and seems to be pretty good.

Why air core?

--
www.wescottdesign.com


It's air core because it interacts with a permanent magnet. But look at
it this way. Suppose you use a single turn of a certian volume and the
resistance was 1 ohm and the voltage is 1 volt for total amp-turns of 1
which produces 1 unit of field and the power input is 1 watt. Now if we
use 2 turns of a smaller wire of twice the resistance with the same
total volume, the current is reduced 50% and the power input is also
reduced 50%
Click to expand...

Nope. If the volume of wire is the same but the length is doubled, then
the area is halved. Doubled length doubles the resistance, halved area
doubles the resistance _again_, and the resistance is quadrupled. The
current is reduced to 25% and the power input stays the same.

This is why manufacturers that sell motors with lots of different
windings on the same basic mechanical arrangement rate the mechanical
arrangement with a "motor constant" that's in units of torque/watt.

It's the same basic issue with solenoids, only you're talking force/watt.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
"Phil Allison" <pallison49@gmail.com> wrote in message
news:fe1f28df-8350-4583-be29-0910195c3fa3@googlegroups.com...
Bill Bowden wrote:


It's air core because it interacts with a permanent magnet. But look at
it
this way. Suppose you use a single turn of a certian volume and the
resistance was 1 ohm and the voltage is 1 volt for total amp-turns of 1
which produces 1 unit of field and the power input is 1 watt. Now if we
use
2 turns of a smaller wire of twice the resistance with the same total
volume, the current is reduced 50%

** No, it is reduced to 25%.

Two turns of double resistance = 4 times resistance.

and the power input is also reduced 50%
but we still get the same field (amp-turns) since amps is cut in half
while
turns has doubled.

** Not correct.

The mag field is now half what it was, you have to double the voltage to
get the same field.

We have been down this path before.


.... Phil
Click to expand...

Ok, so using the same volume with a different wire size just changes the
impedance and does nothing for efficiency. Suppose we use the same wire size
and just add another turn?

In the first case we have 1 turn, 1 ohm, 1 volt, 1 amp, 1 watt, and 1
amp-turn If we add another turn of the same wire, we have 2 turns, 2 ohms,
1 volt, 1/2 amp, 1/2 watt, and the same field. That looks like the
efficiency improved by 50% ? But this doubles the volume, so the question is
how far can you increase the volume before it's not worth the effort?









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On Mon, 02 Nov 2015 18:27:41 -0800, Bill Bowden wrote:

"Phil Allison" <pallison49@gmail.com> wrote in message
news:fe1f28df-8350-4583-be29-0910195c3fa3@googlegroups.com...
Bill Bowden wrote:


It's air core because it interacts with a permanent magnet. But look
at it this way. Suppose you use a single turn of a certian volume and
the resistance was 1 ohm and the voltage is 1 volt for total amp-turns
of 1 which produces 1 unit of field and the power input is 1 watt. Now
if we use 2 turns of a smaller wire of twice the resistance with the
same total volume, the current is reduced 50%

** No, it is reduced to 25%.

Two turns of double resistance = 4 times resistance.

and the power input is also reduced 50%
but we still get the same field (amp-turns) since amps is cut in half
while turns has doubled.

** Not correct.

The mag field is now half what it was, you have to double the voltage
to get the same field.

We have been down this path before.


.... Phil


Ok, so using the same volume with a different wire size just changes the
impedance and does nothing for efficiency. Suppose we use the same wire
size and just add another turn?

In the first case we have 1 turn, 1 ohm, 1 volt, 1 amp, 1 watt, and 1
amp-turn If we add another turn of the same wire, we have 2 turns, 2
ohms, 1 volt, 1/2 amp, 1/2 watt, and the same field. That looks like the
efficiency improved by 50% ? But this doubles the volume, so the
question is how far can you increase the volume before it's not worth
the effort?
Click to expand...

That's where some sort of already-worked solenoid equation, or finite
element analysis, would come in handy.

Have you Googled "solenoid equation" yet?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
"Tim Wescott" <seemywebsite@myfooter.really> wrote in message
news:yr-dna6YmeIbeaXLnZ2dnUU7-bmdnZ2d@giganews.com...
On Mon, 02 Nov 2015 18:27:41 -0800, Bill Bowden wrote:

"Phil Allison" <pallison49@gmail.com> wrote in message
news:fe1f28df-8350-4583-be29-0910195c3fa3@googlegroups.com...
Bill Bowden wrote:


It's air core because it interacts with a permanent magnet. But look
at it this way. Suppose you use a single turn of a certian volume and
the resistance was 1 ohm and the voltage is 1 volt for total amp-turns
of 1 which produces 1 unit of field and the power input is 1 watt. Now
if we use 2 turns of a smaller wire of twice the resistance with the
same total volume, the current is reduced 50%

** No, it is reduced to 25%.

Two turns of double resistance = 4 times resistance.

and the power input is also reduced 50%
but we still get the same field (amp-turns) since amps is cut in half
while turns has doubled.

** Not correct.

The mag field is now half what it was, you have to double the voltage
to get the same field.

We have been down this path before.


.... Phil


Ok, so using the same volume with a different wire size just changes the
impedance and does nothing for efficiency. Suppose we use the same wire
size and just add another turn?

In the first case we have 1 turn, 1 ohm, 1 volt, 1 amp, 1 watt, and 1
amp-turn If we add another turn of the same wire, we have 2 turns, 2
ohms, 1 volt, 1/2 amp, 1/2 watt, and the same field. That looks like the
efficiency improved by 50% ? But this doubles the volume, so the
question is how far can you increase the volume before it's not worth
the effort?

That's where some sort of already-worked solenoid equation, or finite
element analysis, would come in handy.

Have you Googled "solenoid equation" yet?
Click to expand...

Yes, there seems to be a lot of ideas about solenoids. This one seems fairly
simple. It looks like the solenoid length is important since the field
diminishes as the length increases. So, the idea is to add more turns
without increasing the length, which means the resistance of each layer is
greater since the circumference increases rapidly for a short length. So,
there must be some optimum design considering length, diameter, power input,
and volume. u is just a constant and be omitted since I'm just looking for a
maximum. I'll play around with a basic program to see what I get. I only
have two wire sizes to work with #34 and #28, so I have to use some
combination of the two..

B = ľNI/L, where N is the number of loops, I is the current, L is the
length of the solenoid, and ľ is the magnetic permeability of the core
material.






--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
 
"Bill Bowden" <bperryb@bowdenshobbycircuits.info> wrote in message
news:n1k07h$182l$1@adenine.netfront.net...
"Tim Wescott" <seemywebsite@myfooter.really> wrote in message
news:yr-dna6YmeIbeaXLnZ2dnUU7-bmdnZ2d@giganews.com...
On Mon, 02 Nov 2015 18:27:41 -0800, Bill Bowden wrote:

Ok, so using the same volume with a different wire size just changes the
impedance and does nothing for efficiency. Suppose we use the same wire
size and just add another turn?

In the first case we have 1 turn, 1 ohm, 1 volt, 1 amp, 1 watt, and 1
amp-turn If we add another turn of the same wire, we have 2 turns, 2
ohms, 1 volt, 1/2 amp, 1/2 watt, and the same field. That looks like the
efficiency improved by 50% ? But this doubles the volume, so the
question is how far can you increase the volume before it's not worth
the effort?

That's where some sort of already-worked solenoid equation, or finite
element analysis, would come in handy.

Have you Googled "solenoid equation" yet?


Yes, there seems to be a lot of ideas about solenoids. This one seems
fairly simple. It looks like the solenoid length is important since the
field diminishes as the length increases. So, the idea is to add more
turns without increasing the length, which means the resistance of each
layer is greater since the circumference increases rapidly for a short
length. So, there must be some optimum design considering length,
diameter, power input, and volume. u is just a constant and be omitted
since I'm just looking for a maximum. I'll play around with a basic
program to see what I get. I only have two wire sizes to work with #34 and
#28, so I have to use some combination of the two..

B = ľNI/L, where N is the number of loops, I is the current, L is the
length of the solenoid, and ľ is the magnetic permeability of the core
material.jui
Click to expand...

The formula B = uNI/L is apparently correct. I simulated a coil using a
constant length of #34 wire and rearranged the same wire on different core
diameters and lengths. The field increases as the number of turns and length
decreases. I wound an experimental coil using 1/4 length with all #34 wire
and it produced about the same field with only 1/4 of the power of the old
unit since the resistance was 4 times higher for the same result. Quite a
difference.


Diameter Length Turns Field

1 1 8290 61

1.4 0 .5 6214 91

2 0.245 4531 136

2.5 0.155 3657 174







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"whit3rd" <whit3rd@gmail.com> napisal w wiadomosci
news:dda9837d-4506-4d23-a633-055c3ba4ee40@googlegroups.com...
On Saturday, October 31, 2015 at 8:26:49 PM UTC-7, Bill Bowden wrote:

[about an air-core electromagnet]

Adding iron gives a major improvement in field strength (factor of 100).
Air core
magnets aren't very power-efficient, usually only used for high field when
the field desired is higher than the saturation limit for iron, or when
the
windings can be superconductors.
Click to expand...

And: "As ferromagnetic material causes noise or distortion in the signal, it
should be avoided in high frequency application like signal transmission.
Thus air core transformer is introduced, in the application of high
frequency radio transmission." From:
http://www.electrical4u.com/air-core-transformer/
S*
 
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