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More wire improves efficiency of a coil?

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Yanom

Sun Oct 14 2012, 11:01PM

Registered Member #4659 Joined: Sun Apr 29 2012, 06:14PM
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The magnetic power of a coil is determined by amp-turns - how many amps flow through it times how many turns of wire it has (more or less).
Subjecting the following coils to a 1v power supply, and assuming that 1 turn of wire adds 1 ohm of resistance:

1 turn coil has 1 ohm - 1 amp - 1 amp-turns
2 turn coil has 2 ohm - 0.5 amp - 1 amp-turns
4 turn coil has 4 ohm - .25 amp - 1 amp-turns
8 turn coil has 8 ohm - .125 amp - 1 amp-turns

and so on. However, coils with more turns are more efficient because they get the same amp-turns for less power draw

right?

Pinky's Brain

Sun Oct 14 2012, 11:14PM

Registered Member #2901 Joined: Thu Jun 03 2010, 01:25PM
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Well sure, but you use 8 times as much copper on the 8 turns one ... with the same amount of copper you could have 1 turn (or 8 parallel turns) with 1/8 ohm at 1 amp OR 8 turns with 8 ohm at 0.125 amp. So I2R = 1/8 Watts for both of them.

TANSTAAFL.

PS. that said, you'll generally have fewer losses in the driver with lower currents.

radiotech

Sun Oct 14 2012, 11:35PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
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Ampere-turns determines the magnetic flux in a core of a given area. The coil is
sized to produce the flux desired. The thickness/resistance of the wire permits
various voltages to produce flux.

Yanom

Mon Oct 15 2012, 03:36AM

Registered Member #4659 Joined: Sun Apr 29 2012, 06:14PM
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Pinky's Brain wrote ...

Well sure, but you use 8 times as much copper on the 8 turns one ... with the same amount of copper you could have 1 turn (or 8 parallel turns) with 1/8 ohm at 1 amp OR 8 turns with 8 ohm at 0.125 amp. So I2R = 1/8 Watts for both

oh. but assuming they're both equal magnetic performance , and I run both coils off a 9v battery, wont the 0.125amp one drain the battery slower because its lower current?

Steve Conner

Mon Oct 15 2012, 07:22AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

As Pinky's Brain points out, the number of turns washes out. If you are designing an electromagnet of a given size and shape, then if you want to increase the number of turns, you have to use thinner wire, or it won't fit. The required input voltage goes up and the current goes down, but the current density stays the same, hence so does the input power for a given MMF.

If you allow the magnet to get bigger as you add more wire, this doesn't hold. But you'll find that the larger coil doesn't generate as many amp-turns as you'd think. MMF is actually measured in amp-turns per meter, and the bigger coil simply spreads the flux over a wider area rather than increasing it at the measuring point.

It follows that the "efficiency" of a magnet (amp-turns of MMF per watt of input power) is mainly determined by the conductivity of the magnet wire. Superconducting magnets need no input power at all once charged.

Yanom

Mon Oct 15 2012, 12:09PM

Registered Member #4659 Joined: Sun Apr 29 2012, 06:14PM
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Steve Conner wrote ...

As Pinky's Brain points out, the number of turns washes out. If you are designing an electromagnet of a given size and shape, then if you want to increase the number of turns, you have to use thinner wire, or it won't fit. The required input voltage goes up and the current goes down, but the current density stays the same, hence so does the input power for a given MMF.

If you allow the magnet to get bigger as you add more wire, this doesn't hold. But you'll find that the larger coil doesn't generate as many amp-turns as you'd think. MMF is actually measured in amp-turns per meter, and the bigger coil simply spreads the flux over a wider area rather than increasing it at the measuring point.

It follows that the "efficiency" of a magnet (amp-turns of MMF per watt of input power) is mainly determined by the conductivity of the magnet wire. Superconducting magnets need no input power at all once charged.

so adding more wire helps up to the point that the coil gets too big?

and the 8ohm 0.125 amp coil would drain the battery slower, wouldn't it? battery capacity is measured in amp-hours.

Steve Conner

Mon Oct 15 2012, 12:20PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

You need to define "helps" more clearly.

Yes, the 8 ohm coil would drain the battery slower. However, the lower resistance coils would produce the same MMF with a lower battery voltage.

A 9V battery is 6 1.5V cells in series inside. If you used a coil with one-sixth the turns, of wire 6x thicker in cross-sectional area, you could drive it using the same 6 cells in parallel. The resulting coil dimensions, amp-turns and battery life would be identical.

Moral of the story: there is no optimum number of turns for an electromagnet. You choose the number of turns and wire gauge to suit whatever power supply you want to use, and whatever current density your cooling system can handle with a safe temperature rise.

Pinky's Brain

Mon Oct 15 2012, 01:18PM

Registered Member #2901 Joined: Thu Jun 03 2010, 01:25PM
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Yanom wrote ...

oh. but assuming they're both equal magnetic performance , and I run both coils off a 9v battery, wont the 0.125amp one drain the battery slower because its lower current?

My point was that they are at equal magnetic performance at equal power consumption for the same amount of copper ... they are however not at equal power consumption at the same voltage and thus not at equal magnetic performance either.

If you just want to make a solenoid to be powered from a 9V battery I would follow the following steps:

- Determine how much current you want to pull and calculate the resistance necessary for that
- Determine how much money you want to spend on wire
- Find the thickest gauge wire from which you can wind enough turns within your budget to get to the resistance needed

This is assuming thicker gauge will be cheaper per weight of copper, which won't always be true ... so keep that in mind if you end up looking at exotic gauges. Also assuming that since it's a 9V battery you probably won't use enough power to burn up the wire.

Yanom

Mon Oct 15 2012, 02:45PM

Registered Member #4659 Joined: Sun Apr 29 2012, 06:14PM
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ok. I think I sort of get it....

Let's take an equal amount of copper and make 2 coils:

#1: 1 turn, (1/8) ohm
and
#2: 8 turn, 8 ohm
and apply both coils to a 1v power source

so for coil #1:
8 amps * 1 turn = 8 amp turns
8 amps * 1v = 8 watts
1 Aturn / watt

so for coil #2:
(1/8) amps * 8 turn = 1 amp turns
(1/8) amps * 1v = (1/8) watts
8 Aturn / watt

so the coils aren't equal magnetically. but i still don't quite understand - isn't coil #2 more efficient, with 8 amp turns per watt?

thanks for the help.

Pinky's Brain

Mon Oct 15 2012, 03:20PM

Registered Member #2901 Joined: Thu Jun 03 2010, 01:25PM
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Posts: 837

Magnetic potential does not have a linear relation with electrical power ... so Aturn/Watt is not a useful metric (try swapping the currents for both). At the same electrical power they have the same magnetic potential.

The force a solenoid exerts on another magnet or ferromagnetic object Force does have linear relation with electrical power though ... and a quadratic relation with magnetic potential.

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