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[GavJ]

I'm trying to design some homemade solenoids for my pipe organ I'm making, and I don't think I'm understanding the concept well enough to make efficient decisions. Maybe there are some electricity folks out there in the forums.

So Ohm's law is I = V / R
And magnetic flux = wraps * I

Let's say I have 1 volt and effectively infinite current capacity from my power source, and I wrap a wire 10 times, with a gauge of wire that = 1 ohm for ten wraps.  My current will be 1 amp, and my flux will be 10 amp-wraps
If I now wrap it 20 times instead, the resistance doubles since the wire is now twice as long, and with my same supply voltage, my current drops to 0.5 amps. 0.5 * 20 = flux is now 10 amp-wraps

So it doesn't seem like it should have any effect on magnet/solenoid strength...? But when i test it empirically, it clearly does have an effect. As in, just winding more times with the same power source and same wire type very noticeably increases the power. Why?

[Telgin]

It's been a long time since I took a circuits class, but increasing the windings should change the inductance of the system if it's AC current.  Resistance would still be pretty negligible and has no direct effect here anyway.  I'm pretty sure increasing the windings increases the strength of the magnetic field because of the inductance, probably at the cost of increased current draw.

[Vector]

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[GavJ]

If length doesn't matter, then why does every single table and calculator for resistance online ask for length? Such as:
http://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table
and
http://www.powerstream.com/Wire_Size.htm (note: ohms per kilometer. Just "ohms" per diameter is not a thing here)


Inductance I don't know anything about. Does it help to do several layers of wraps versus one long line? Etc. etc. anything like that?

And if wraps add something beyond just "amps * wraps" then am I better off using the tiniest wire possible and huge numbers of wraps until it's hot but not a fire hazard, then, for most power? Or what?

[Arx]

You should be working with B = μ*((I*N)/l) as well, I think, rather than simply I*N. I'm not sure how that would change it though - I have a fairly limited understanding of solenoids myself, having focused more on motors and generators.

[Leafsnail]

The length does matter when it comes to calculating the resistance of the wire.  However, unless your system is kilometers long this factor is too small to matter.

Why?  Because batteries have their own internal resistance.  Normally this is too small to matter much when you're considering circuit design, but if you only have wires and no other components then the thing that determines the resistance of your system is pretty much the internal resistance of your battery.

http://en.wikipedia.org/wiki/Internal_resistance#Batteries
Here Wikipedia suggests that the internal resistance of a typical AA battery at room temperature is 0.15ohms.  The calculator you've provided above says that even 10 feet of typical 12 gauge wire (which seems like more than you're likely to use) has a resistance of just 0.016 ohms.  So doubling the length of your wire to 20 feet would take your system from a total resistance of ~1.165 to ~1.18 ohms - not a big change really, so your current will stay similar.

Note that this is with just one small battery.  If you're using multiple batteries or larger batteries the internal resistance is likely to dwarf the resistance of your wires even more.

[GavJ]

Currently I'm using 36 gauge enameled wire, and rectifiers from mains electricity (spare orphaned power adapters for unknown electronic devices). Several of them have proven quite capable of catching my solenoids on fire, so wire resistance is definitely not negligible.

My most successful so far 500-turn solenoid I calculate at about 15 ohms. It can hold a few grams when driven by a 12 volt, 2.8 amp rated adapter, while only getting fairly hot. I assume the adapters are roughly 1-2 ohms themselves, judging by their V/A ratings.

(the only other wire I have on hand is a ridiculously huge 10 gauge that I can't even wrap around the pencil-size solenoids I need because it's too thick. Even when I tried it was much weaker than the 36 gauge wire in available space)


Ampere's law is nice and all, but anything with double and line integrals in it draws a big fat "Nooooope" from me. Would rather keep trialing and erroring than touch that with a ten foot pole, if necessary.

[Sergarr]

One does not simply become an engineer without calculation some nice big integrals 

Otherwise, be prepared for fire. Lots and lots of fire. Inductivity is a pretty dangerous thing in big amounts, that's why engineers tend to avoid it, preferring to opt for capacitors instead, when possible.

[da_nang]

Let's say you have a voltage source u going over the solenoid. The solenoid can be modeled as an ideal inductor serially coupled with a resistor.

Let the resistor have a resistance of R = ρN where ρ is the resistance per turn and N is the number of turns.

The ideal inductor has a self-inductance L ≈ μ₀r²N²π/d , where μ₀ is the permeability constant, r is the radius of the solenoid and d is the length of the solenoid. This is an approximation for r/d << 1 .

Kirchhoff's voltage law states that u = V_R + V_L = R i + Ldi/dt where i is the current going through the solenoid. This ODE has the solution i = (i₀ - i_p(0))*exp(-Rt/L) + i_p where i_p is the particular solution and i₀ is the initial condition.

The magnetic flux density inside a solenoid is defined as B =  μ₀Ni/d and the magnetic flux Φ = BA = Bπr² = μ₀Niπr²/d = (L/N)i .

Now if we assume we've been running a voltage through the solenoid for a long time, the transient part of the current vanishes i.e i ≈ i_p for large t.

For DC, i_p = u/R = u/(ρN) .
For AC, assuming u = u₀ sin(ωt), then i_p = (u₀/Z)sin(ωt - φ) where Z = √(R² + (Lω)²) and tan(φ) = Lω/R.

Inserting in the AC case gives you
Φ = (L/N)(u₀/Z)sin(ωt - φ) = (L/[NZ])u₀sin(ωt - φ)

and L/(NZ) = 1/√((RN/L)² + (Nω)^2).

Now let k = μ₀πr²/ρd which can be assumed to be constant thus L/R = Nk.

Therefore L/(NZ) = k/√(1 + (Nωk)²) and Φ = (u₀k/√(1 + (Nωk)²))sin(ωt - atan(Nkω)).

Φ_RMS = u₀k/√(2 + 2(Nωk)²).

For the DC case, Φ = (L/N)(u/R) = uk .

[GavJ]

Okay, still not understanding why turns matter then.

If flux is Li/N and L includes relevant terms N^2 and 1/d...
Then going from 10 to 20 turns should increase L by a factor of 2. but then the 1/N in flux divides it by 2 again.
So flux is just the same still?

Is it the time factor I'm missing? What is "large t" in an inductor? Are we talking like 5 minutes, or a millisecond...? You don't mention any units for reference of whether to consider more of an initial condition or a limit condition, and I don't grok it well enough to fill in my own whole suite of units.

(This is a pipe organ, so activation needs to be prompt to within a fraction of a second and sustain for sometimes several seconds)

[Arx]

Look at Lagslayer's post. I'm increasingly sure he's got a point for a set-up the scale of yours, where the battery will have a significant fraction of the resistance.

[GavJ]

Right, so the power supply is 12V, 4A, which suggests it has 3 ohms itself.
50 turns of my wire is about 1.5 ohms.  500 turns is 15 ohms.
So total should be going from 4.5 to 18 ohms = 4 times more resistance, versus 10 times more turns.

So the solenoid should get 2.5x stronger?  Maybe, but it really seems like it's getting more like 10-15x stronger. I don't have it hooked up to a scale or anything though. I could be wrong.

[da_nang]

Okay, still not understanding why turns matter then.

If flux is Li/N and L includes relevant terms N^2 and 1/d...
Then going from 10 to 20 turns should increase L by a factor of 2. but then the 1/N in flux divides it by 2 again.
So flux is just the same still?

I suppose one thing that should be mentioned is that the constant k assumes the dimensions of the solenoid remain the same. If you're already winding the wires tightly then obviously that doesn't hold true anymore. If we assume the wires are wounded right next to each other with no overlapping wires, then d = wN where w is the width of the wire.

So let k = k₀/N where k₀ = μ₀πr²/ρw.

Also, since you're obviously not working in a vaccuum, you'd want to multiply the permeability constant with the effective constant i.e. μ = μ₀μ_eff.

Is it the time factor I'm missing? What is "large t" in an inductor? Are we talking like 5 minutes, or a millisecond...? You don't mention any units for reference of whether to consider more of an initial condition or a limit condition, and I don't grok it well enough to fill in my own whole suite of units.

(This is a pipe organ, so activation needs to be prompt to within a fraction of a second and sustain for sometimes several seconds)

The "large t" typically means that t approaches infinity, but in practice it means that in this case t > t₀ where exp(-Rt₀/L) << 1. As for units, it's SI units.


EDIT:

Spoiler: More readable (click to show/hide)

[GavJ]

So in practical terms, I just need a larger wire but not one so large that I can't wrap it effectively, and then I won't hit the efficiency brick wall with fire. Okay, that's easy, will order some 24 gauge