Voice coil wire diameter relation to BI

hey, i want to design a subwoofer and i am thinking about the voice coil now. i have this question:
i have read on this forum that for paralel connection, for discusiion, 0.5mm wire x 100 meters is 8.7 ohms. 0.7mm x 200m is 8.9 ohm

parallel connection increase curent while lowering resistance in half, if 2x100m wires used, and in series increse the resistance wile BI stays the same
this seems counterproductive, it has the same effect as driving the single voice coil more without the added mass. maybe i am wrong.
the 0.7 wire would be a better solution ? but why? and how? this i dont know. i undertand that BI doubles with the increase in lenght
 
BL = B(The flux density of the magnet) x L (The Length of voice coil wire in the magnetic gap).

The length of the entire coil won't be in the gap unless the coil is underhung, pretty rare for subwoofers.
A longer coil length is needed for more excursion, and thicker wire for more power handling, whether the thinner or thicker is a better solution would depend on the application.
 
A couple notes about the length.
The resistance of the wire will be your DCR and not impedance. An 8 ohm woofer will likely have something like a 6 or 7 ohm DCR. The increase in resistance is mostly driven by back EMF. As the coil is allowed to move, resistance goes higher.
On the higher frequency side, voice coil inductance causes an impedance increase.

Also you are filling out a coil former as far as layers and coil height. You cannot really start off saying you will use a given number of meters of wire.
 
You do not decide firsthand how many meters to use, even less 100 meters which is an absurd number.

To use somewhat realistic numbers (fwiw I commercially design and build speakers, wind own coils, etc.) , a "buildable" subwoofer might have a 2" voice coil, winding (track) length 20mm which with an 8 mm plate will give you quite good 6mm X-Max

With a standard 2 layer VC, you can fit 80 turns total (40 turns per layer) if using 0.50mm wire, average turn length will be 16cm=0.16 meters so total wire length will be 80*0.16=12.8 meters (actually somewhat less because we are ignoring enamel thickness but hey, this is a basic example) so we are getting 8.7*0.128=1.11 ohms DCR, call it a 1.5 ohm speaker.

Not sure you want that,so clearly 0.5mm diameter enamelled wire is already way too thick.
So won´t waste time calculating for 0.7mm wire which will give you around 0.4 ohm DCR .... unusable.

Actual design goes the other way: you start with geometric conditions (VC diameter and track length), then solve for what wire fits there giving reasonable DCR, then find how many meters you got.

Small variations of wire diameter have huge impact on final DCR, since it is inversely proportional to the cube of diameter. 😱

Quite counterintuitive until you do the full Math.
 
I was thinking underhung with 4 inch former with a 10cm plate, thick. Got the B at around 0.41 tesla, gap 6-7mm, for the 0.7mm wire, less for 0.5.
For 0.5 would be 5cm height 3 layers to achive 100 meters, which woupd be 8.7 ohms. I dont quite understand DCR