Relationship between inductance and number of turns in a coil in loudspeaker

Question

I am in the process of designing a loudspeaker and have a question regarding the number of turns in the multilayered solenoidal coil and the speaker impedance of $8\:\Omega$. I understand that the Impedance is equal to the real part $R$ (resistance) added to the inductance and that the inductance is partly a function of the number of turn of the copper wire. I am having difficulty with understanding which equation describes this relationship between inductance and number of turns. Thanks.

Answer 1

The one that comes to mind:

$L= μN^2 A/s ∴N= \sqrt{\frac{Ls}{μA}}$

$\text {where A is the cross sectional area, s is the length, and μ is the permeability }(4\pi \times 10^{-7} \frac {H} {m} \text{of air}) $

Answer 2

a fairly nice read....wiki

In short, impedance $X$ is expressed as $$X=X_R+X_I+X_C$$ where the resistive load, $$X_R = R$$, the inductive load, $$X_I=j\omega L$$ the capacitive load, $$X_C=\frac{1}{j \omega C}$$

where, $L,C,R$ are inductance, capacitance and resistance.

Now since you have the complex impedance, find out the $|X|$ to get net impedance in Ohms.

So, to get $L$, we have a formula which as you correctly suspect is dependent on $N$, number of turns.

$$L=\frac{\mu_r \mu_0 N^2A}{l}$$ where, $A$ is circular cross section of solenoid, $l$ is length of solenoid (not of wire), $\mu_0$ permeability of air (a constant) and $\mu_r$ relative permability of core (iron, in your case; probabably; you can find the value here).

you can calculate it here itself.