Production of the magnetic field from 3-phase currents

[PhysicsTest]

TL;DR Summary

Magnetic field equation based on the 3 phase currents.

This is not a homework problem, I want to calculate the equation of the magnetic field intensity from the 3 phase currents separated by 120Degrees. The 3 currents are
$I_{{aa}^{'}} = I_M\sin \omega t ; -> eq1 \ I_{{bb}^{'}} = I_M\sin(\omega t - 120); -> eq2\ I_{{cc}^{'}} = I_M\sin(\omega t -240); -> eq3$
The current in coil ${{aa}^{'}}$ flows into the $a$ end of the coil and out the ${{a}^{'}}$ end of the coil. It produces the magnetic field intensity $H_{{aa}^{'}}(t) = H_M\sin(\omega t) \angle 0$ -> eq4;
I wanted to derive eq1. The equation from the book about the magnetic field intensity
$\oint H.dl = I_{net} -> eq5;$
My attempt is
$H = \frac{dI_{{aa}^{'}}} {dl} ->eq6$ substitute eq1 into eq6.
$H = \frac{dI_M\sin \omega t} {dl} -> eq7$
But i don't see the equation 6 in terms of $l$, how to solve this and derive equation 4?

 

[Delta2]

You have misinterpret Ampere's Law (eq5), the integral that appears there is not a "simple" integral but a line or path integral around a closed curve. So (eq6) simply doesn't hold (it would hold if that integral was a simple integral as i said).
Check here for more info about line integrals.
https://en.wikipedia.org/wiki/Line_integral
More specifically check the section about "Line integral of a Vector field" because H in eq5 is a vector field.

Anyway if you want to find H at the center of the three coils, then you have to make some extra assumptions about the geometry of the coils (for example are they solenoids?), the information from the currents only is not enough.