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- 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?

##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?