Reasoning transformer identity for currents When looking for reasons for the Transformer Identity for Currents with a short secondary circuit
$$\frac{I_2}{I_1}=\frac{N_1}{N_2}$$
I find two common answers:


*

*Both coils create the common magnetic field together. Its magnetic flux density $B=\mu N\frac{I}{l}$ is proportional to $I$ and $N$ in each coil separately:  


*

*$B\sim I_1\cdot N_1$  

*$B\sim I_2\cdot N_2$
Consequently, it must be
$$I_1\cdot N_1=I_2\cdot N_2$$
and the identiy follows by algebraic transformations.


*No, it's not all like 1. The secondary current creates it's own, antiparallel magnetic field, which reduces the overall field lowering the induction voltage in the primary circuit, letting the primary current get bigger, …. If you want to solve the problem, you have to use Ampère's Law (or similar):
Take a rectangular path through the iron of the transformer and you get [blabla] 
$$I_1\cdot N_1=I_2\cdot N_2$$
My question is now:
Why renders the second version the first version not only obsolete but also wrong? The formula for the magnetic flux density in a coil is also retrieved from Ampère's Law. The Length $l$ doesn't stem from the law itself, but from the model, that the magnetic field density is nearly zero outside the coil.
So, the only grave difference between version 1 and 2 is the length and material of the coil. But material and effective length of the magnetic field is indeed equal for both coils, as the transformer core is homogenous and a complete circuit core.
 A: TL;DR  The first common answer adds up wrongs to make a right. The flux density is not proportional to $I$ and $N$ separately, but it is the result of the superposition of the two windings. It simply isn't how it phisically happens.

With a simplified Amperes law you get:
\begin{align}
H\cdot\ell=N_1I_1-N_2I_2
\end{align}
In this case $H$ is actually the magnetising field, meaning the difference between the fields of each of the windings.
\begin{align}
H_1\cdot\ell&=N_1I_1\\
H_2\cdot\ell&=N_2I_2\\
H\ell&=H_1\ell-H_2\ell=N_1I_1-N_2I_2
\end{align}
The difference is therefore lost as magnetisation current. You should note that for a magnetic core with $\mu\rightarrow\infty$ the value of $H$ is zero (perfect magnetic conductor). The other effect would be that you shouldn't have $H_1$ or $H_2$ either, they would actually dissapear and the only thing oposing the magnetisation would be the secondary winding.

(Digression) Question is why do they dissapear? Lets look at the law of induction (without the integrals):
\begin{align}
U=N\frac{\mathrm{d}\Phi}{\mathrm{d}t}
\end{align}
That combined with Gauss's law of magnetism(for an non-closed area)
\begin{align}
B\cdot A=\Phi
\end{align}
gives:
\begin{align}
U=NA\frac{\mathrm{d}B}{\mathrm{d}t}
\end{align}
That means that the value of $B$ is fixed (a result of the above equation) and we need to calculate the value of $H$ based on this $B$. 
\begin{align}
H=\lim_{\mu \rightarrow \infty}\frac{B}{\mu}=0\\
\end{align}

In an ideal Transformer you have perfecty coupled circuits (either $\ell=0$(makes not much sense) or $\mu\rightarrow 0$):
\begin{align}
N_1I_1&=N_2I_2\\
H\ell&= 0
\end{align}
In the non-ideal transformer you have that:
\begin{align}
H_1\cdot\ell&\approx H_2\cdot\ell\\
N_1I_1&\approx N_2I_2\\
H\ell&\neq 0
\end{align}
