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In a transformer, let's say we have:
$I_1, I_2$ - currents through the primary and secondary winding
$V_1, V_2$ - voltages
$N_1, N_2$ - number of turns
$F_1, F_2$ - magnetic fluxes through core, produced by the currents $I_1$ and $I_2$ (they are opposing...)
$R$ - the reluctance of the core

We have $$\frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}.\\ F_1=N_1*\frac{I_1}{R}, \hspace{2mm} F_2=N_2*\frac{I_2}{R}$$ I think $I_1$ and $I_2$ are in phase too.
That means $F_1$ and $F_2$ are practically equal.
How can we have a nonzero resulting flux through the core? $F=F_1-F_2$
If the flux through the iron core is zero, then the cause that produces $V_2$ (variation of the flux) does not exist, then $V_2=0$?

Transformer

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The currents you mention (I1 and I2) are due to the load on the secondary and these currents produce equal and opposite magneto motive forces. This means that the magnetic fluxes cancel (as you quite rightly point out) however, there is another magnetic flux and this is due to the inductance of the primary taking a small current in the absence of any load current.

It is this current (and flux called magnetization flux) that works with Faraday's law of induction to produce an open circuit secondary voltage in proportion to the turns ratio. Any secondary load current will produce an extra current in the primary and these load-currents together produce NO NET FLUX, leaving just the magnetization flux i.e. it is constant under all load conditions.

OK that's a slight exaggeration - given that a real transformer has leakage inductance and dc resistance, these components produce small volt drops in the primary and this slightly lowers the magnetization current.

So, magnetic flux lowers (a little bit) as load currents increase.

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