No-load and loaded operation, constant flux

An ideal transformer has no winding resistance, no leakage flux, no core loss and infinite core permeability, so it needs no magnetising current.

No-load operation

The secondary is open. The primary is connected to a sinusoidal supply $V_1$.

  I1=0      +------+      I2=0
 o--->---+  |||||||  +--- o
  V1     ) core     (  E2   open
 o-------+  |||||||  +--- o
         N1         N2

• The primary draws only the magnetising current $I_\mu$ needed to set up the core flux $\phi$. Because the core of an ideal transformer has infinite permeability, $I_\mu$ is taken as zero. The flux itself is not zero: it has the finite value $\phi_m = V_1/(4.44fN_1)$ needed to balance the supply voltage.
• The flux induces $e_1 = -N_1\,d\phi/dt$ in the primary. For an ideal transformer $E_1$ is equal in magnitude and opposite in direction to $V_1$.
• The same flux links the secondary, so $E_2 = V_2 = -N_2\,d\phi/dt$ and $\dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}$.

Loaded operation

A load is connected to the secondary and a current $I_2$ flows.

• $I_2$ sets up a secondary mmf $N_2I_2$ which, by Lenz's law, tends to oppose the core flux.
• The primary draws an extra current $I_1'$ so that its mmf $N_1I_1'$ cancels the secondary mmf:

$$N_1I_1' = N_2I_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1}$$

• The input power equals the output power, $V_1I_1 = V_2I_2$.

Proof that the net flux remains constant

The primary equation gives $V_1 \approx E_1 = 4.44fN_1\phi_m$. Since $V_1$ and $f$ are fixed by the supply, $\phi_m = V_1/(4.44fN_1)$ is fixed, whatever the load.

Also, from the mmf balance, the net mmf acting on the core on load is:

$$\begin{aligned} F_{net} &= N_1I_1 - N_2I_2 \\ &= N_1(I_\mu + I_1') - N_2I_2 \\ &= N_1I_\mu + (N_1I_1' - N_2I_2) = N_1I_\mu \end{aligned}$$

The net mmf is $N_1I_\mu$, the same as at no load. The core flux is $\phi = N_1I_\mu/S$, so the flux is constant. Any load only adds the balancing component $I_1'$ in the primary, which cancels the secondary mmf and leaves the flux unchanged. (In the ideal limit both $I_\mu$ and the core reluctance $S$ tend to zero, but their ratio still gives the same no-load flux $\phi_m$.)