Assume that there is a transformer as in the image above. The applied primary side voltage \$E_1\$ is always AC. We design the transformer so that the magnetizing flux \$\Phi_m\$ does not saturate the core. Any \$I_2\$ current that will be drawn by the load on the secondary side will generate \$\Phi_2\$ flux in the transformer core and \$I_1\$ current at the primary side. And this \$I_1\$ current will generate a \$\Phi_1\$ flux, which will be equal to \$\Phi_2\$ in both magnitude and phase, but in opposing directions. Therefore, \$\Phi_1\$ and \$\Phi_2\$ will cancel out each other, and the net flux in the core will always be \$\Phi_m\$ at most.
(This is what I know about working principle of transformers. Please correct me if I did any mistake.)
What limits the practical load current in a system like this? According to what I explained above, we can theoretically draw infinite current from the secondary side, because \$\Phi_1\$ and \$\Phi_2\$ will always cancel out each other and the core will never saturate or excessively heat up. I understand that the copper losses will be a limiting factor. But my question is rather about the magnetics of the system. So, please assume that the wires are perfect conductor, and the AC source is ideal.
What physical factor does limit the maximum transferable power through a transformer like this?