Infinite EMF produced in transformer If I have an ideal lossless transformer, where the ratio of voltages is proportional to the ratio of turns of wire, what, theoretically would stop infinite amplification of the EMF in one coil, and therefore infinite power being drawn from the circuit, if I increase the number of turns in the secondary coil towards infinity? Edit: I know that it's common knowledge that $V \cdot I$ is conserved from one coil to the other. I don't, however, see how this comes about from Faraday's law.
 A: When you increase the induced EMF in the secondary winding of a transformer, the induced current in the winding is reduced by the same factor. So in an ideal transformer the product $V_j I_j$ (where the index $j$ labels the primary or secondary winding, and $V$ and $I$ are amplitudes of the voltage and current) is the same in the two windings. This product gives the power, and so rather than the power increasing towards infinity in the situation you consider, and thereby breaking the conservation of energy, it in fact remains constant.
To see the physical reason for this, consider the magnetic fields created by the primary and secondary. The magnetic field generated by the primary $B_p \propto N_p I_p$, and in the secondary $B_s \propto - N_s I_s$, where $N_j$ is the number of turns in the winding. For an ideal transformer the magnetic flux threading the primary all passes through the secondary, as shown in the figure.

If for simplicity we take the cross-sectional areas of the windings to be the same, then the constancy of the magnetic flux implies that $I_p N_p = - I_s N_s$. For a step-up transformer, $N_s > N_p$, and so $I_s$ is reduced in the same ratio to which $V_s$ is increased. This keeps the product $V I$ constant. 
In practice the power in the secondary will always be less than the power in the primary due to losses in eddy currents, flux leakage, the resistance of the windings, and so on.
Taken from Wikipedia https://en.wikipedia.org/wiki/Transformer#/media/File:Transformer3d_col3.svg licensed under CC BY-SA 3.0.
A: You can have a large number of turns on the secondary side, leading to a large secondary side voltage, if you keep the secondary open. But the moment you add a load to the secondary and close the secondary side circuit, a current will start flowing through the secondary, which will cause a large back-emf, $v=L{\cdot}\frac{di}{dt}$, or $V=Z{\cdot}I$. The larger the current in the secondary, the larger the back-emf, and the larger the current that has to be supplied on the primary side to maintain the original magnetizing flux through the core. This is the fundamental energy balance.
