For an Ideal transformer where permeability of the iron is infinity, why is the excitation current 0 if the secondary winding is open circuit?
My intuition tells me that when a voltage is applied, a current will begin to flow, causing an induced magnetic field in the coil. However, since the permeability is infinity, that will result in an infinite B field, and an infinite flux.(this is apparently not true, and I am quite confused)


Intuition and infinity don't play nicely together.

Field = inductance x current.

If the inductance is infinite, then you can get any field with zero current, as \$0 \times \infty \$ is undefined, it can be any real number.

I'm an engineer, so I don't know what mathematicians mean by infinity, but whatever it is, it's not really relevant to real life. Engineers generally mean it in the terms of a limit, as something becomes so big that its reciprocal becomes negligible.

Going back to field = inductance x current, and ignoring any constants of proportionality, let's say we get unit field with one 1A and 1H. If we had 10H, then we'd need only 100mA. With 100H, we need only 10mA. As we let the inductance get larger and larger, the required current gets smaller and smaller. It gets far smaller than any load current, it gets smaller than anything we can measure. 

This is what engineers mean by infinite inductance, that the current required to create any finite field in it is negligible for all practical purposes.