If it is a very very large inductance, as "Mark M" stated, then the reactance will be very large (reactance is angular frequency times inductance) then the current will be very small.
This inductance exists if the secondary is loaded or not, so an open secondary would have less current on the primary, than one with a load connected to it, and your transformer would not heat up as much.
If there was no load in the secondary the transformer will act like an inductor and will draw current based on it's impedence. In an ideal transformer as there are no losses the in phase current will be absent and the current will be lagging by 90 degrees.
A unloaded transformer is a giant inductor. The current will be there, but phase shifted from the voltage.
If the core had infinite permeability then the magnetic fields would be huge, and it would be a very large inductor and draw lots of current.
"So, assuming we have perfect transformer...
...if you don't connect secondary coil...
...would it mean that there would be no current going through primary coil?"
Yes.
However, even a fairly simple model of a real transformer includes a shunt inductance across the idealised circuit element, and thus it always draws current.
Investigate how an MRI machine works. It is an air-core (I guess the human body modifies that) transformer with no internal resistance (cooled by liquid helium). I think that the secondary coils have very sensitive circuitry to measure tiny changes of voltages and currents during the scans.
IF NO IRON AND COPPER LOSS, NO CURRENT WOULD BE PASSING PRIMARY WINDING IF NO LOAD ON SECONDARY WINDING !
== BUT SUCH CONDITION WAS EVER EXISTED ==
So, assuming we have perfect transformer (with no wire or core looses, and with core of infinite permeability) it would mean that current going through primary coil is equal to Ip=Vp/(a*a*Zl); This means that current through primary (Ip) is equal to voltage on primary (Vp) devided by square transformer identity (a=Np/Ns or ratio between number of windings on primary and secondary) multiplied by impedance on secondary coil (which is basically all resistance elements combined)!
My question is if you have very high impedance on secondary coil, for example if you don't connect secondary coil wires to any circuit or to insulator (plastic or ceramic for example) would it mean that there would be no current going through primary coil?
