You're measuring what is probably perfectly normal capacitive leakage, but without a well defined impedance (your voltmeter). You need to measure leakage current by connecting resistor across each thing you want to measure and then measure the voltage across that. A 1K resistor will give you 1mv/uA, but I'd recommend you try a 100K first, then if the voltage is low, go to 1K. Obviously not across the mains....
Right.... measured with a 980 ohm resistor connected to mains earth at one end, and the other end connected to a) - gives b) mV across the resistor.
a) Laminations of T1 b) 27.3mV AC
a) Laminations of T2 b) 19.1mV AC
a) Hot output of T2 b) 32.5mV AC
a) Neutral output of T2 b) 17mV AC
a) one side of LV link between T1 & T2 b) 30.8mV AC
So, that means the largest leakage current in the whole thing is 32 uA - which I'm guessing is absolutely fine ?
As an aside, I'm a member of another forum (not electronics related) where I've mentioned that I've built this isolation transformer and one of the guys on there, who I've previously regarded as knowledgeable about electronics, has gone crazy saying I'm going to kill myself if I use this, that it defeats the RCD (I know that !) and that if I were to touch either of the isolated output leads, I would get a potentially fatal shock due to the "capacitance of the human body". I can't see the logic in his claim, as unless I complete the circuit in some way (impossible if the output isn't earth-referenced and I'm not stupid enough to grab both output leads!), I can't see how any current could flow through me, let alone give me a potentially fatal shock.
He's stated that "the circuit is actually completed by your capacitance to earth (all your surroundings, in fact) and the capacitance of the transformer secondary to earth in series - which you have also made far worse (approximately doubled, I think) by earthing the low-voltage link between the two transformers".
Is he talking out of his *** ?