Winding one wire round another doesn't couple them magnetically. You have to
wind both wires round the same magnetic core (which is how current transformers work)
Putting a wire through a toroid counts as 1 turn exactly, even if the wire is straight, note.
However your problem may not need to measure current at all - most mains appliances
are switched on the live side (they really should be!) Thus the voltages on the wires
are approx zero when its off (not quite true, neutral can be 10's of volts sometimes),
but when it is on the live wire is at full mains voltage, the other wire(s) are approx 0,
so the average is about 1/2 of mains.
Capacitively couple a wire to the whole lead (NO ACTUAL CONNECTION, NOTE WELL),
and it will pick up a small AC current due to the average AC voltage thus described.
Take that wire to a high-impedance input and it will exhibit a large enough voltage
You need to have the wire coupled to the mains supply after the switch of course, and not significantly coupled to anything else (detector box nearby?).
The impedance of a few pF at mains frequencies is of the order of 1000M, so you'll need
to avoid loading the wire with less than about 100M. A digital pin used as an input is
fortunately far higher input resistance than that so will see the mains signal easily, and
the input protection diodes will clip it nicely to 0..5V range.
You could run the wire alongside the mains cable for a few feet, or wrap it around a bit,
just be sure the insulation is sound since you only want capacitive coupling... Stray
capacitance of a few pF isn't hard to get this way.
So basically one digital input and a piece of wire may be enough - the hard part will be
keeping interference from other sources out perhaps. If really paranoid about safety add
an inline mains-rated capacitor between the wire and the arduino of 1nF or so.