Both of Nedd's diagrams are correct for what they aim to portray BUT your green single turn winding is the one that most portrays the concept of what is involved. A current transformer is "just" an ordinary 1 turn to N turn transformer, even though this fact is usually not obvious because of the way we use it.
There is considerable misunderstanding of what a CT (current transformer) is and how it works. What is not usually realised is that a CT is entirely a "normal" transformer for operating purposes. It will usually be optimised physically and electrically for this special role but it acts as a transformer with a one turn (or few turn) primary and an N turn secondary. If you apply normal transformer 'rules' and electrical analysis the result will be what you expect for a CT or for a "normal transformer (as they are one and the same).
For the ideal case and a 1 to N turn ratio, Rout output "burden resistor" :
Standard transformer equations:
Vin x Iin = Vout x Iout ... 1
Vout = Vin x N and ... 2
Iout = Iin / N ... 3
V = I x R (from Ohms law) so
Vout = Iout x Rout ... 4
But Iout = Iin/N ... 3 above so from 3 & 4
Vout = Iin x Rout / N ... 5
Iin = Vout x N / Rout ... 6
ie we can determine Iin from Vout & turns ration for a given burden resistor Rout. This is in fact trivial (but has snuck up on us) as in 6. Vout/Rout = Iout so 6. reduces to a rearrangement of 3.
The reason that the CT appears to walk funny is that we usually deal with Voltage ratios and are interested in the power and current into a given load.
With a CT we are interested in Iin and Vout. We set the load resistor but do not seek to measure the current in it or the power dissipated. We never ask at all about Vin (the AC drop in the single input turn.
OK - so what? - what has all that got to do with answering the question?
It's relevant because, once we establish that it is an ordinary every day transformer and nothing magic we need to establish that "proper" transformer requirements are met. In the ideal case (and as much as possible in practice) we generate magnetic flux in the primary (one turn) and "link" all that flux with the secondary (N turns). So if we have a single input turn it MUST be a "real" turn. It cannot be half a turn (which in fact cannot exist) or 75% or 99% of a turn. We may run the wire straight through the core physically but if we looked at the magnetic field it will "complete the path" around the core.
So, in your examples the whole single turn shown bu the green line comes closest to showing the single turn to N turns transformer action.
Thanks for your answer, do you think you could explain this graphically as well? That was my main question, how to show the circuit, not just describe it.
As noted above - your greenline version is, in my opinion, the most correct of any I've seen (anywhere ! :-) ) (even though "a little untidy" :-) ).
@Nedd 's left hand diagram effectively shows the same thing BUT is not so obvious. ie As we know that the N turns side goes completely round the core with N turns it implies the 1 turn does this too BUT the implication of this is probably lost on most people.
Your red version is how it usually APPEARS in practice and how it is usually shown BUT not how it really is electrically.
Your blue version is closer to reality electrically but also not how it usually appears OR is shown and is actually probably more confusing even though more correct :-).