In my opinion The other answers I saw in this post accurately define the practical results of your question, I very much liked @Norm's answer since he likened it to a cutting the internal wire of a co-axial cable. I will attempt to explain a perhaps more theory based solution by starting with a transient discussion of waves to which steady state could then be interpreted from.
Consider the simple schematic at the end of this post as a reference for my answer, the "Mutual C" represents the break point.
Please pardon the fact that I didn't model the transmission line using all circuit parameters, it is enough just to place that parasitic capacitance to get my point across.
In traveling wave theory, you'd expect a wave to propagate to the end of the conductor (just like water would in a pipe), but to really represent this moving voltage and current wave we have to use inductors and capacitors to show that there is a "time delay" in the voltage and current signals. This is a transfer of energy.
I know that a transmission line is essentially two wires with some series resistance and inductance and some parallel capacitance
You nailed the point right here, and if you use an entirely ideal model the theory will fall apart here. Notice that by simply adding in some parasitic and non-ideal devices we can get a better idea as to what will happen.
the voltage and current in the extra section of long line never sees any change - is that right?
Not quite, consider my follow up question; if there was no voltage or current change at that longer line, then what do you think the voltage would be at the end of that longer line at steady state?
Let's look into this:
A voltage impulse will travel from the source to the break point and charge up the line. There will be reflection at the break point as if it was an open circuit (slightly different but safe to assume).
The beauty of transients is that the circuit does not "know" that that wire "goes nowhere" until the energy reflects and cancels. So there will be an energy wave in the form of a voltage impulse that then travels to the end of the longer line. From here you would find a partial reflection and partial dissipation by an "antenna-like" effect and by parasitic impedances. The current in this path will be negligible from the other sources but it's existence is what validates our schematic. This will continue until there is a standing voltage across all points.
Now remember that question I asked up there? Well now you can say that at steady state the voltage at the end of both wires will approximately be equal in magnitude to the source voltage (given some losses). This agrees with our open-circuit theory of voltage.
simulate this circuit – Schematic created using CircuitLab