So supposedly there is a potential difference between those two rods when the circuit is open. Does this mean theres an electric field present here such that ∫Edl≠0 ?
If I put some random loop of wire in the vicinity (I'm not thinking of Faraday's law, instead just a simple line integral acting on charges), will current start flowing "because theres a potential difference"?
Why does KVL apply to open circuits and how exactly does it?
Is it only true for paths close to the circuit?
Can I take trajectory that goes all the way to infinity and back and still apply KVL?
Can I pick a random trajectory that ends where it started and even if theres a gap, assume that the total voltage drop was the one provided by the voltage source?