I was studying Electromagnetic topics and there's something that I don't understand. If we have a cell connected to an ideal circuit (like the figure)
The textbooks say that the $\Delta V_{ab}=0$, and I understand that despite the voltage is zero it is able to establish current. But why $\Delta V_{ab}$ is zero? I mean
$\begin{equation*} \displaystyle \Delta V_{ab}=-\int_A^B \vec{E}\cdot \text {d}\vec{r} \neq 0 \end{equation*}$
Or am I wrong?
The description of the circuit is talking about steady-state. Prior to steady-state, there is a field between A and B. This field is sufficient to establish a current. But once established, no field is required to maintain the current in an ideal wire.
In a real wire, there will be a minimal field sufficient to maintain the current against the minimal resistance.
You have draw the electric field that would surround a charge dipole in a vacuum. But the wire is not a vacuum. It has mobile charges that rearrange themselves in response to the field. This rearrangement excludes (or nearly excludes) the electric field between A and B. Therefore your equation holds.
Note that in electric circuits, there is surface charge accumulation along the wire that maintains an electric field along the wire so that charge can move through it, so $V_{AB}\ne 0$.
But in electric circuits (as is probably the case here) when we also assume the wire to be a perfect conductor, we can consider the points A and B to have the same voltage so that the points A and B have no potential difference. In other words, in that case $V_{AB}=0$.
