If I have a trace on a PCB that is wider on the left side and then tapers down to a narrower trace on the right side, why is it that the inductance is greater on the narrower side? I realize there are equations that describe this behavior but I'm just trying to get a qualitative understanding of this. I can see how the capacitance on the wider side would be greater than the capacitance on the narrower side (due to area) but I can't quite see why the inductance is greater on the narrower side. I know that the current density J = I / A and given that we are pushing the same amount of current through the trace, the current density would be greater in the trace with less area. Will a higher current density result in a higher indutance -- is this because it is able to store more energy?