Current in a Nonuniform Diameter Wire

Question

According to the text Current depends on Area, but then in problems with a "nonuniform diameter" we say that the Current does not change even though the Area changes.

I think I understand why there is some confusion around this, but was hoping someone could verify if the following is an accurate interpretation of Current:

I = dq/dt

dq = (nAdx)q where n = # of charge carriers/Vol

if we assume the wire is cylindrical (not exactly true for a nonuniform diameter) then Adx = Vol right? So then:

dq = q * # of Charge Carriers and does not depend on Area. Therefore I does not depend on A.

It makes sense why do not cancel Area (or volume) because it also cancels dx which is needed to substitute Drift Velocity into the equation.

If this isn't right please let me know. I just find it confusing that current is a constant even if the area is not.

Thanks!

Answer 1

Consider the formula you provide: dq = (nAdx)q. dx has nothing to do with the speed of the electric charges. It is just helps define the boundary within which you are measuring conducting charge. The charges do not have to be moving in this situation. There does not have to be any current. If instead you use dx as the average distance a charge carrier travels in time dt, then dx=vdt with v=drift velocity. As area decreases, drift velocity or charge density will increase. As charge density will often depend on the material, drift velocity is more likely to be the factor that changes. This is very much like the flow rate equation in fluid dynamics. If $\rho$Av is constant for a fluid moving through a closed tube. If the fluid has constant density (in-compressible liquid), then Av is constant. This is why a more narrow wire has a larger resistance. The same current requires that the electrons be moving at a faster speed. This in turn requires a greater force and electric field to maintain, and thus a greater voltage per unit length. These faster electrons transfer greater energy to the atoms with which they collide, thus making the more narrow wire warmer with the same current.