Due to path independence the line integral of a conservative vector field for a closed loop is 0–assuming this is over some simply connected region.
In my calculus text book it says that "conservative" implies that it follows the principles of conservation of energy, hence does this imply the inward and outward flux over the same closed loop is also 0?
Best Answer
No a vector field being conservative has more to do with circulation (or work) of the field.
Consider the example $\textbf{F}=x\textbf{i}+y\textbf{j}$ on the path $C$ parameterized by $\textbf{r}(t)=\cos t \textbf{i}+\sin(t)\textbf{j}$ for $0 \le t \le 2\pi$.
The mixed partials are all zero so $\textbf{F}$ is conservative. That is $\int_C \textbf{F}\cdot \textbf{T}\, ds=0$ (circulation or work zero)
But if you look at the outward flux
$$\int_C \textbf{F}\cdot \textbf{n}\, ds=\int_0^{2\pi} \cos^2t+\sin^2t\, dt=\int_0^{2\pi}1dt=2\pi$$