If the work done by a force along a closed path is zero, is it necessarily conservative? I just had a simple doubt. If a force is conservative, we know that the work done by it around a closed path is zero. I believe the converse should also be true. I can't think of any counterexamples in which the work done by the force along a closed path is zero but the force is non conservative. 
However, when asked how to check whether a force is conservative or not by my professor, I suggested the method of checking the work done along a closed path, but he rejected it saying it wasn't always true. Can you give me a counterexample, or explain why it may not always be true? 
 A: A field is conservative if and only if the work around any closed path is $0$. Therefore, if a field is conservative then the work around a single chosen path is guaranteed to be $0$, but this does not mean if we have a field and a single path has a work of $0$ that the field is conservative, as we have only checked one path, not all paths$^*$.
A simple yet contrived example is a a field described by
$$\mathbf F(x,y)=
\begin{cases}
F\,\hat y,  & \text{for $x\geq0$} \\
-F\,\hat y, & \text{for $x<0$}
\end{cases}$$
You could look at the work done around a closed path where the sign of $x$ does not change and find that the work is $0$. However, if you look at the work done along a closed path where the sign of $x$ does change then you could get paths where the work is not $0$. An example of such a path would be a square path that is bisected by the $x=0$ line. Since we have found a closed path where the work is not $0$ the field is not conservative, even though there do exist closed paths where the work is $0$.

$^*$Of course, there are other ways to check if a field is conservative besides explicitly checking the work along every possible path.
