(ANSWER ASSUMPTION: In this answer, I have assume that the stick is not hinged to the rotor, but actually properly fixed. Also, I've interpreted $\tau_B$ as an internal torque/moment in the rotor-stick system because, unless there is some external effect that is causing a torque to be applied, like someone twisting with their fingers at B, there is nothing to cause an external $\tau_B$ to exist. Feel free to correct me if I have misunderstood your question)
So, to analyse this problem, you have to have a pretty good grasp on exactly what system it is you are analysing. You could look at the whole rotor and stick thing a few ways: you could consider the rotor and stick as two rigid bodies, with equal and opposite reaction forces and moments at the point of attachment, or you could consider it as one whole rigid body, and that the moment you are looking for is an internal force (i.e. if you find the net force/moment a rigid body, internal forces are not involved in the analysis. This is to do with how internal forces and moments have equal and opposite pairs that cancel with each other from an external analysis, like finding net force. Newton's 3rd Law is to blame for this cancelling.) A problem seems to arise in your analysis as you appear to be treating the system as a single rigid body, and you are treating $\tau_B$ as an external force, when it is an internal force for the rotor+stick body.
For the sake of analysis, let's treat the stick and rotor as individual bodies with equal and opposite reaction forces/moments at the point of attachment. The value of the reaction moment will be equal to $\tau_B$. Here is a diagram depicting the two separate bodies:
If you were to consider the rotor and stick as one rigid body for analysis, then the moments and force with purple arrows would be the internal forces in that system. This makes sense as these forces and moments occur in equal and opposite pairs that cancel for external analysis.
Now, all we have to do is to apply the equations of equilibrium for each rigid body.
Taking moments about, say, the left end of the stick:
$$-\tau_B + FL = 0$$
$$\therefore \tau_B = FL$$
Whether the value is positive or negative depends on the sign convention useful for inter-body reaction forces/internal forces or moments.
Just to check that the calculation is consistent, take moments about the centre of the rotor:
$$-\tau_A + \tau_B + Fr = 0$$
$$\therefore \tau_B = \tau_A - Fr$$
Substitute the result of $\tau_A$ you derived earlier and you get the same result. The best point I can make for this issue is to be careful and aware of what system you are analysing.