A bar moves to the right with a velocity $v$ in a uniform magnetic field. What is the induced current? I know that a change in a magnetic field produces a current, as $-\frac{\partial}{\partial t}\int B\text{d}a = \int E \text{d}l$. In this case however, the magnetic Flux stays constant through the loop so there shouldn't be a net electric field. 
Looking at $F = IL \times B$, I get that there should be an electric field directed downwards in the bar, causing a counterclockwise current. If the direction of movement would be switched, a clockwise current. Is that correct? Does that mean Faradays Law does not apply to this situation, or should I also look at the wires/resistor and find an opposing electric field direction? 

 A: This is to expand on @Jelly_Strawberry 's answer. You wrote "[...]a change in a magnetic field produces a current, as $-\frac{\partial}{\partial t}\int_\mathcal{A} B\text{d}a = \oint_{\partial \mathcal{A}} E \text{d}l$ " (I added the integration bounds.) 
This is Faraday's law in integral form over an arbitrary simple surface, say $\mathcal {A}$, whose boundary is $\partial{\mathcal{A}}$. The quantities $B, E$ are the flux density over the surcae $\mathcal{A}$ and the electric field intensity along the boundary $\partial \mathcal{A}$ of the surface, but notice there is no current mentioned in this formula. What is induced here is a non-zero line integral of the electric field taken along the loop (not wire, loop = arbitrary simple curve, a mathematical curve). 
So if the flux changes in time across a surface then the contour integral of the electric field along the surface boundary is not zero. You get a current flowing if the integrating loop is taken along a conductor, because then the tangential component of the electric cannot be zero everywhere, hence charges are moved along the wire, but the formula is more general and hold even when there is no current flowing. If the integration loop contains an open circuit wire while the rest of the loop is just an arbitrary curve which closes the loop, charges will accumulate at the ends of the wire and forming a dipole whose field inside the conductor will be equal and opposite to the induced field so the charges do not move.
A: I see that I misinterpreted the question. For some reason I thought of the whole circuit moving, while actually only the bar moves. Thus the flux through the circuit does indeed change, and because of Lenz's Law there will be a current flowing clockwise (opposite to the B field). If  is negative, the flux will decrease and the current counterclockwise, as it always opposes the change in lux, somewhat analogous to inertia in mechanical systems. Please correct me if I'm wrong.
