EMF due to inductance

A loop of wire, with the ends not touching nor connected to anything else, is this:

^{simulate this circuit – Schematic created using CircuitLab}

If you look at the schematic, it even makes intuitive sense, graphically. As you can see at the gap between the capacitor "plates", the "wires" are not touching. That the plates (which are really just the wires that make up your loop) are really small and far apart doesn't make it not a capacitor: it just makes the capacitance very low, some picofarads at most.

Normally it's so low we can ignore it, but in this case, there's nothing else, so it's pretty key. When the magnetic flux through the loop changes, an EMF is induced just as you'd expect. Some current does flow, according to the governing laws of a capacitor:

$$ I(t) = C\frac{\mathrm{d}V(t)}{\mathrm{d}t} $$

Of course, with C = 1pF, the current will be very low indeed, and if the EMF isn't changing (\$\mathrm{d}V(t)/\mathrm{d}t = 0\$), then there is no current.

There are two approaches of the same phenomena. For an observer located inside the wire, the sources of magnetic field are moving, and therefore, an electric field appears according to Faraday's Law, generating an EMF \$\epsilon\$. On the other hand, an observer located outside the wire (in the same inertial frame of the magnetic field source) will only see the charges (electrons and protons) on the wire moving at the velocity \$\overrightarrow{V}\$. Those charges will experiment a Lorentz force \$\overrightarrow{F}=e\left(\overrightarrow{E}+\overrightarrow{V}\times\overrightarrow{B}\right)\$ and will accumulate at the tips of the wire which creates a potential difference \$\epsilon\$ between the tips, same as the EMF calculated by the other approach.

Therefore, since that potential difference \$\epsilon\$ is proportional to the component of the Lorentz force parallel to the wire, the inclination of it will decrease \$\epsilon\$.

In order to obtain an equation, first consider the wire perpendicular to \$\overrightarrow{V}\$. It can be demonstrated that \$\epsilon = \left|\overrightarrow{V}\right|\left|\overrightarrow{B}\right|L \$ (as you said). Now, consider the wire inclinated an angle \$\theta\$ with respect to \$\overrightarrow{V}\$. The component of the Lorentz force in the direction of the wire will be \$\left|\overrightarrow{F}\right|\sin(\theta)\$, so, the resulting potential difference between tips is \$\epsilon = \left|\overrightarrow{V}\right|\left|\overrightarrow{B}\right|L \sin(\theta)\$