Induced-EMF due to the induced magnetic field of a wire

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)\$

Yes, it would be zero. According to Faraday's Law the induced EMF would be zero because by moving the loop in that way the magnetic flux would not change.

For an EMF to be induced the magnetic flux must change. This could happen if you spinned the loop so that the angle between \$ \vec{B} \$ and the normal to the loop would change.

But if the magnetic field itself was nonuniform then by moving the loop through areas with different intensities you could induce an EMF.

Edit:

A single wire moving in a magnetic field could have a current, which is explained by Lorentz force: because the wire is being moved so are its charges with velocity \$ \vec{v} \$, and so a force is applied on those charges, hence the current.

The same principle applies to the loop in the picture. Since \$ \vec{v} \$ is to the left and \$ \vec{B} \$ points out of the screen, \$ \vec{v}\times\vec{B} \$ will point down. Thus the current should have that direction (the electrons move in the opposite direction, i.e. up). But now notice both the wire on the left and on the right have currents that cancel out (they have the same absolute value but point opposite ways). So the resulting current is null.