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)\$
1.
No, there is no interaction between fields. It is always treated this way, that there is an external field, which acts on the local current, while the field generated by this current does not do anything.
It's the same for E-fields: A charge generates a field
$$\vec{E}=\frac{1}{4\pi \varepsilon_0}\frac{Q_1}{|\vec{r}|^2}\frac{\vec{r}}{|\vec{r}|}$$
If you place a second charge, the field of the first results in a force on the second:
$$\vec{F}=\vec{E}\cdot Q_2$$
Due to Newton's actio et reactio, there must be a second force of same strength but opposite direction. You can say that the second force is applied by the second charge to the field, and the field transfers it to the first charge. Or you can say, that the second charge also generates a field, and the first charge is placed in it.
Back to your magnetic fields: The external field applies a force on the wire, and as consequence, the wire also applies a force to the field. And the field transmits this force back to its origin, e.g. a coil or permanent magnet. But you can also say that your wire generates a field, and calculate the force applied to the current in the coil by it. But this is difficult.
2.
Your drawing already shows that the field of the wire is not parallel to the external field, except on the vertical line in the right drawing.
3.
The external field and the field of the wire sum up, so the flux density is higher in the upper half of the right drawing and lower on the lower half. This is the principle of superposition and can be used to calculate e.g. the force on a second wire in your setup. I'm not sure what exactly you are asking for, but in principle, the inductivity does not change. The higher flux density on the one side and lower flux density on the other sum up as if there is no external field, and there is no effect on the inductivity of the wire.
Please note that this is valid in vacuum. If you have a coil with iron yaw in an external field, the external field causes a constant flux in the yaw. If it is too large, you may get saturation effects, which do change the inductivity.
Best Answer
The rate of change of flux is directly proportional to the voltage induced in the coil. Here you say have a solid conductor c connected with terminal hence some current is flowing through it,this current will be opposed by the conductors internal resistance(which will be more in case of the sheet than a wire assuming area of cross-section is fixed),knowing the current in the sheet(derived from the battery) and using Amphere circuital law the magnetic field radiating from the conductor can be calculated.
As you have stated the Emf induced is directly proportional to the product of velocity and the Magnetic field of conductor c will give you the EMF.
The current will flow through wire as load lies in the way it will flow through the load also as it is the only way a closed path can be formed.
Assuming that you are inducing magnetic fields by using the conductor C, the voltage difference across it must be what is across the battery terminals.
Resistance in the magnetic field?
It is called reluctance and it a property of the conductor which depend upon the physical parameters, you can google to find the formula for various shapes.
The concept usually depends on flux coupling the more flux you can get through a surface the more voltage will be induced, but the flux introduced must be differential with respect to time.
For more information I suggest you can watch lectures of Sir Walter Lewin of MIT 8.01 on electromagnetics.