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.
Similar but not equal. All the losses that subtract from 100% efficiency will work in the opposite direction as a dynamo.
So if the motor is 75% efficient under those conditions, then its 100% efficiency speed (unloaded) might be 4000rpm and it would draw 0A. You can roughly crosscheck by monitoring its stall current at the same voltage : in this case, it may draw 24A at 0 rpm. (The motor regulation would be 1000 rpm for 6A, or 167rpm/A, or 24A for 4000 rpm)
If that's the case, then you have motor constants of 4000/20 = 200RPM/volt, and a winding resistance of 20/24 ohms = 0.8333 ohms.
Given these values - and they are based on my guess of 75% efficiency for your motor - the open circuit voltage would be 3000rpm /(200rpm/v) = 15V.
And if you drew 6A from it, you would drop 5V across the winding resistance, so you would see 15-5 = 10V across your load, instead of 20V.
The bridge rectifier would then drop 2* the diode drop (say 1.4V for silicon diodes) giving 8.6V DC. Schottky diodes or synchronous rectifiers can improve on this, but probably at more cost than is justified for such low power.
Measure the motor regulation by running your motor at different loads and measuring both speed and current, and cross-check by measuring stall current, and you can arrive at the likely dynamo performance for your actual motor.
Best Answer
The force that causes motion is due to the interaction of two magnetic fields, at least one of which is created by a current flowing through a conductor. Force is given by F=BIL, which is, essentially, the Lorentz force.
In a generator, a voltage (and consequent current if there's a circuit) is created by moving a conductor in a magnetic field. Mechanical energy is thereby converted to electrical energy (if a current flows). If there is no circuit, there will be zero current, and zero force is required to move the conductor. The induced voltage (actually, an EMF) is E=BLv, where v is the velocity of the conductor wrt the magnetic field.