First a few lines of basics. If you put a loop into the magnetic field and this loop turns within it, the magnetic flux through loop shall change according to the formula
$$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\phi = B A \cos\omega t,$$
where $\vec{B}$ is magnetic field strength, $\vec{A}$ is area of the loop and $\phi$ is angle between $\vec{A}$ (perpendicular to loop) and $\vec{B}$, while $\omega$ is angular velocity of the rotation of the loop.
If you use coil with $N$ loops, then induced voltage on the coil shall be
$$\mathcal{E} = N \frac{\text{d}\Phi_B}{\text{d}t} = - N B A \omega \sin\omega t.$$
Therefore, yes, if you use twice larger magnetic field, you get twice larger voltage.
However, twice larger magnetic field does not necessarily mean it is twice as hard to turn. For example, if you have no electric load on the generator, there is practically no current in the coil and there is practically no Lorentz force! However, if you have completely ohmic load, voltage twice larger means current twice larger and yes it becomes twice as hard to turn.
In practice, you should also consider the internal friction of the generator. So even if there is no load, some muscular power will be required in order to overcome friction. When you increase the electrical load, required power in order to turn generator increases.
Of course, it is difficult to keep rotation constant, which means that with larger angular velocity $\omega$ voltage increases. To keep generator's output voltage constant, you need some electronic circuit, of which the simplest possible includes zener diode.
Does the same voltage results in bigger charges in bigger radii?
If the charges (about some coulombs) numerically equalize or compensate (like doubling both the values simultaneously) the radius (in meters) of the sphere, then the potential indeed remains the same. $$V=\frac{kQ}r$$
The equation simply gives the relationship in $C/m$. Now, doubling the value of coulombs and meters would cancel each other out and so - you'd get the same value of potential...
Best Answer
If the voltmeter is able to measure such high voltages and has a very high internal resistance you can connect it between the sphere and ground and it will display 50kV.
The thickness of the wall doesn't matter at all. So r is the external radius.