Turning of a paddle wheel when exposed to light
Not the complete answer, but a short analysis of expectable forces:
The energy and momentum of light are related by equation $E=p \cdot c $.
Then force caused by light $F=\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}E}{c \cdot \mathrm{d}t} = \frac Pc$
Therefore the light power 30 W (some cca 50-60W fluorescent bulb or cca 35-40 W LED bulb ) would cause direct  force cca  100 nN, equivalent of cca 0.01 mg.
Or twice as much if reflected.
Crookes radiometer shows the wheel is turning in the opposite direction than it should be, if caused by the light pressure.
The motion is caused by uneven heating and expansion of gas around the wheel, causing pressure differences in the opposite sides of paddles.
When pressure is too low, the gas there is not enough and there remains just light pressure, that is too weak to turn the wheel.

Quoting Effect observations from article.
The effect begins to be observed at partial vacuum pressures of several hundred pascals (or a few torr), reaches a peak at around $\pu{1 Pa}$ ($\pu{7.5e−3 torr}$) and has disappeared by the time the vacuum reaches $\pu{1e−4 Pa}$ ( $\pu{7.5e−7 torr}$) (see explanations note 1). At these very high vacuums the effect of photon radiation pressure on the vanes can be observed in very sensitive apparatus (see Nichols radiometer) but this is insufficient to cause rotation.

Nichols radiometer can register the true light pressure by mirrors on torsion balance.

For cathode rays = electrons
If there is
potential difference $U$
current $I$
elementary charge $e$
electron mass $m_e$
charge $Q$
Area $A$
pressure $P$ (avoiding confusion)
momentum of 1 Coulomb  $p$
electron momentum $p_e$
then for kinematics:
$$eU=\frac 12 m_e.v^2$$
$$v=\sqrt{\frac{2eU}{m_e}}$$
for momentum:
$$p_e=m_e.v=\sqrt{2m_eeU}$$
$$p=\frac 1e \cdot p_e=Q \sqrt{\frac{2m_eU}{e}}$$
for force and pressure
$$F=dp/dt= I \sqrt{\frac{2m_eU}{e}}$$
$$P=F/A= \frac IA \sqrt{\frac{2m_eU}{e}}$$
for $\pu{380 V}$ and $\pu{1 A}$ it gives force about $\pu{66e-6 N}$, what is equivalent to weight of the mass about $\pu{6.6 mg}$.
This is a model case. For a particular scenario, the geometry and electrostatic forces would play significant role.