Radio Waves and the Lorentz Force? Is the Lorentz Force involved when the EM wave intersects the receiving antenna? I.e. induces EMF on electrons in the antenna conductor?
I use the RHR with index finger $\vec v$ pointing in the direction of wave propagation towards antenna, middle finger $\vec B$ perpendicular and my thumb $\vec F$ points in the direction of EMF along the antenna wire. Is this correct?
 A: Why have you put V as the velocity of the wave? You're misunderstanding what the right hand rule represents.
The right hand rule, is a tool that is used to visually represent the cross product.
The most common application is to find the magnetic  force on a charge,
$\vec{F}_{B} = q (\vec{v} × \vec{B})$
v is the velocity of the charge, not the wave, and B is the magnetic  field. From your other posts, you seem to think B and V are always perpendicular, this is not true. The cross products mathematical definition just means that B and V are both perpendicular to $\vec{F}_{B}$, which says nothing about the angle between v and B
For EM waves:
To answer your question more specifically, Let's say that I have an electromagnetic  wave moving with a wave vector $\hat k$. this specifies the direction of propagation.
From EM theory, we know that
$\hat E × \hat B = \hat k$
$\hat E \cdot \hat B = 0$
If a EM wave is travelling out of the page, B and E must be in the plane of the screen, perpendicular to another.
Once this EM wave reaches the wire, the charges  are going to feel a force
$\vec{F} = q ( \vec{E} + \vec{v} × \vec{B})$
Let's say I have a wave coming out of the screen, such that the E field at t=0 is horizontal pointing left, and the B field is vertical pointing down.
I then place a conductive wire horizontally.
Each charge is going to then feel the lorentz force. From my setup, the E fields contribution to the emf is going to be a negative emf, measuring from the left hand side to the right of my wire. And then the  magnetic  contribution is going to be zero  as the charges are not initially moving.
This E field is going to ocsscilate, back and forwards, changing the emf across the horizontal wire as time goes on.
Is the B fields contribution to the emf important?
In reality, we need to include the B fields contribution to the emf. Which means we need to use the right hand rule, with the pointer finger in the direction of the velocity of the CHARGE, and B  field as the middle finger, and thumb the force (per unit charge). For this setup however, the force is perpendicular to the wire at all points, so contribute nothing.
In general however, we also know from EM theory, that for plane waves atleast,
$|B| = \frac{|E|}{C}$
meaning $(\vec{v} × \vec{B}) << \vec{E}$, so contributes barely anything anyway.
So in general, the direction of the E field is the thing that determines the EMF.
A: The usual way to model a small dipole receiving antenna is as an electrostatic probe sensing the electric field. Small loop antennas are modeled as sensing the EMF due to the magnetic field.
However, you should note that the way we understand the incoming wave is that the electric field excites the magnetic field, and vice-versa. So, I believe you may model it your way. Try it and see if you can quantitatively match the textbook formulae.
