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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?

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  • $\begingroup$ Short answer: no $\endgroup$
    – jensen paull
    Commented May 27, 2022 at 21:18

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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.

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  • $\begingroup$ I used $\vec v$ as the velocity of the particle through the $\vec B$ field because a moving wave passes though the antenna, the same as moving it in a field like a generator. Is this thinking wrong? $\endgroup$
    – Nick
    Commented May 27, 2022 at 22:52
  • $\begingroup$ As the wave passes through the antenna, the charges experience the full lorentz force, this force determines the emf. This situation is slightly different than a generator as the B field is varying $\endgroup$
    – jensen paull
    Commented May 27, 2022 at 23:15
  • $\begingroup$ I am fairy confident, that the magnetic component of the lorentz force does not contribute much to the emf anyway. The reason being, is that when the charges move as a result of the E field, the added emf due to the motion of charges parrallel to the wire, contribute nothing to the emf. The only contribution to the emf that the magnetic component bring, is the emf due to the component of velocity perpendicular to the wire. This, combined with the fact that the magnitude of the v × B field is much smaller than the E field, renders the magnetic contribution insignificant, (I think) $\endgroup$
    – jensen paull
    Commented May 28, 2022 at 0:29
  • $\begingroup$ This is ofcourse because the velocity of the charges in the wire, are mainly due to the Efield in the wire( which is parrel the wire) instead of a forced motion manually. $\endgroup$
    – jensen paull
    Commented May 28, 2022 at 0:31
  • $\begingroup$ The proof for this, is $$\vec{V} = \vec{V}_{par}+ \vec{V}_{per}$$ $$ (\vec{V}_{par} + \vec{V}_{per})× \vec{B} \cdot \vec{dl}$$ $\vec{V}_{par} $ can be written as $|\vec{V}_{par}| \hat dl $ $$(|\vec{V}_{par}| \hat {dl} × \vec{B}) \cdot \vec{dl} = 0 $$ a vector perpendicular to dl, dotted with dl is by definition 0, leaving only the perpendicular component to contribute to the emf $\endgroup$
    – jensen paull
    Commented May 28, 2022 at 0:35
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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.

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  • $\begingroup$ I want sure if Lorentz Force is the right law to apply. I see it is a basis for Faraday induction laws but that is for a close loop $\endgroup$
    – Nick
    Commented May 27, 2022 at 18:51
  • $\begingroup$ Antenna theory is subtle, but there are many paths through it. You might be able to make your formulation work. "God is in the details". $\endgroup$
    – John Doty
    Commented May 27, 2022 at 19:48
  • $\begingroup$ Re: $\vec B$ exciting the $\vec E$, this article says there is a 90 degree phase difference between the two. Not sure if it’s accepted. sciencedirect.com/science/article/abs/pii/S0030402621014431 $\endgroup$
    – Nick
    Commented May 27, 2022 at 22:56
  • $\begingroup$ Theoretically, in the "plane wave" idealization, the magnetic and electric fields are in phase. In the "standing wave" idealization, the fields are 90º out of phase. Mathematically, you may combine plane waves to make standing waves or vice-versa. There are many paths through the mathematical landscape, and antenna theory is difficult. If you really want to learn antennas, they are much simpler as physical objects than as mathematical abstractions. Get a $60 nanoVNA, build some antennas, and see if you can understand what's happening. Then tackle the math. $\endgroup$
    – John Doty
    Commented May 29, 2022 at 16:25