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In AC due to the fact that electrons are accelerated back and forth they emit electromagnetic radiation.

Electromagnetic radiation must depend on the value of AC because more accelerated charges emit more radiation than less accelerated charges

Electromagnetic radiation must also depend on the frequency because higher frequency means that the rate of acceleration of charges is higher. Is there a formula for calculation of the amount of EM radiation from AC?

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In AC due to the fact that electrons are accelerated back and forth they emit electromagnetic radiation.

Ah, hm, no. You can't really explain EM waves with a particle model of current (these concepts clash), but let's just say,

In AC, due to the derivative of current to time, there's geometries in which a EM wave might propagate away from the conductor

because for e.g. perfect transmission lines, or termination resistors, no wave is formed. It's not generally so that whenever you have AC, you have a propagating wave (or "radiation", which, if I had to nitpick, are not the same thing).

Electromagnetic radiation must depend on the value of AC because more accelerated charges emit more radiation than less accelerated charges

Again, forget about the charges; you'll want to read up on Maxwell's equations, as they describe the relation between current (density), its time derivative (and potentially even changes in charge density, but these are not what you mean here).

But yes, Maxwell's equations are linear through and through: you get proportionally stronger fields with growing current.

Electromagnetic radiation must also depend on the frequency because higher frequency means that the rate of acceleration of charges is higher.

This is one of the points where your model with electrons and radiation instead of currents and waves falls flat: You get an EM wave of a different frequency, not a stronger one.

Is there a formula for calculation of the amount of EM radiation from AC?

Maxwell's equations do exactly that.

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  • \$\begingroup\$ But accelerated particles emit radiation we dont need any of the Maxwell equation to prove that we can use the Larmor formula for a single particle. \$\endgroup\$ Oct 26 at 14:43
  • \$\begingroup\$ But: Larmor believed in the Aether. His model of reality describes a lot of the reality we can observe, but it's in direct conflict with a lot of other observations, and it's a bad model that was disproven ~ 100 years ago.... so we rather not stick with any formula from him when it comes to describing things that are not individual charged particles, as the conclusions we draw from that might be wrong! \$\endgroup\$ Oct 26 at 15:09
  • \$\begingroup\$ The approximation is still good in non-relativistic speeds but i think we have started to be missing the point here,dont you agree? \$\endgroup\$ Oct 26 at 15:11
  • \$\begingroup\$ (this would very quickly evolve into a discussion on how electrons are actually pretty slowly moving, and how this is the same whether they change direction 10⁶ or 10⁹ times a second) \$\endgroup\$ Oct 26 at 15:11
  • \$\begingroup\$ no, I don't agree! again, speed of electron \$\ne\$ frequency of current \$\endgroup\$ Oct 26 at 15:11
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Lamor demonstrated that an individual accelerating charged particle emits radiation, i.e. loses energy that is radiated away. So we would say that the fields produced by an accelerating charge have a radiation component.

Maxwell's equations are linear, so when you have a large collection of charges, the total field is the sum of the fields produced by the individual charges. These fields add in amplitude, and so it is possible to have cancellation. It's perfectly possible to have the radiation fields produced by one accelerating charge to be cancelled by the radiation fields produced by the other charges in the collection.

As an example, consider a 2 wire transmission line, we know that there is a solution to Maxwell's equations that is a pure TEM travelling wave along the wires. There is no radiation component - all the energy is directed along the wires. The individual charges on the wires will be moving and accelerating, but their radiation fields MUST cancel in this pure TEM mode. If you isolated the movement of one individual charge, and then removed all the other charges in the system, you would likely find that that moving charge radiated (in the absence of all other charges).

Now, of course, it is impossible to set this up as the TEM fields go off to infinity all around the wires and the wires need to be infinitely long, but in many practical situations this field distribution is a (very) useful approximation, and the deviations from it - including some radiation loss - are minimal.

In antenna and circuit work it is not normally necessary to consider the movement of individual charges, instead to use the current distribution to calculate the cumulative field of all the charges moving in the current.

Added in response to: "Yes okay the radiation fields may cancel out but the energy loss of the electrons will be the same in every part of the circuit "

This is not true, Maxwell's equations are linear in fields quantities, not energy.

Consider a +ve charge moving in simple harmonic motion. It will radiate as it is accelerating, locally you would observe this as a force making it hard to accelerate the particle, and the work you did on the particle to make it follow SHM would equal the radiated energy.

Now consider the same scenario, but this time a -ve charge (equal in magnitude to the first), same outcome. Now place the two charges close together and move them, as a pair, in SHM. In the far field the radiation fields are much smaller as the radiation fields from the two charges tend to cancel, and locally you would find them easier to move as less power is radiated. So cancellation of far fields can allow charged particles to accelerate without (or more probably with minimum) radiation.

If you could just add all the energies, you could radiate EM simply by waving a block of wood around as it contains all those protons and electrons.

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  • \$\begingroup\$ Yes okay the radiation fields may cancel out but the energy loss of the electrons will be the same in every part of the circuit \$\endgroup\$ Oct 27 at 22:12

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