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The starting point is a wire with no current.

We now turn on a DC power source and a current starts flowing.
There was an accelerating/changing current during 'power on'. This causes 1 electromagnetic wave to propagate outward. Correct?
If yes, will there also be an EM wave propagating outwards when the current flow stops?

If a moving charge causes an EM wave to propagate outward, does it mean that every moving electron in a direct current wire is emitting an EM wave?

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  • $\begingroup$ may bee you calculate the velocities of the e! when you turn on or of current you are right , you have a very short and weak wave. $\endgroup$
    – trula
    Mar 9 '20 at 21:46
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    $\begingroup$ Moving charge causes an EM wave to propagate outward. No, accelerating charge causes this. But when there is a loop of accelerating charges, their radiative fields cancel. $\endgroup$
    – G. Smith
    Mar 9 '20 at 23:20
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    $\begingroup$ Does this answer your question? If you run an electric current through a wire loop, do the accelerated charges radiate? $\endgroup$
    – G. Smith
    Mar 9 '20 at 23:21
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Let us begin with a definition, because arguing things without definition may lead to confusion. An electromagnetic wave is defined as an energy wave that has a velocity in vacuum that coincides with the velocity of light in vacuum. According to this definition, gamma rays, X-rays, ultraviolet rays, visible light rays, infra red rays, radio waves, and gravitational waves are known electromagnetic waves. Microwaves, alpha rays, and beta rays are not em waves as per this definition. To produce electric field waves there should be a strain in electric fields, and to produce magnetic field waves there should be a strain in magnetic fields. If the voltage (DC or AC) is increased in a resistance, two successive electrons in serial try to come closer because of voltage, and two electric fields of two electrons should repel each other. So, there is a strain in electric field. If the current is increased in a conductor, two successive electrons in parallel try to come closer because of increased current, and two magnetic fields of moving electrons should repel each other. So, there is a strain in magnetic field. But, for production of magnetic field waves AC is required for variation of directions. These are all only situations related with voltage and current in production of electric field waves and magnetic field waves. In practical situations one need not go for voltage and current. In a fire, electrons of two successive molecules may bring a strain in electric field and it may produce light (or electric field wave). When there is a flow of gases in sun, the moving electrons moving in parallel may bring a strain in magnetic field and it may produce radio waves (or magnetic field waves). DC is sufficient to produce the light waves which may also be considered as electromagnetic waves.

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Yes and no.

Why "no"?
Technically direct current describes a stationary situation, when the current is constant in time. This is obviously not the case just after the switch is turned on, i.e., when the current grows from zero to its steady value. This is also not the case when the switch is turned off and the current decays. To describe the current growing from zero to steady value one would need to use a more refined model than just a resistor conencted to a battery - including wire capacitance (and possibly also its inductance) would do. (Note that any wire has capacitance and inductance, it is just that for short wires these are often neglected, but it is not the case for the electric transmission lines or the microwires on an electronic chip).

When "yes"?
If we consider a straight infinitely long wire, parallel to the $z$-axis and carrying current $I(t)$, we can write the current density as $$ \mathbf{J}(\mathbf{r},t) = \delta(x)\delta(y)I(t)\mathbf{e}_z. $$We can now substitute this current to the Maxwell equations (or better into the electromagnetic wave equations that follow from the Maxwell equations) and solve for the resulting field. Substituting the expressions above into the general solution for the vector potential we have: $$ \mathbf{A}(\mathbf{r}, t)= \int\frac{\delta\left(t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}-t\right)}{|\mathbf{r}-\mathbf{r}'|} \frac{\mathbf{J}(\mathbf{r}', t')}{c}d\mathbf{r}'dt' = \frac{\mathbf{e}_z}{c}\int \frac{I\left(t+\frac{\sqrt{x^2+y^2+(z-z')^2}}{c}\right)}{\sqrt{x^2+y^2+(z-z')^2}}dz' $$ If the current is time-independent, i.e., $I(t)$I_0$, the vector potential is idnependent on time and no electromagnetic waves generated. However, in case of time-dependent current we can expand the equation above itno Fourrier integral, i.e., into a spectrum of electromagnetic waves.

Remarks:
The solution above is simplistic. In real situations there are other factors to account for, such as:

  • the wire is not necessarily straight, and not infinite - the other parts fo teh circuit are also emitting
  • Whether the EM wave actually propagates and is detectable depends on its intensitya nd the environment (typical circuits emit waves that are too weak and quickly decay).
  • One needs to account for the feedback - the change of the current in the circuit due to the rpesence of teh wave, i.e., one really has to solve a self-consistent problem.
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