Why is the amount of power radiated by an oscillating current source much smaller at lower frequencies?
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$\begingroup$ Can you provide some equations which are causing your confusion? And try to quantify "that much," i.e. show the relationship of power emitted vs. electric current frequency? $\endgroup$– Carl WitthoftCommented Jul 31, 2020 at 11:00
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$\begingroup$ My question is not based on some equations I worked with. It is based on applications we see every day. For instance tv or radio antennas vs power chords at home at 60 Hz. $\endgroup$– AriaCommented Jul 31, 2020 at 15:55
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$\begingroup$ Antennas are trying to radiate. Your power cords are not designed to radiate effectively at household frequencies, or you would be wasting a lot of money. $\endgroup$– Jon CusterCommented Apr 11 at 19:45
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2 Answers
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Microscopically, this can be traced back to a basic relationship between the power emitted and the square of the acceleration of charged particles - the radiation terms of the electromagnetic fields are proportional to the acceleration and the power is proportional to the square of these fields - see Larmor's formula.
If a charged particle undergoes a sinusoidal displacement such as $$ x = a\sin (2\pi f t),$$ where $a$ is an amplitude and $f$ is the frequency, then differentiating twice gives the acceleration $$\ddot{x} = -4\pi^2 af^2 \sin(2\pi f t).$$ Squaring this and taking its time average gives $$\langle \ddot{x}^2 \rangle = 8\pi^4 a^2 f^4.$$
There is thus a very strong dependence on the frequency of oscillation and the emitted power.
If expressed in terms of a current, then since current is already proportional to charge multiplied by $\dot{x}$, then $$\ddot{x} \propto f I$$ and so the power emitted is proportional to $f^2 I^2$.
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$\begingroup$ Why the emitted power is proportional to the electron acceleration? $\endgroup$– AriaCommented Jul 31, 2020 at 6:59
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1$\begingroup$ It isn't. It's proportional to the square of the charged particle acceleration. For that you have to prove Larmor's formula @Aria. $\endgroup$– ProfRobCommented Jul 31, 2020 at 7:08
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$\begingroup$ Very minor squabble: $\ddot{x}$ should be negative. +1 from me. $\endgroup$– GertCommented Jul 31, 2020 at 10:53
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$\begingroup$ The question is about power radiated by an oscillating current, the Larmor formula is for single extended charged particle. One can't apply Larmor's formula to all the charged particles in the current separately, because it is only valid for isolated charged particle; synchronously oscillating particles in a line current radiate more strongly than Larmor's formula would give, due to constructive interference of their fields. $\endgroup$ Commented Apr 12 at 12:21
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$\begingroup$ @JánLalinský the proportionality is correct. $\endgroup$– ProfRobCommented Apr 12 at 15:23
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Yes,electric current at lower frequency do not radiate e.m.r that much as compared to higher frequency current. Freq ∝ temp. A rise in temperature of a body increases the intensity (brightness or amount of radiation)of the frequency. There is a difference between intensity and frequency.I*ntensity means amount of radiation while frequency means the rate of vibration.
You can comment me if you are not satisfied with this answer.
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$\begingroup$ This is a bit too unfocussed to be a good answer. You're mixing heating effects (delta resistance with temperature of the wire, perhaps) with the direct radiation caused by accelerating charged particles, which does not require resistance to happen. $\endgroup$ Commented Jul 31, 2020 at 11:03