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How many joules --> e.m. waves?



The capacitance C=30 pF, used in the previous version of this problem, was
too small to validate the formula for the radiation resistance:

Rrad=19200*(d/lambda)^4 only when (d/lambda)<<1.

Here is the new version (C=3000 pF) for which d/lambda is equal to 0.006.
...........................................................................
We are discharging a capacitor in the LCR circuit. Suppose that
V=1 volts, C=3000 pF (1500 pJ of energy is available) and that
the ohmic resistance of the wire is 0.01 ohms/meter. To simplify
assume that the wire through which the current is flowing is a
circular loop whose diameter d=6 m. According to Alex's formula
r=300 cm corresponds to L=0.083 mH. How many pJ are taken away
in the form of electromagnetic waves? The setup is in the vacuum.
..........................................................................

The method of the solution was described two days ago. The efficiency is
defined as Rrad*100%/(Rrad+Rohmic) because an assumption is made that
Rrad and Rohmic share the same current . The new answers are shown below.

d(m) L(mH) Rohmic f(MHz) lambda (m) R_rad(ohms) effic (%)
------------------------------------------------------------------
6 0.083 0.19 0.32 943 3.1e-5 0.0167
2 0.028 0.063 0.55 544 3.5e-6 0.0056
0.6 0.0083 0.019 1.01 298 3.1e-7 0.0017
0.2 0.0028 0.0056 1.74 172 3.5e-8 0.00056
0.06 0.0008 0.0019 3.18 94.3 3.1e-9 0.00017
------------------------------------------------------------------

Note that (d/lambda) decreased by 10, from the first to last line,
while Rrad decreased by 10000. The efficiency drops less rapidly
becuase Rohmic is proportional to d.
Ludwik Kowalski