🚀 go-pugleaf

>If there were an inductor inserted in the circuit, no energy would be carried
>off in EM radiation, because that required perpendicular electric and magnetic
>fields to create a propogating wave. Anyway, an ideal inductor won't lose any
>energy either, it is stored in a magnetic field. I believe the equation that

>Any comments?

Yep.  As long as we remain solely in the realm of *circuit* theory, and
ignore *field* theory, you are correct in the assumption that the circuit
would oscillate "forever", given ideal components, and not lose energy via
EM radiation.

However, your statment above which supposedly eliminates EM radiation as a 
factor is flawed.  Consider the fact that any time-varying magnetic field
will generate a time-varying electric field, and vice-versa.  If there
is a time-varying current in a conductor, there will be a time-varying
magnetic field surrounding that conductor, regardless of whether or not
the conductor is "ideal".  *Some* energy will be lost to radiation; how much
depends on the size, shape, and orientation of the components in the circuit,
as well as the frequency of the signal involved.  The classic circuit-theory
assumption that an inductor (or any "purely reactive" component) loses (or
dissipates) no energy ignores this fact, and rightly so.  With typically-sized
components and frequencies, the effect is probably several orders of
magnitude below the unavoidable resistive losses in real components.  However,
it *does* exist, and this is, after all, a "ideal components and take the
answer to the Nth degree" kind of problem.  The oscillation will not continue
at the same amplitude "forever".

(If you still don't believe that energy can be lost through EM radiation,
then please explain where the ubiquitous 60Hz hum comes from.) 



Bob Myers  KC0EW   HP Graphics Tech. Div.|  Opinions expressed here are not
                   Ft. Collins, Colorado |  those of my employer or any other
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