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

[Bob Sciamanda <Sciamanda@worldnet.att.net>]

LUDWIK KOWALSKI wrote:
>                                     Three questions:
> On 22 Apr 1997 Bob Sciamanda wrote:
>
> > Your "rig" is putting out a broad band signal; I don't expect much of a
> > tuning effect over a broad range of the receiver "dial"!
>
> We have a well defined LC circuit with low R. Thus the frequency band of
> the oscillating current is very narrow. Why would the e.m. wave emitted by
> the wire loop have "a broad band"?
If your circuit were driven by a sinusoidal source, its steady state 
signal would be narrow band.  But you are simply giving it a single 
starting pulse and letting it damp out.  If I read your figures
correctly, 
your amplitude damps out with a time constant of the order of 1 msec. 
A spectral analysis of such a wave has an appreciable spread (even 
a truncated sine wave without damping will have a spread in freq.
space).

> > The E field has a null along the normal to the plane of the coil.
>
> That was a big surprize to me. For some reason I used to think that the
> NORMAL to a wire loop antenna must be oriented toward the source to maximize
> the response from a station (same as for the NORMAL to the dipole TV antenna
> on a roof). If this were correct then shouldn't the power emitted be also
> maximized ALONG THE NORMAL? Does E=0 imply zero power?
The radiation E field of both magnetic and electric dipoles go as the 
sine of the angle measured from the direction of the dipole moment
vector.
The two kinds of dipoles have identical fields, except for an
interchange 
of E and H fields, and one change of sign.

PS: There are 5 terms in the H (E) field of the oscillating electric
(magnetic) 
point dipole. Three of them (including the radiation term) have a
sin(theta)
factor; the other two (near field) have a cos(theta) factor.  There are
two 
terms in the E (H) field of the oscillating electric (magnetic)
point dipole.  One of them is the radiation field; both of them have a 
sin(theta) factor.  For both dipoles, the Poynting vector involves a 
{sin(theta)}^2  factor.   Cf Corson & Lorrain or others.

> > Don't under-estimate the sensitivity of (even simple) radio receivers.
>
> Do you really think that a small fraction of 84 pJ (a small fraction of
> less than 1e-10 joules !), intercepted by a tiny antenna of a cheap radio
> set, will not be lost below the noise level? Does anybody know what a
> typical noise level of a radio set is, in term of microwatts?
>                                                          Ludwik Kowalski

I don't think you're asking the right question.  We are looking at a
signal 
source, not a power source.  The pertinent parameter is the E and/or H
field 
at the receiver. (Like a high impedance scope measuring V without
drawing power.) 
You are discharging a cap charged to 100 volts across a fraction of an
ohm.  
The initial current will be huge (but brief) and will generate an E
field well 
in excess (at one meter)of the .9mV/M threshold (plug the numbers into
the equation 
I gave you and see).  And this is counting only the radiation E term in
the 
total field.

The energy in a visible photon is of the order of 10^(-24) J and is not 
difficult to detect. (I know, that's a whole different field, but still
.. . .)
While we're looking at energy scales,  1 ev is only 1.6 X 10^(-19) J so
that 
1 mev = only 1.6 X 10^(-13)J.  My only point is that your 1e-10 J is a
lot 
of signal energy.  Again, here the bottom line is the FIELD strength at
the 
receiver.

Reminds me: as a kid on a farm in the boonies, I used to get miniscule
signals 
from very distant radio stations to palpably activate the headphones of
my 
crystal set (no batteries!).
-- 
Bob Sciamanda		sciamanda@edinboro.edu
Dept of Physics		sciamanda@worldnet.att.net
Edinboro Univ of PA	http://www.edinboro.edu/~sciamanda/home.html
Edinboro, PA		(814)838-7185