I do like G3GVB's kind of reasoning. But I think the situation is simplified a
bit too much. Assume as an example that the coil is wound on a toroid of
high permeability material (e.g. for use with a QRP transmitter....).
The magnetic flux in the core is proportional to the current in the
winding. When the current increases from one end of the winding to the
other end this would mean thatt he flux in the core would rise as well
along the core. But this is impossible. The flux must be uniform. So
current in must be equql to current out.
73, Dick, PA0SE
Dear Dick, LF Group,
This is not actually true – there is
no requirement for the current to be equal in all parts of the winding; if the
magnetic flux is the same in all parts of the winding, the voltage across each
turn will be identical ( = d(phi)/dt), but the only requirement on the current
is that the vector sum of all the (current x turns) products flowing in all the
windings will be that required to generate the magnetic flux phi. As an
example, if you divide the winding on your high permeability core into two halves,
and connect an AC source across one half while the other half is left open
circuit, obviously no current will flow in the open circuit half, while the
other half will carry a current determined by the source voltage and the
winding inductance. The voltage across each half of the winding will be equal,
but the current in each half will be different.
The inductance and capacitance of the loading coil depend a lot on how
they are measured and at what frequency – as a rough approximation think of
the total distributed capacitance as small, equal, capacitors between each turn
and ground. If you apply a signal far below the self-resonant frequency of the
coil to one end of the coil, and leave the other end open circuit, the impedance
at that point will nearly be the sum of the small capacitors in parallel. But
as the frequency increases, the inductive reactance of the coil partly cancels
the reactance of the capacitors, reducing the overall reactance, so the overall
effective capacitance is larger. Because the capacitors at the “hot”
open end of the coil have more inductance in series with them, their effective reactance
is reduced more, and so more current flows in them than the capacitors at the
driven end of the coil – this also means that the voltage across the hot
end capacitors is higher than at the driven end. The limiting case is where the
coil and the distributed capacitance becomes self resonant, and most of the
current flows through the capacitance at the hot end, where the voltage is only
limited by the Q. Of course in practice the distributed capacitance along the
coil will not normally be uniform, and there will be extra capacitance attached
to the “hot” end in the form of the antenna, but qualitatively the
same thing will happen – with most of the displacement current flowing to
ground from the high-voltage end of the coil. The difference in current
measured at the two ends of the coil is due to the presence of this
displacement current. The smaller the antenna is, (and so the bigger the coil
and its distributed capacitance is) the greater the fraction of the total
current flows through the distributed capacitance of the coil.
Cheers, Jim Moritz
73 de M0BMU
