Will displacement current form a close loop ?

[ieee std]

Folks:

I have one question. We know normal conductor current form a close
loop,
is this also true for displacement current ?

In the following diagram, assume a an b are two long conductor wires,
form a close loop and source AC current flows in them, c and d are
another two long conductor wires, form a close loop too. There is no
source current in this loop. We want to investigate the induced
displacement due to source current in ab loop.

a b
+ -


+ -
c d


If we denote displacement current
from a to c as I_ac
from b to c as I_bc
from a to d as I_ad
from b to d as I_bd

Will I_ab = - I_bc ? (I do not think so, am I right ? )
Does this depend on the shape of abcd ?

Let I_c = I_ac + I_bc, total displacement current to c,
Let I_d = I_ad + I_bd, total displacement current to d,

will I_c = - I_d ?
Does this depend on the shape of abcd ?

If displacement current must form a close loop, I think I_c = - I_d
must true, but is this the case ?


Thanks.

Bow && Bow && Bow

 

[Rene Tschaggelar]

ieee std wrote:

Folks:

I have one question. We know normal conductor current form a close
loop,
is this also true for displacement current ?

Does normal conductor current form a close loop ?
No, it does not. You can have current without a closed loop.
Not a DC though. A dipole antenna and a capacitor come to
my mind.

In the following diagram, assume a an b are two long conductor wires,
form a close loop and source AC current flows in them, c and d are
another two long conductor wires, form a close loop too. There is no
source current in this loop. We want to investigate the induced
displacement due to source current in ab loop.

a b
+ -


+ -
c d


If we denote displacement current
from a to c as I_ac
from b to c as I_bc
from a to d as I_ad
from b to d as I_bd

Will I_ab = - I_bc ? (I do not think so, am I right ? )
Does this depend on the shape of abcd ?

Let I_c = I_ac + I_bc, total displacement current to c,
Let I_d = I_ad + I_bd, total displacement current to d,

will I_c = - I_d ?
Does this depend on the shape of abcd ?

If displacement current must form a close loop, I think I_c = - I_d
must true, but is this the case ?

The induced voltage is the derivative of the field. Whether the field
changes due to geomety changes, area changes, or source current is
meaningless. And the current follows the voltage with respect
resistances, capacitances, inductances and geometry.
I wouldn't worry too much about the sign of the induced current.

Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net

 

[Kevin Aylward]

Rene Tschaggelar wrote:

ieee std wrote:

Folks:

I have one question. We know normal conductor current form a close
loop,
is this also true for displacement current ?

Does normal conductor current form a close loop ?
No, it does not. You can have current without a closed loop.

The whole point of the "discovery" of displacement current is that it
does precisely, this, close the loop, where a loop didn't apparently
exist. So, no you can't have a "current" without a closed loop.
"Current" must be extended in its definition to anything that is an
effective flow of charge.

Not a DC though. A dipole antenna and a capacitor come to
my mind.

Again, charging the capacitor is what displacement current was
introduced to solve.

Have you never looked at derivations of displacement current? Without it
ampere laws runs into trouble, as does conservation of charge. Just a
simple browse picks up
http://farside.ph.utexas.edu/teaching/em1/lectures/node41.html, of
thousands.

All E&M effects are attributed to the motion of photons. So, in in some
way, displacement current is a measure of how photons are flowing.
However, not necessarily in the same direction, e.g. a field can have
curl even when the flow is in one direction.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Rene Tschaggelar]

Kevin Aylward wrote:

Rene Tschaggelar wrote:

ieee std wrote:


Folks:

I have one question. We know normal conductor current form a close
loop,
is this also true for displacement current ?

Does normal conductor current form a close loop ?
No, it does not. You can have current without a closed loop.


The whole point of the "discovery" of displacement current is that it
does precisely, this, close the loop, where a loop didn't apparently
exist. So, no you can't have a "current" without a closed loop.
"Current" must be extended in its definition to anything that is an
effective flow of charge.

I was somewhat unclear and commented on the normal conductor current.
As you correctly mention : the normal conductor current plus the
displacement current form the loop.
When the displacement current is without a conductor current, then
we have a wave, bounded or unbounded.

Rene