magnetic field

"Dave" <davidqanta@yahoo.com> wrote in message
news:1116103520.136133.144900@g47g2000cwa.googlegroups.com...
Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave

If the currents were not identical where the difference current go?

Radiated.
 
In article <Xns9656B81A0AFDE831041831041@130.133.1.4>,
Joe Soap <me@privacy.net> wrote:

| >> but only because the flicker frequency is actually 100Hz - the lamp
| >> brightens and dims twice per cycle.
|
| So try feeding a lamp via a diode, and report back on the flicker.

If you feed an incadescent lamp with a 50 Hz signal, it'll flicker at
100 Hz. It won't flicker much, however, as the filament won't have
enough time to really heat up and cool down.

If instead you feed it through a diode, it'll flicker at 50 Hz, and it
will be much more noticable. During half of the 50 Hz cycle, it'll
get no power at all, where without the diode it's power would go up
and down again during that part of the cycle.

Now, if you fed it through a 4 diode bridge rectifier, the flicker
would be back at the level without any diodes, assuming that the
voltage drop caused by the diodes was small compared to the total
voltage.

Flourescent lights and LEDs and other lights flicker more than
incadescent lights because they don't rely on something getting hot to
create light, and so the light emitted is a function of how much power
it's getting *now*, not a function of how much power it got in the
immediate past. (The filament of the bulb cannot cool off much each
cycle, so that decreases the amplitude (but not the frequency) of each
`flicker'.

I'm not sure if that's what you were getting at, Mr Soap, or not.

--
Doug McLaren, dougmc@frenzy.com
Death before dishonor. But neither before breakfast.
 
Doug McLaren wrote:
In article <Xns9656B81A0AFDE831041831041@130.133.1.4>,
Joe Soap <me@privacy.net> wrote:
|
| So try feeding a lamp via a diode, and report back on the flicker.

snip a bunch of correct stuff

I'm not sure if that's what you were getting at, Mr Soap, or not.
I'm not sure we have evidence yet that Mr. Soap knows what he was
getting at...
 
In response to what Doug McLaren <dougmc@frenzy.com> posted in
news:uduhe.87076$hu5.34820@tornado.texas.rr.com:

I'm not sure if that's what you were getting at, Mr Soap, or not.
A diode in the feed is the easiest way of getting 50Hz flicker; full wave
rectification is irrelevant to the discussion. If you feed a sine wave
oscillator at 50Hz to a power amplifier driving a bulb, you will also get
noticeable flicker. The flicker effect will vanish at about 70Hz.

--
Joe Soap.
JUNK is stuff that you keep for 20 years,
then throw away a week before you need it.
 
Joe Soap wrote:
A diode in the feed is the easiest way of getting 50Hz flicker; full wave
rectification is irrelevant to the discussion. If you feed a sine wave
oscillator at 50Hz to a power amplifier driving a bulb, you will also get
noticeable flicker. The flicker effect will vanish at about 70Hz.
Reason for the flicker at 50Hz with a diode is more a function of the
50% off time than the frequency per se.

--
John Miller
email domain: n4vu.com; username: jsm(@)
Surplus (For sale or trade):
Besson International Trumpet by Kanstul
 
On 14 May 2005 13:45:20 -0700, "Dave" <davidqanta@yahoo.com> wrote:

Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave

Current in is same as current out. That is the LAW. The magnetic fields do
not cancel exactly because they are not occupying same space.

Try this for fun: http://motionmountain.com/



Regards,

Boris Mohar

Got Knock? - see:
Viatrack Printed Circuit Designs (among other things) http://www.viatrack.ca
 
On 14 May 2005 16:29:00 -0700, "Dave" <davidqanta@yahoo.com> wrote:

Boris Mohar wrote:
On 14 May 2005 13:45:20 -0700, "Dave" <davidqanta@yahoo.com> wrote:



Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave


Current in is same as current out. That is the LAW. The magnetic
fields do
not cancel exactly because they are not occupying same space.

Try this for fun: http://motionmountain.com/



Regards,

Boris Mohar

I see. In a typical electrical cord, there is a few distance in mm
between the 2 parallel wires with opposing magnetic field. So
not all magnetic field cancel as you say. Do you know of a drawing
or illustration in a site of the residue magnetic field produced
or an illustration of the graphics of the magnetic field being
in the act of being canceled. Thanks.

Dave
I do not wish to sound terse but when you get through that book at the
website I recommended you will be able to make your own drawing. If you are
really thorough with the book you will not need the drawing.

--

Boris Mohar
 
On 14 May 2005 13:45:20 -0700, "Dave" <davidqanta@yahoo.com> wrote:

Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave
If one wire has a leak to ground, the current after the leak is lower.
Generally it woiuld be the hot wire; the other one is at ground
already.
You can buy a GFI ground fault interrupter. It takes both currents
through opposed windings in a relay, but with a leak, the current
difference opens the relay to prevent shock.
John Polasek
http://www.dualspace.net
 
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116073801.168301.293620@g49g2000cwa.googlegroups.com...
John Miller wrote:
Dave wrote:
How can I create a 8Hz Sine Wave 220 volts signal. If
I convert the AC 220 volts to DC. Will I still get 220
volts? So what's the smallest circuit or device that
can make 8Hz sine wave out of the 220 volts DC. Thanks.

At how many amps?

--
John Miller
email domain: n4vu.com; username: jsm(@)
Surplus (For sale or trade):
Besson International Trumpet by Kanstul

Any amperage will do. I guess the steps I must take
is to put some diode to make the 220 volts 60 hz
becoming dc 220 volts. Then put some circuit to
make it produce 8Hz. Right?

What's the schematics of the circuit. As simple
as possible. Thanks.

D
You can use a phase shift oscillator to get the 8 Hz sinewave. Then
amplify it to your power level. At this frequency, a transformer will
be quite large even for low power. It might be easier to use a 220VDC
power supply and a high voltage transistor.
 
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116109787.318838.325760@o13g2000cwo.googlegroups.com...
R.Lewis wrote:
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116103520.136133.144900@g47g2000cwa.googlegroups.com...


Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave

If the currents were not identical where the difference current go?

Radiated.

At what portion of the circuit does it have possibility of
radiating? The circuit is just composed of an electric
bulb and wires plug to AC. You mean the bulb radiate the
missing current as EM?

Dave
Apologies.
My response was rhetorical in that if the currents were not equal the
difference current must be radiated else there must be a mechanism to
'store' an infinite amount of current in your load (electric fan).
Since if the fan was left on a long time this would store all the electrons
in the universe we would have a problem.
Since it isn't radiated the currents must be the same.


..
 
"R.Lewis" <h.lewis@connect-2.co.uk> wrote in message
news:d67h3p$uoh$1@domitilla.aioe.org...
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116109787.318838.325760@o13g2000cwo.googlegroups.com...

R.Lewis wrote:
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116103520.136133.144900@g47g2000cwa.googlegroups.com...


Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave

If the currents were not identical where the difference current go?

Radiated.

At what portion of the circuit does it have possibility of
radiating? The circuit is just composed of an electric
bulb and wires plug to AC. You mean the bulb radiate the
missing current as EM?

Dave


Apologies.
My response was rhetorical in that if the currents were not equal the
difference current must be radiated else there must be a mechanism to
'store' an infinite amount of current in your load (electric fan).
Since if the fan was left on a long time this would store all the
electrons
in the universe we would have a problem.
Since it isn't radiated the currents must be the same.
the currents would be the same even if it was radiated. as you said, the
electrons are not destroyed, so they have to be the same unless there is a
third path for them to get back to the source... such as a ground fault in
the device.
 
Dave schrieb:
....

the currents would be the same even if it was radiated. as you said, the
electrons are not destroyed, so they have to be the same unless there is a
third path for them to get back to the source... such as a ground fault in
the device.
Nevertheless the current which flows into the fan is equal to the
current which leaves the fan. It has nothing to do with the _way_
the current takes when flowing out of the fan, e. g. partially along
a leakage path.

HTH

Reinhard
 
In article <118cb8mf6f7ga4c@corp.supernews.com>, RST Engineering \(jw\) wrote:
That doesn't happen to be true. The so-called flicker rate for the eye is
24 Hz.
Not true generally.

I believe this figure comes from movies having 24 frames per second.
But with movies the dark period is short, the light period is long, and I
am told that projectors interrupt the light twice per frame to make the
flicker 48 Hz.

ANything faster than that is perceived to be steady light. Witness
the standard TV set with a frame rate of 30 Hz. (2 interlaced sets of video
at 60 Hz.). No flicker there that I can see, nor anybody else for that
matter.
60 Hz - I occaisionally see flicker.

56 Hz - done by some monitors in 800x600 mode - I often see flicker.

- Don Klipstein (don@misty.com)
 
In article <%Nphe.1351567$6l.1107625@pd7tw2no>, Dwayne wrote:
Higher frequencies have higher loss.
You body has more resistance to higher frequencies.
You eyes can perceive light flicker at 40 Hertz.
You nerve system runs at ~40 Hertz.

So low frequency is bad for lighting and electrocution. High frequencies are
bad for motors. Thus the compromise.
Electrocution has nothing to do with choice of 50-60 Hz, since these
frequencies are actually close to as bad as it gets for electrocution.
The way I hear it, 50 and 60 Hz are good for motors. Otherwise the best
frequency overall is higher.

- Don Klipstein (don@misty.com)
 
Don Klipstein wrote:
Not true generally.

I believe this figure comes from movies having 24 frames per second.
But with movies the dark period is short, the light period is long, and I
am told that projectors interrupt the light twice per frame to make the
flicker 48 Hz.
In most cases projectors expose each frame twice making the fundamental of
the flicker frequency 48hz (and incidentally meaning that pulldown has to
be complete in 1/96th of a second).

Even with this, a common indication that the screen brightness is too high
is the perception of flicker as the flicker fusion frequency is higher at
greater light levels.
Note that this is distinct from the lower frequency beat effects you can get
between an poorly filtered lamp supply and the second harmonic of the
shutter.....

60 Hz - I occaisionally see flicker.

56 Hz - done by some monitors in 800x600 mode - I often see flicker.
Try turning the brightness down, flicker is MUCH more apparent at high
brightness.

Regards, Dan.
 
Dave wrote:
Hi,

How easy or hard it is to construct a kit that can
produce 12 volts with variable sine wave frequency
from 8Hz to 60Hz? Can I do it with only 6 components
or parts? I forgot many things in electronics so can't
construct complex kit.
My first tought was a RC-stage with a potmeter, feeding to the base of a
transistor...

Also if I have to extend the range to 5Khz to 5Mhz
using the same circuit wherein I can dial the sine
wave from 8Hz to 5Mhz. Is the circuit much more
complex than the first one.

Many thanks.

Dave

--
MVH,
Vidar

www.bitsex.net
 
On Sun, 15 May 2005 18:57:38 +0000 (UTC), don@manx.misty.com (Don
Klipstein) wrote:

I believe this figure comes from movies having 24 frames per second.
But with movies the dark period is short, the light period is long, and I
am told that projectors interrupt the light twice per frame to make the
flicker 48 Hz.
True.


60 Hz - I occaisionally see flicker.
56 Hz - done by some monitors in 800x600 mode - I often see flicker.
75 Hz - I often see flicker (CRT screen).
 
Dave schrieb:
...

I see. So if one wire has 10 mG, the second one has 10 mG opposite
in vector, because they are not in the same position in space, there
is a residual 0.8 mG left. ...
It depends upon _where_you_measure_ the residual magnetic field.
There are points in the space around the wires where a complete
cancellation occurs: with a punctiformly field sensor you'll find
those points. In other points where the measuring device isn't
equidistant to the wires you'll measure the difference between
the "to-field" and the "fro-field" at that point. Okay?

HTH

Reinhard
 
rgregoryclark@yahoo.com wrote:

Then for a diameter of 25mm, you would only need a focal length of
40mm to get a spot diameter of 0.1mm. How are 40mm focal length, 25mm
diameter UV lenses pricewise?
What about using a reflector?


Thomas
 
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116223459.522417.13060@g14g2000cwa.googlegroups.com...
Don Kelly wrote:
"Dave" <davidqanta@yahoo.com> wrote in message
news:1116103520.136133.144900@g47g2000cwa.googlegroups.com...


Suppose you have 220 volts AC and a load such as an
electric fan. Is the current before and after the
load exactly 100% similar in value? Are there loads
wherein the current can become different before and
after the load. If they are all 100% exactly the
same. How come the two parallel electric cord glued
together commonly used in appliances still has residue
magnetic field (like 1.5mG) and they don't cancel 100%.
They theoretically should since the current before and
after the load is exactly the same so the magnetic field
is exactly 180 degrees out of phase so 0 mG output should
suppose to occur.

Thanks.

Dave
-The currents are the same. The reason that there is a residual field
is
because the two current "filaments" are not exactly in the same
location
with respect to anything beyond the wires. The field at a distance r
from
one wire is proportional to I/r while that due to the other is due to
I/(r+d) where d is the distance between the wires. Complete
cancellation
only occurs at points equidistant from both conductors.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
--------


I see. So if one wire has 10 mG, the second one has 10 mG opposite
in vector, because they are not in the same position in space, there
is a residual 0.8 mG left. Do you know a url that has illustration
what part of the waveform are cancelled and what part of the
waveform survives to create the 0.8 mG.

I wonder if the 0.8 mG left in the above configuration is 100%
the same in waveform shape, etc. than a single wire that emit
0.8 mG. What do you think?

Dave
-------------
The above question is partially meaningless- field at a point has no "shape"
It has direction and magnitude- At any given point, the direction of the
field and its magnitude will not be the same as that due to a single wire.

I don't have a site but it is easy to work it out for a particular
situation.
For example consider a 0.5 mm wire centered at 0,0 and another at 2, 0 with
the currents equal and opposite.
B1=mu0I/r for r>0.5 (muoI*r for 0<=r<=0.5 due to internal flux linkages -
lets not get into this detail now)
B2 =-mu0I/(r-2) for 1.5<r and r>2.5 ( muoI*(r-2) for 1.5<=r<=2.5) Same
distribution as B1 except shifted by 2 units.
Assume muoI =0.5 to simplify
Position: B1 B2 Total
0 0 0.25 0.25
0.5 1 0.33 1.33
0.75 0.67 0.4 1.07
1.0 0.5 0.5 1.0
1.25 0.4 0.67 1.07
1.5 0.33 1.0 1.33
2.0 0.25 0 0.25
2.5 0.2 -1 -0.8
3 0.17 -0.5 -0.33
4 0.125 -0.25 -0.125
10 0.05 -0.0625 -0.0125
In this case the fields are + directed in the -y direction.In between the
conductors, the fields add but beyond this region they cancel. The smaller
the conductors and the closer they are, the smaller the field will be
outside the region to left or right of the conductors (i.e. less residual).
Complete cancellation occurs only at infinity but practical cancellation is
much closer
In a general case it is necessary to evaluate the direction as well as the
magnitude of the field due to each current and sum the results as vectors.
This is the same problem as that of finding the fields due to line
electrostatic charges. I have a program for both but the programs are in
APL- which is another learning curve.

If a point under consideration is at a distance R1 at an angle A1 from the
center of conductor 1 and R2 and A2 from the center of conductor 2 (angles
measured ccw), then the fields will have components
By =muI{(cos A1)/R1 -(cos A2)/R2}
Bx =muI{(sin A1)/R1 -(sinA2)/R2}
Total field at this point will be root (Bx^2 +By^2) in the direction
arctan(By/Bx)

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

>
 

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