| Topic: |
Science > Physics |
| User: |
"qwerty" |
| Date: |
10 May 2007 01:23:27 PM |
| Object: |
Stupid electromagnetics question |
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
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| User: "Androcles" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 03:04:57 PM |
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"qwerty" <qwerty@ihopethisisnotadomain.com> wrote in message =
news:Xns992CD9A0172C6p3ifw90nsdek@193.92.150.76...
I'm having some trouble getting my head around this:
=20
A current flowing through a (static) conductor produces a (static)=20
magnetic field around it, as described by Ampere's Law.
=20
But the reverse appears not to be true; A magnetic field around a=20
conductor doesn't induce current to it, unless the conductor is=20
moving or the magnetic field changes (or both.)
=20
Isn't this a paradox?
A waterfall makes a pool of water, but the reverse appears not to be =
true;
a pool doesn't make a waterfall.=20
Isn't this a paradox? :-)
What would you need to do for the pool to make a waterfall?
Answer: collect the pool in a rubber bulb and squeeze it to make a =
fountain,=20
collecting the water that rose up (as you might do at a drinking =
fountain).
A fountain is an upside down waterfall.=20
If you collapse the magnetic field, then you'll push the current back
the other way. That is essentially what a generator does.
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| User: "Anthony Fremont" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 08:10:53 AM |
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Androcles wrote:
If you collapse the magnetic field, then you'll push the current back
the other way. That is essentially what a generator does.
I'm pretty dumb about these things, but doesn't a collapsing magnetic field
(such as around an inductor) attempt to push the current in the direction it
was going when the field was created? IOW, doesn't it become EMF in the
same direction as the current was flowing already?
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 11:56:54 AM |
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Anthony Fremont wrote:
Androcles wrote:
If you collapse the magnetic field, then you'll push the current back
the other way. That is essentially what a generator does.
I'm pretty dumb about these things, but doesn't a collapsing magnetic field
(such as around an inductor) attempt to push the current in the direction it
was going when the field was created? IOW, doesn't it become EMF in the
same direction as the current was flowing already?
Yeah you are pretty dumb, but apparently not dumb enough! :-) It's
called Lenz's Law and the principle is that when you set up or tear
down a magnetic field it always tends to oppose the action you are
undertaking. Therefore if you are trying to put a current into an
inductor (which sets up the external field) setting up that field will
cause the inductor to resist the current you are trying to push
through it. Similarly, if the inductor already has a current flowing,
the collapsing field will try to keep the current going, namely resist
your actions of trying to STOP the current. so yeah, you are correct.
The basic idea is that in shoving current through an inductor you are
in essence shoving energy into the magnetic field about it where it is
stored. Later when you try to stop the current, all that stored energy
comes back out of the field and does so in a direction that tends to
keep the current flowing in spite of your efforts to stop it.
Benj
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 04:06:38 PM |
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On 10 mai, 14:23, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
Not really.
You sort of assume that the magnetic field you visualize around
the conductor suddenly appears full strength about it.
In reality, it will have to build up to full strength as it is
generated
by current flowing in some other nearby conductor or from the
magnetostatic field of a permanent magnets being moved towards
the conductor .
As this magnetic field intensifies about your conductor, current
will flow in it, but will stop flowing when the magnetic field is
made to stop increasing. Current will start flowing again it
you start decreasing the magnetic field.
Moving current induces magnetic field.
Moving magnetic field induces current.
No paradox.
Andr=E9 Michaud
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| User: "The Great Attractor" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 07:36:25 PM |
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On 10 May 2007 14:06:38 -0700, wrote:
As this magnetic field intensifies about your conductor, current
will flow in it, but will stop flowing when the magnetic field is
made to stop increasing.
No. whenever it stops *moving*. It doesn't have to change strength,
just position.
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| User: "Rock Brentwood" |
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| Title: Re: Stupid electromagnetics question |
16 May 2007 01:17:56 PM |
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On May 10, 1:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
The fields produced by the charge and current are the D and H fields.
The fields *experienced* by the charge and current are the E and B
field.
In a vacuum or near-vacuum, the two sets are linearly related (D =
epsilon_0 E, B = mu_0 H); but not necessarily so -- even in a vacuum
when very close to a charge, to the point of practically being on top
of it. Confusion may result from failing to distinguish sufficiently
these two sets of fields.
If there were such things as magnetic charges and currents, it would
be other way around: the E and B fields would be those produced by
magnetic charges and currently, while the D and H fields would be
those experienced by the magnetic sources.
*That* is the symmetry you're alluding to! It's not there, because
magnetic sources are not known to exist.
In this light, this will also help make more sense of the units in
electromagnetism. The units of electric and magnetic charge are,
respectively, Coulomb and Weber. The corresponding units of their
currents would be Coulomb/second and Weber/second, which are
respectively termed the Amp and Volt.
Electric potential has the same units as magnetic current (and vice
versa, magnetic potential is equivalent in units to electric current).
Older technology, in fact, used such things as magnetic circuits.
Though there are no magnetic monopoles, it's still meaningful to talk
about effective magnetic currents and the like.
This, in turn, will help further make sense of the three fundamental
linear circuit elements and their corresponding units. A resistor and
conductor are measured, respectively, in Ohms and Mhos (and a Mho
which is Ohm spelled backwards is now called a Siemens). An Ohm is a
Volt/Amp or Weber/Coulomb; a Mho a Amp/Volt or Coulomb/Weber.
A capacitor stores electrical energy and has a law of the form I = C
dV/dt (or Q = CV). The units of the capacitance (C) are Farad =
Coulomb/Volt or, equivalently, Amp-second/Volt or Coulomb-second/
Weber. An inductor stores magnetic energy and has a law of the form V
= L dI/dt. If there had been an actual magnetic analogue of a
capacitor, it would have an analogous behavior to an inductor and
satisfy the same relation, only now with V interpreted as magnetic
current, and I as magnetic potential. So, the units of inductance (L)
are Henri = Weber/Amp = Volt-second/Amp = Weber-second/Coulomb.
In the presence of magnetic sources (with charge density s, current
density K) and electric sources (charge density r, current density J),
the field laws would read:
div D = rho; curl H - dD/dt = J
div B = sigma; curl E + dB/dt = -K
the force density would be
F = r E + J x B + s H - K x D
and the power density
P = J.E + K.H
where (x) denotes vector cross product, and (.) scalar product, and d/
dt the partial derivative with respect to time.
In terms of fluxes and circulations, let V denote a volume, and dV the
surface bounding the volume associated with an outward orientation.
Let S denote a surface, with a given orientation attached to it, and
dS the boundary of the surface equipped with an counter-clockwise
orientation around the surface, when the surface is oriented with its
"top" side up.
Denote by D(S), B(S), I(S), K(S) respectively, the fluxes of D, B, J
and K with respect to a given surface S. Let Q(V), P(V) denote the
total charges (corresponding, respectively, to electric charge density
r and magnetic charge density s) contained in a volume V. Let E(C) and
H(C), respectively, denote the circulations of E and H along an
oriented curved C.
Then the 4 equations above read:
D(dV) = Q(V) -- Gauss' Law for electric charges
B(dV) = P(V) -- Gauss' Law for magnetic charges
H(dS) = d/dt D(S) + I(S) -- Ampere's Law for electric currents
-E(dS) = d/dt B(S) + K(S) -- "Ampere's Law" for "magnetic currents"
In the absence of magnetic sources, you have the asymmetry your
original question referred to:
B(dV) = 0, while D(dV) = Q(V)
E(dS) = -d/dt B(S), while H(dS) = d/dt D(S) + I(S).
There's nothing "static" relating to E(dS), as I(S) does to H(dS).
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| User: "RP" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 12:51:12 AM |
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On May 10, 1:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
The magnetic field is just one perspective of the superposed fields of
the drifting charged particles. The B field is just a perspective of
the current that we associate that B field with, rather than a cause
of the current. Only E fields can accelerate charges, which implies
that currents are produced solely by E fields.
Conservation of energy prohibits a current from generating itself.
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 08:30:50 AM |
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On 11 mai, 01:51, RP <no_mail_no_s...@yahoo.com> wrote:
On May 10, 1:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
The magnetic field is just one perspective of the superposed fields of
the drifting charged particles. The B field is just a perspective of
the current that we associate that B field with, rather than a cause
of the current. Only E fields can accelerate charges, which implies
that currents are produced solely by E fields.
Conservation of energy prohibits a current from generating itself.
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
In an electric wire, the E field is contained within the wire. The
E field operates between the electrons moving at the surface
and the atomic nuclei inside.
A distinction must be made between the mathematical
theoretical Gaussian general description of an E field
and an actual E field, and the same for the B field.
In a transformer, you feed an alternating current into the
primary winding wire, meaning that the flow of electrons will
reverse cyclically in the primary wire. But it is the associated
fluctuating B field about the primary wire that induces motion of
electrons into the secondary wire winding.
To produce an real E field external to a wire, you need to charge
surfaces, one with atoms maintained into an electron depleted
state, that will then be electrically positive and the other with
excess electrons that will then be electrically negative.
An E field will then be present between these plates.
In a Betatron, you accelerate charged particles with such
an E field, but you need a very precisely calibrated B field
(separately produced by electromagnets) to contain them
on the very precise circular path.
In linear accelerators, you need to calibrate both E and B
fields such that v=3DE/B for the path to be straight
Ref "Principles of Charged Particle Acceleration" by
Stanley Humpries.
Andr=E9 Michaud
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 11:36:05 PM |
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wrote:
To produce an real E field external to a wire, you need to charge
surfaces, one with atoms maintained into an electron depleted
state, that will then be electrically positive and the other with
excess electrons that will then be electrically negative.
Correct. It's often called a "capacitor" if configured to conform to
circuit theory approximations.
An E field will then be present between these plates.
Yes.
In a Betatron, you accelerate charged particles with such
an E field, but you need a very precisely calibrated B field
(separately produced by electromagnets) to contain them
on the very precise circular path.
No not correct. In a betatron (which as Salmon egg pointed out is a
relativistic device capable of accelerating electrons very close to
the speed of light) electrons are NOT accelerated by an electrostatic
E field. Allow me to quote the freshman physics text, Resnick and
Halliday, 1960 ed. p 756.
" The induced electric field that are set up by the induction process
are not associated with charges but with a changing flux. Although
both kinds of electric fields exert forces on charges there is a
difference between them."
And at p 757.
" Electric fields associated with stationary charges are
*conservative*, but those associated with changing magnetic fields are
*nonconservative*. Since electric potential can only be defined for a
conservative force, it is clear that it has no meaning for electric
fields produced by induction as in a betatron."
Dare I say it? So much for the famous "one E field" dogma!
Of course there are lots of side engineering issues in betatron design
including focusing the beam, keeping the beam circulation in its
"doughnut" vacuum tube at a constant diameter, and injection and
extraction of the beam at the right time. But these are details that
do not impinge on the statements above.
Benj
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 12:16:22 AM |
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On 12 mai, 00:36, Benj <bjac...@iwaynet.net> wrote:
s...@microtec.net wrote:
To produce an real E field external to a wire, you need to charge
surfaces, one with atoms maintained into an electron depleted
state, that will then be electrically positive and the other with
excess electrons that will then be electrically negative.
Correct. It's often called a "capacitor" if configured to conform to
circuit theory approximations.
An E field will then be present between these plates.
Yes.
In a Betatron, you accelerate charged particles with such
an E field, but you need a very precisely calibrated B field
(separately produced by electromagnets) to contain them
on the very precise circular path.
No not correct. In a betatron (which as Salmon egg pointed out is a
relativistic device capable of accelerating electrons very close to
the speed of light) electrons are NOT accelerated by an electrostatic
E field.
My blunder, thanks. I was thinking generally.
I noticed Salmon egg comment after I answered.
I also have this quote regarding the 100 Mev GE Betatron
from Halliday and Resnick (I have the 1967 edition),
page 884
"The magnetic field in the betatron has several functions:
(a) it guides the electrons in a circular path; (b) it accelerates
the electrons in this path; (c) it keeps the radius of the orbit in
which the electrons are moving a constant; (d) it introduces
the electrons into the orbit initially and removes them from
the orbit after they have reached full energy; and finally (e)
it provides a restoring force that resists any tendency for the
electrons to leave their orbit, either vertically or radially. It
is remarkable that it is possible to do all these things by
proper shaping and control of the magnetic field."
It is a pity that the GE betatron is no longer in operation.
I would very much have liked that more experiment would
have been carried out regarding the unexplained energy
loss reported by John P. Blewett. in his paper "Radiation
Losses in the Induction Electron Accelerator", Phys. Rev.
69, 87 (1946).
Allow me to quote the freshman physics text, Resnick and
Halliday, 1960 ed. p 756.
" The induced electric field that are set up by the induction process
are not associated with charges but with a changing flux. Although
both kinds of electric fields exert forces on charges there is a
difference between them."
What I have page 885 (possibly the same quote revised ? )
"The electrons are accelerated by electric fields set up by the
changing
flux [The current in the GE Betatron coils was made to reverse 60
times
per seconds] "
And at p 757.
" Electric fields associated with stationary charges are
*conservative*, but those associated with changing magnetic fields are
*nonconservative*. Since electric potential can only be defined for a
conservative force, it is clear that it has no meaning for electric
fields produced by induction as in a betatron."
Couldn't locate your quotes in my 1967 edition. Could you
refer me to chapters and section numbers ?
Dare I say it? So much for the famous "one E field" dogma!
Of course there are lots of side engineering issues in betatron design
including focusing the beam, keeping the beam circulation in its
"doughnut" vacuum tube at a constant diameter, and injection and
extraction of the beam at the right time. But these are details that
do not impinge on the statements above.
Right. All issues apparently dealt with by the magnetic field
as H&R quote.
Andr=E9 Michaud
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 11:30:29 AM |
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wrote:
Couldn't locate your quotes in my 1967 edition. Could you
refer me to chapters and section numbers ?
It's Chapter 35; Section 35-7 "The Betatron"
My guess is that as time went on and Betatrons fell out of favor as
accelerators, the betatron stuff in the book was dropped or much
reduced.
Heh. Here's a great story! At one point many years ago I was offered
the Betatron at Case Western Reserve University FREE!!! It was a cute
little machine quite a bit smaller than the GE monster in the H&R
photo, but still weighed tons due to the iron. All I had to do was
haul it away! Happily I came to my senses in time and turned them
down!
Benj
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 05:10:28 PM |
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On 12 mai, 12:30, Benj <bjac...@iwaynet.net> wrote:
s...@microtec.net wrote:
Couldn't locate your quotes in my 1967 edition. Could you
refer me to chapters and section numbers ?
It's Chapter 35; Section 35-7 "The Betatron"
Cool, I got it!
In my edition, it became section 35-6 (3 pages worth)
Your quote about conservative-nonconservative is the last
sentence of the previous section (35-5 Time-Varying Magnetic
fields)
What I gather from nonconservative bit is simply that E fields
associated with changing magnetic fields transmit energy
to the test charge, causing it to change its state of motion.
changing magnetic field
My guess is that as time went on and Betatrons fell out of favor as
accelerators, the betatron stuff in the book was dropped or much
reduced.
It is a fact that not much can be found on this wonderful device.
Humphries book "Principles of Charged Particle Acceleration" that
I gave a reference to earlier does pretty much cover all aspects
of betatron operation.
Heh. Here's a great story! At one point many years ago I was offered
the Betatron at Case Western Reserve University FREE!!! It was a cute
little machine quite a bit smaller than the GE monster in the H&R
photo, but still weighed tons due to the iron. All I had to do was
haul it away! Happily I came to my senses in time and turned them
down!
No kidding ! What a loss! Do you remeber how many tons ? Would
you know if they still have it ?
Darn! I must be losing my own senses right now!!!
Andr=E9 Michaud
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| User: "RP" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 10:51:59 PM |
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On May 11, 8:30 am, wrote:
On 11 mai, 01:51, RP <no_mail_no_s...@yahoo.com> wrote:
On May 10, 1:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
The magnetic field is just one perspective of the superposed fields of
the drifting charged particles. The B field is just a perspective of
the current that we associate that B field with, rather than a cause
of the current. Only E fields can accelerate charges, which implies
that currents are produced solely by E fields.
Conservation of energy prohibits a current from generating itself.
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
In an electric wire, the E field is contained within the wire. The
E field operates between the electrons moving at the surface
and the atomic nuclei inside.
A distinction must be made between the mathematical
theoretical Gaussian general description of an E field
and an actual E field, and the same for the B field.
The distinction is that the actual field is not an E field, it is an
em field. Wether it is describable in terms of E and/or B is a matter
of relative motion, meaning that E and B are frame dependent fields
and by virtue of that fact they cannot be representative of reality.
It is only those things that are invariant that even have hope of
being described as real causes.
In a transformer, you feed an alternating current into the
primary winding wire, meaning that the flow of electrons will
reverse cyclically in the primary wire. But it is the associated
fluctuating B field about the primary wire that induces motion of
electrons into the secondary wire winding.
The B field is a composite E field. Again, see Purcell.
To produce an real E field external to a wire, you need to charge
surfaces, one with atoms maintained into an electron depleted
state, that will then be electrically positive and the other with
excess electrons that will then be electrically negative.
The E field is a mathematical representation of the force that a group
of charged paticles exerts on point charges. It is no more than that,
and no less. The B field consists of E fields with opposing vectors
producing a net lateral force on point charges in motion with respect
to the sources of those crossed E fields. All conductors consist of
both negative and positive charges, thus at least two E fields are
associated with any conductor. In current carrying conductors those
two charge components are moving relative to each other, giving rise
to the velocity dependence of the effect known as the B field. This
effect is in turn produced via relativistic changes to the apparent
charge densities of those two components in the conductor with respect
to the external moving charge.
An E field will then be present between these plates.
In a Betatron, you accelerate charged particles with such
an E field, but you need a very precisely calibrated B field
(separately produced by electromagnets) to contain them
on the very precise circular path.
Which can be translated as "a very precisely calibrated set of E
fields (separately produced by electromagnets) to contain them on the
very precise circular path".
In linear accelerators, you need to calibrate both E and B
fields such that v=E/B for the path to be straight
All of the E fields.
Ref "Principles of Charged Particle Acceleration" by
Stanley Humpries.
Don't give a damn about such references :)
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| User: "Salmon Egg" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 11:33:16 AM |
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On 5/10/07 10:51 PM, in article
1178862672.426610.12860@o5g2000hsb.googlegroups.com, "RP"
<no_mail_no_spam@yahoo.com> wrote:
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
If that is the case, explain how betatrons work. Betatrons, for all you you
youngsters, were electron accelerators using a magnetic field to drive them.
At the guts of this is that the electromagnetic field is a consolidated
concept of electric and magnetic fields. This combination is a 4 by 4 tensor
that is invariant under the Lorentz transformations of special relativity.
The components may change values in different uniformly moving coordinate
systems, but represents the sam electromagnetic field in all these
coordinate systems.
Bill
-- Fermez le Bush--about two years to go.
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| User: "Autymn D. C." |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 11:25:08 PM |
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On May 10, 10:51 pm, RP <no_mail_no_s...@yahoo.com> wrote:
The magnetic field is just one perspective of the superposed fields of
the drifting charged particles. The B field is just a perspective of
the current that we associate that B field with, rather than a cause
of the current. Only E fields can accelerate charges, which implies
that currents are produced solely by E fields.
boost, not accelerat
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 01:25:38 AM |
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RP wrote:
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
Allow me to point out a few commonly and widely accepted fallacies
with regard to this matter. One is that a changing E field produces a
B field. I recently checked here on the latest experimental results
attempting to prove the existence of Maxwell's displacement current.
So far as anyone here knows no definitive direct proof of the
existence of a displacement current has EVER been forthcoming!
Point two is that Amperes law is somewhat less than fundamental! It
only works part of the time. Calculations by "flux linkages" will not
give a correct result in all cases.
Point three is that famous "one E field" dogma is completely bogus! An
electrostatic E field is a TOTALLY different animal from the
"supposed" "E" field accelerating electrons in a Betatron. Proof is
that if you place an electron on the pole piece of an electromagnet
and then start increasing the current, the DIRECTION the electron will
be accelerated will depend on it's tiny initial velocity. Electro-
static E fields always provide a force direction!
Point four is that Maxwell's equations, are incomplete as they do not
include the relation, qV x B. Thus Maxwell's equations in spite of
commonplace wild-***** assertions on PBS, do NOT "explain" "All electro-
magnetics" anymore than Newton (or Einstein for that matter)
"explained" gravity.
Benj
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 08:35:31 AM |
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On 11 mai, 02:25, Benj <bjac...@iwaynet.net> wrote:
RP wrote:
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
Allow me to point out a few commonly and widely accepted fallacies
with regard to this matter. One is that a changing E field produces a
B field. I recently checked here on the latest experimental results
attempting to prove the existence of Maxwell's displacement current.
So far as anyone here knows no definitive direct proof of the
existence of a displacement current has EVER been forthcoming!
The proof has been around all along. Just check the fundamental
LC relation.
Andr=E9 Michaud
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 11:17:44 PM |
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wrote:
The proof has been around all along. Just check the fundamental
LC relation.
Andr=E9 Michaud
Sorry Andre, but you cannot use circuit concepts to prove field
theory!
Now if you wanted to say that the existence of EM waves proves the
existence of "displacement current" by inference, Ok. That's what
everyone has done since the 19th century! But the question is if
displacement current is a real thing and not some mathematical error,
then one should be able to come up with some clever experiment to
demonstrate it directly. Apparently nobody has.
Benj
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 12:47:45 AM |
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On 12 mai, 00:17, Benj <bjac...@iwaynet.net> wrote:
s...@microtec.net wrote:
The proof has been around all along. Just check the fundamental
LC relation.
Andr=E9 Michaud
Sorry Andre, but you cannot use circuit concepts to prove field
theory!
Not my intention really. But simply to mention that displacement
current is by structure part and parcel of the LC oscillation.
Now if you wanted to say that the existence of EM waves proves the
existence of "displacement current" by inference, Ok. That's what
everyone has done since the 19th century!
Well, since it is this intuition by Maxwell in 1865 that caused him
to properly define em wave treatment afterwards, it seems to me
that there must be some ground for its validity. For em waves to
exist at all, in the absence of conduction current in vacuum, only
displacement current could possibly be at play as a source of
magnetic field and that the associated changing electric field in
a region of space will induce a magnetic field in neighboring
regions, even when no conduction current and no matter are
present.
Sears, Zemansky & Young do discuss this in their "University
Physics", 6th Edition, Addison Wesley, (1984), page 625.
But the question is if
displacement current is a real thing and not some mathematical error,
then one should be able to come up with some clever experiment to
demonstrate it directly. Apparently nobody has.
To demonstrate it at our macroscopic level, I see only the
LC circuit as an option.
At a more fundamental level, I think there would be need
to experiment with individual photons (is it even possible ?)
De Broglie had an hypothesis in this regard that seems
logical enough to me and that, according to him, satisfied
at the same time Bose-Einstein's statistic and Planck's
Law; and perfectly explained the photoelectric effect
while obeying Maxwell's equations and conforming to
the properties of Dirac's theory of complementary
corpuscles symmetry.
His hypothesis was that discret photons would be
made up not of one corpuscle, but of two corpuscles,
or half-photons, that would be complementary like the
electron is complementary to the positron.
According to him, "Such a complementary couple of
particles (that he perceived as unsigned charges) is
liable to annihilate at the contact of matter by
relinquishing all of its energy, which perfectly
accounts for the characteristics of the photoelectric
effect. "
Furthermore, he said "The photon being made up of
two elementary particles of spin h/4pi, it must obey
the Bose-Einstein statistic as the precision of Planck's
law for the black body requires."
Finally, he concluded that "this model of the photon
allows the definition of an electromagnetic field
linked to the probability of annihilation of the photon,
a field that obeys Maxwell's equations and has all
of the characteristics of electromagnetic light waves."
His two half-photons had to cycle towards and away
from each other (displacement current), inducing a magnetic
field that would increase as the two half-photons close
in on each other, and would decrease as they receide
from each other,
This would by the same token take care of the
magnetic monopole issue, since in this configuration,
both magnetic phases cannot be present at the same
time. Meaning that while the charges symmetrically
separate in space, the magnetic phases symmetrically
separate in time.
But his hypothesis did not really take at the time.
Andr=E9 Michaud
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| User: "Benj" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 11:42:14 AM |
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wrote:
At a more fundamental level, I think there would be need
to experiment with individual photons (is it even possible ?)
Yes, it is.
De Broglie had an hypothesis in this regard that seems
logical enough to me and that, according to him, satisfied
at the same time Bose-Einstein's statistic and Planck's
Law; and perfectly explained the photoelectric effect
while obeying Maxwell's equations and conforming to
the properties of Dirac's theory of complementary
corpuscles symmetry.
Whoa! I've been hearing about this theory for quite some time and was
interested but could never seem to find any definitive information
beyond a description like that above!
His hypothesis was that discret photons would be
made up not of one corpuscle, but of two corpuscles,
or half-photons, that would be complementary like the
electron is complementary to the positron.
According to him, "Such a complementary couple of
particles (that he perceived as unsigned charges) is
liable to annihilate at the contact of matter by
relinquishing all of its energy, which perfectly
accounts for the characteristics of the photoelectric
effect. "
Furthermore, he said "The photon being made up of
two elementary particles of spin h/4pi, it must obey
the Bose-Einstein statistic as the precision of Planck's
law for the black body requires."
Finally, he concluded that "this model of the photon
allows the definition of an electromagnetic field
linked to the probability of annihilation of the photon,
a field that obeys Maxwell's equations and has all
of the characteristics of electromagnetic light waves."
Photon-pair annihilation to explain the photo-electric effect! That is
totally cool!
Do you have a reference to this paper?
Benj
(Who has to admit that he isn't sure just what a half-photon might be!)
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
12 May 2007 05:30:29 PM |
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On 12 mai, 12:42, Benj <bjac...@iwaynet.net> wrote:
s...@microtec.net wrote:
At a more fundamental level, I think there would be need
to experiment with individual photons (is it even possible ?)
Yes, it is.
I'd very like to know how, since they move at c.
De Broglie had an hypothesis in this regard that seems
logical enough to me and that, according to him, satisfied
at the same time Bose-Einstein's statistic and Planck's
Law; and perfectly explained the photoelectric effect
while obeying Maxwell's equations and conforming to
the properties of Dirac's theory of complementary
corpuscles symmetry.
Whoa! I've been hearing about this theory for quite some
time and was interested but could never seem to find any
definitive information beyond a description like that above!
Yes. Well, there is a very special problem with de Broglie's
work. Although he was instrumental with his 1924 thesis
in giving Schr=F6dinger his lead to develop his famous
equation, the community always treated him as some
sort of weird cookie, because he was a causalist.
Meaning that he was convinced that elementary particles
were at all times localized, even as they are moving,
which includes photons and also electrons (and
positrons), and this despite his understanding of
QM as a proper mathematical description (Einstein
agreed with him also).
This idea was discussed at the time, but to my knowledge
no formal paper ever came to be published on this issue.
To my knowledge, no causalist work was accepted for
formal publication since the early 1960's, whoever
promising.
He even had to set up a foundation to make certain
that his work would not be lost.
His hypothesis was that discret photons would be
made up not of one corpuscle, but of two corpuscles,
or half-photons, that would be complementary like the
electron is complementary to the positron.
According to him, "Such a complementary couple of
particles (that he perceived as unsigned charges) is
liable to annihilate at the contact of matter by
relinquishing all of its energy, which perfectly
accounts for the characteristics of the photoelectric
effect. "
Furthermore, he said "The photon being made up of
two elementary particles of spin h/4pi, it must obey
the Bose-Einstein statistic as the precision of Planck's
law for the black body requires."
Finally, he concluded that "this model of the photon
allows the definition of an electromagnetic field
linked to the probability of annihilation of the photon,
a field that obeys Maxwell's equations and has all
of the characteristics of electromagnetic light waves."
Photon-pair annihilation to explain the photo-electric effect! That is
totally cool!
It certainly is.
Do you have a reference to this paper?
As I said, no formal paper was published, to my knowledge.
He was however a very prolific writer and published a number
of books, some of which are still being constantly re-edited
in French. I don't believe there is anything available in English
on this.
The one ref I can give you still in print is this well known book
of his
"La physique nouvelle et les quanta", Flammarion, France 1937,
Second Edition 1993, with new 1973 preface by L. de Broglie
Page 277.
But if you are not convinced yourself of the permanent localization
of elementary particles, I am afraid, even if you can read French,
that what he says will remain just about meaningless to you.
Benj
(Who has to admit that he isn't sure just what a half-photon might be!)
Welcome to the club !
Andr=E9 Michaud
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| User: "RP" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 01:59:24 AM |
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On May 11, 1:25 am, Benj <bjac...@iwaynet.net> wrote:
RP wrote:
Though it is common to speak of changing B fields as generating
currents, this is not technically correct either. A B field can only
change a particle's direction, not its speed, and this is why your
static B field cannot produce a current. The so-called changing B
field that was mentioned by others is in fact literally and exactly an
E field.
Allow me to point out a few commonly and widely accepted fallacies
with regard to this matter. One is that a changing E field produces a
B field. I recently checked here on the latest experimental results
attempting to prove the existence of Maxwell's displacement current.
So far as anyone here knows no definitive direct proof of the
existence of a displacement current has EVER been forthcoming!
Point two is that Amperes law is somewhat less than fundamental! It
only works part of the time. Calculations by "flux linkages" will not
give a correct result in all cases.
Point three is that famous "one E field" dogma is completely bogus! An
electrostatic E field is a TOTALLY different animal from the
"supposed" "E" field accelerating electrons in a Betatron. Proof is
that if you place an electron on the pole piece of an electromagnet
and then start increasing the current, the DIRECTION the electron will
be accelerated will depend on it's tiny initial velocity. Electro-
static E fields always provide a force direction!
Point four is that Maxwell's equations, are incomplete as they do not
include the relation, qV x B. Thus Maxwell's equations in spite of
commonplace wild-***** assertions on PBS, do NOT "explain" "All electro-
magnetics" anymore than Newton (or Einstein for that matter)
"explained" gravity.
Benj
E fields don't come in different types. They're just a mathematical
way of saying that a unit point charge Q will experience such and such
force if located at such and such point in space relative to some
other charge(s). Both the E and B fields are artifices, but there are
real em fields. E and B are just different mathematical perspectives
of the em field.
But there is a problem with common conceptions. To correct those: A
changing B field doesn't cause an E field, it is the changing current
that causes the E field, and that change in current can in turn be
traced to changes in charge densities caused by yet other E fields, or
in other words the induced E field is produced by other E fields, and
all are precisely what you just said that one of them wasn't, Coulomb
forces in action. Even the force associated with the B field is just
superposed Coulomb forces. See Purcell if you want a mainstream
derivation of these conclusions, or see Weber for a previous version
of pretty much the same ideas.
.
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| User: "Randy Poe" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 01:43:36 PM |
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On May 10, 2:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
Is it an asymmetry where in a beautiful and perfect universe
we might expect complete symmetry between the equations
that E and B follow? Yes, it's an asymmetry. The equations
don't have the same form. That's just the universe we live in.
Static charge distributions produce static E fields with
nonzero divergence (for instance, everywhere radially outward)
and zero curl (they can't form closed loops).
Static current distributions produce static B fields with
zero divergence (no purely radial B fields) and nonzero
curl. Not the same.
However, there is a deeper symmetry, in that B and E can
be unified into one "electromagnetic tensor" and the four
Maxwell's equations combined into one:
http://simple.wikipedia.org/wiki/Maxwell's_equations
- Randy
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| User: "Androcles" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 03:25:02 PM |
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"Randy Poe" <poespam-trap@yahoo.com> wrote in message =
news:1178822616.805210.326860@y80g2000hsf.googlegroups.com...
On May 10, 2:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
I'm having some trouble getting my head around this:
A current flowing through a (static) conductor produces a (static)
magnetic field around it, as described by Ampere's Law.
But the reverse appears not to be true; A magnetic field around a
conductor doesn't induce current to it, unless the conductor is
moving or the magnetic field changes (or both.)
Isn't this a paradox?
=20
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
=20
Is it an asymmetry where in a beautiful and perfect universe
we might expect complete symmetry between the equations
that E and B follow? Yes, it's an asymmetry. The equations
don't have the same form. That's just the universe we live in.
=20
Static charge distributions produce static E fields with
nonzero divergence (for instance, everywhere radially outward)
and zero curl (they can't form closed loops).
=20
Static current distributions produce static B fields with
zero divergence (no purely radial B fields) and nonzero
curl. Not the same.
=20
However, there is a deeper symmetry, in that B and E can
be unified into one "electromagnetic tensor" and the four
Maxwell's equations combined into one:
http://simple.wikipedia.org/wiki/Maxwell's_equations
=20
- Randy
I'd have brought my violin if I'd known you were going to sing.
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| User: "qwerty" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 02:23:42 PM |
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Randy Poe <poespam-trap@yahoo.com> wrote in
news:1178822616.805210.326860@y80g2000hsf.googlegroups.com:
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
But, Ampere's Law states that:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html
So if we have B around a conductor, surely we have to have I as well.
That's what the Law states. I know I'm missing something, but what do
I miss?
.
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| User: "Salmon Egg" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 05:11:19 PM |
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On 5/10/07 12:23 PM, in article Xns992CE3D7689CFp3ifw90nsdek@193.92.150.76,
"qwerty" <qwerty@ihopethisisnotadomain.com> wrote:
Randy Poe <poespam-trap@yahoo.com> wrote in
news:1178822616.805210.326860@y80g2000hsf.googlegroups.com:
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
But, Ampere's Law states that:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html
So if we have B around a conductor, surely we have to have I as well.
That's what the Law states. I know I'm missing something, but what do
I miss?
You have to think of Maxwell's equations.
Ampere's law is the equivalent of
Del X H = J + dD/dt
where J is current flow per unit area.
The corresponding equation for E is
Del X E = -dB/dt.
The derivatives should be partial derivatives.
There is no magnetic equivalent for current. That is, there are no magnetic
monopoles although there are electric charge monopoles. Find magnetic
monopoles and you will have a secure place in the history of science.
Fortune too.
Bill
-- Fermez le Bush--about two years to go.
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| User: "" |
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| Title: Re: Stupid electromagnetics question |
11 May 2007 02:29:28 PM |
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On 10 mai, 18:11, Salmon Egg <salmon...@sbcglobal.net> wrote:
On 5/10/07 12:23 PM, in article Xns992CE3D7689CFp3ifw90ns...@193.92.150.7=
6,
"qwerty" <qwe...@ihopethisisnotadomain.com> wrote:
Randy Poe <poespam-t...@yahoo.com> wrote in
news:1178822616.805210.326860@y80g2000hsf.googlegroups.com:
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
But, Ampere's Law states that:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html
So if we have B around a conductor, surely we have to have I as well.
That's what the Law states. I know I'm missing something, but what do
I miss?
You have to think of Maxwell's equations.
Ampere's law is the equivalent of
Del X H =3D J + dD/dt
where J is current flow per unit area.
The corresponding equation for E is
Del X E =3D -dB/dt.
The derivatives should be partial derivatives.
There is no magnetic equivalent for current. That is, there are no magnet=
ic
monopoles although there are electric charge monopoles. Find magnetic
monopoles and you will have a secure place in the history of science.
Fortune too.
Bill
-- Fermez le Bush--about two years to go.
If they existed at all they must always have been around.
Maybe we just don't recognize them yet for what they
are (if they do exist).
Andr=E9 Michaud
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| User: "PD" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 02:51:51 PM |
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On May 10, 2:23 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
Randy Poe <poespam-t...@yahoo.com> wrote innews:1178822616.805210.326860@y80g2000hsf.googlegroups.com:
Paradox? No. Why should they behave the same? What is
the apparent contradiction? E behaves one way, B another.
But, Ampere's Law states that:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/amplaw.html
So if we have B around a conductor, surely we have to have I as well.
That's what the Law states. I know I'm missing something, but what do
I miss?
One is the cause of the other. The source (the current) is the cause;
the field is the effect.
It isn't appropriate to switch cause and effect, any more than it is
appropriate to say that just because F=ma then acceleration causes a
force -- it doesn't.
Now what you *can* say is that if there is a magnetic field in a
certain region of space, then there is likely a current *somewhere*
that is responsible for that field. But if you just happen to put a
wire in that field doesn't mean that the current responsible for the
field has to live in that wire.
A simple example: Take a H-shaped transformer, something you can pick
up at Radio Shack. Apply a DC current to the primary with a 9V battery
in series with a reading lamp bulb (to confirm the current). This will
generate a magnetic field in the transformer core, which you can
confirm by putting a paper clip next to the core. However, you will
register no current in the secondary, as you can confirm with another
reading lamp bulb.
[Note in passing: the *correct* form of Ampere's law notes that there
are two possible sources of a magnetic field: a current, and a time-
dependent electric field.]
PD
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| User: "qwerty" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 03:22:52 PM |
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PD <TheDraperFamily@gmail.com> wrote in
news:1178826711.404256.30170@l77g2000hsb.googlegroups.com:
Now what you *can* say is that if there is a magnetic field in a
certain region of space, then there is likely a current
*somewhere* that is responsible for that field. But if you just
happen to put a wire in that field doesn't mean that the current
responsible for the field has to live in that wire.
I understand it now. If there's a magnetic field somewhere then that
magnetic field is *always* caused by a current. So there's no way to
create the magnetic field of a current-carrying conductor without
using one.
Thanks.
.
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| User: "PD" |
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| Title: Re: Stupid electromagnetics question |
10 May 2007 05:18:24 PM |
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On May 10, 3:22 pm, qwerty <qwe...@ihopethisisnotadomain.com> wrote:
PD <TheDraperFam...@gmail.com> wrote innews:1178826711.404256.30170@l77g2000hsb.googlegroups.com:
Now what you *can* say is that if there is a magnetic field in a
certain region of space, then there is likely a current
*somewhere* that is responsible for that field. But if you just
happen to put a wire in that field doesn't mean that the current
responsible for the field has to live in that wire.
I understand it now. If there's a magnetic field somewhere then that
magnetic field is *always* caused by a current. So there's no way to
create the magnetic field of a current-carrying conductor without
using one.
Thanks.
Keep in mind the weirdness that I added as a footnote. Even a changing
*electric* field can produce a magnetic field. Though this may seem
like a little oddball thing that doesn't happen very often, it is in
fact half responsible for your being able to see anything. (Light.)
PD
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