E=F/q=vB Magnetic Force does not work on the charge
gu...@hotmail.com
Quote:"It is ****ERRONEOUS**** to think that the magnetic force does
work on the charges to produce an EMF. We recall that the magnetic
force is PERPENDICULAR to the velocity and hence to the displacement
of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
*****MAGNETIC FORCE********."
If one cannot figure out that the reverse is also true and that a
magnetic force also does not affect a perpendicularly moving charge,
absolutely no interaction occurs.
gu...@hotmail.com
the current and emf.
Further Quote:
"From this it follows immediately that the magnetic field of the
current produced by the induced emf strengthens the field on the side
toward which the wire is moving and weakens it on the opposite side (=
direction of deflection). Once we know the direction of the magnetic
field associated with the induced current, we can let the fingers of
the right hand curve along the lines of force; the right thumb then
points in the direction of the induced emf.
Randy Poe
Which way is the current flowing in this wire? Which way
is the wire moving?
>
> Quote:
Who are you quoting?
> "It is ****ERRONEOUS**** to think that the magnetic force does
> work on the charges to produce an EMF. We recall that the magnetic
> force is PERPENDICULAR to the velocity and hence to the displacement
> of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> *****MAGNETIC FORCE********."
>
> If one cannot figure out that the reverse is also true and that a
> magnetic force also does not affect a perpendicularly moving charge,
> absolutely no interaction occurs.
There is a simple relationship between velocity, magnetic
field, and force experienced by a charge. It is:
F = qv x B
That says that, contrary to your statement, there is indeed
a force when v and B are perpendicular.
There is no force when v and B are parallel.
These are empirically validated rules. Charges moving
through magnetic fields curve in just the direction we'd
expect from qv x B. Look at any bubble-chamber photograph.
>From the radius of such paths you can calculate the
charge to mass ratio.
Take it up with Mother Nature if you disagree.
- Randy
Sue...
> Model: A wire moving perpendicularly through a N&S magnetic field
>
> Quote:"It is ****ERRONEOUS**** to think that the magnetic force does
> work on the charges to produce an EMF. We recall that the magnetic
> force is PERPENDICULAR to the velocity and hence to the displacement
> of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> *****MAGNETIC FORCE********."
Right.
http://teachers.web.cern.ch/teachers/archiv/HST2002/Bubblech/e+%20annihilation.png
On a background of + and - siblings, with orthongonal
magnetic field it just goes in a death spiral.
>
> If one cannot figure out that the reverse is also true and that a
> magnetic force also does not affect a perpendicularly moving charge,
> absolutely no interaction occurs.
Not the individual charge but the *current*.
Beam current or conductor current.
See pictures and answers 6 and 7
http://courses.science.fau.edu/~rjordan/busters_28/answers_4.htm
Sue...
T.M. Sommers
Your quotation does not say there is no interaction, it says no
work is done. Work is the integral of the vector dot product of
the force and the differential displacment. Since the force and
the displacement (which is parallel to the velocity) are always
perpendicular, the dot product is always zero, and no work is
done. If work were done, the speed of the charge would have to
change, and it does not.
--
Thomas M. Sommers -- t...@nj.net -- AB2SB
gu...@hotmail.com
> On Jul 12, 4:39 pm, "g...@hotmail.com" <g...@hotmail.com> wrote:
>
> > Model: A wire moving perpendicularly through a N&S magnetic field
>
> Which way is the current flowing in this wire? Which way
> is the wire moving?
>
>
>
> > Quote:
>
> Who are you quoting?
>
> > "It is ****ERRONEOUS**** to think that the magnetic force does
> > work on the charges to produce an EMF. We recall that the magnetic
> > force is PERPENDICULAR to the velocity and hence to the displacement
> > of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> > and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> > *****MAGNETIC FORCE********."
>
> > If one cannot figure out that the reverse is also true and that a
> > magnetic force also does not affect a perpendicularly moving charge,
> > absolutely no interaction occurs.
>
> There is a simple relationship between velocity, magnetic
> field, and force experienced by a charge. It is:
>
> F = qv x B
>
> That says that, contrary to your statement,
McGrawHill "Elements of Physics" book.
> there is indeed
> a force when v and B are perpendicular.
>
Well they say a magnetic field cannot interact with motion
perpendicular to it, thus it's the charge's magnetic field that is
interacting with B.
> There is no force when v and B are parallel.
>
That is because the charge's magnetic field is perpendicular.
Should the forces be very different if we remove one of the two N&S
magnets....answer = NO
IF we remove one of the two magnets the charge begins to "rotate"
perpendicular to the remaining magnet's magnetic field, the cause of
this rotation is not related to the charge, but the charge's own
magnetic field interacting with the remaining magnets magnetic field.
http://courses.science.fau.edu/~rjordan/busters_28/answers_4.htm (look
at answer6 for charge rotating)
Eric Gisse
> On Jul 12, 4:48 pm, Randy Poe <poespam-t...@yahoo.com> wrote:
>
>
>
> > On Jul 12, 4:39 pm, "g...@hotmail.com" <g...@hotmail.com> wrote:
>
> > > Model: A wire moving perpendicularly through a N&S magnetic field
>
> > Which way is the current flowing in this wire? Which way
> > is the wire moving?
>
> > > Quote:
>
> > Who are you quoting?
>
> > > "It is ****ERRONEOUS**** to think that the magnetic force does
> > > work on the charges to produce an EMF. We recall that the magnetic
> > > force is PERPENDICULAR to the velocity and hence to the displacement
> > > of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> > > and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> > > *****MAGNETIC FORCE********."
>
> > > If one cannot figure out that the reverse is also true and that a
> > > magnetic force also does not affect a perpendicularly moving charge,
> > > absolutely no interaction occurs.
>
> > There is a simple relationship between velocity, magnetic
> > field, and force experienced by a charge. It is:
>
> > F = qv x B
>
> > That says that, contrary to your statement,
>
> McGrawHill "Elements of Physics" book.
Open up a real electrodynamics text.
[...]
Randy Poe
Remember when I said that every time you report on
what somebody else said, you get it wrong?
The quote you provided does not say that. You got
it wrong.
What they said is no WORK is done. If a planet is
in circular orbit around the sun, no work is done
there either for the same reasons (motion is perpendicular
to the force). Would you say the sun has no effect
on the planet? Would you say the sun can not interact
with the planet?
You are now fixated on the equivalence that "no work"
equals "no interaction" and you are flat out wrong.
The authors could sue you for libel for claiming that
they made such an idiotic statement as "the field
can not interact with the charge".
- Randy
Randy Poe
- Randy
gu...@hotmail.com
No but really, what a childish reply, if now you are saying that the
BOOK is incorrect. Then you need to start school again.
> [...]- Hide quoted text -
>
> - Show quoted text -
gu...@hotmail.com
Unfortunately you are correct. A net displacement of zero (such as
delta_x = +5 then delta_x =-5) = zero work even if takes a million
years to do it, mind you energy is lost/depleted.
Work = force x displacement.....in this case the force applied is zero
and thus no displacement and no net displacement or work will occur.
The point is that there is no FORCE at all. The Force is proportional
to the angle, at angle zero the force is 100% at cos 90 the FORCE IS
ZERO.
LIKEWISE
(The words in parantenthesis are my own input)
Quote:"if the wire (or a single moving charge or MULTIPLE moving
charges) lies along the lines of force, there is NO FORCE AT ALL".
> The authors could sue you for libel for claiming that
> they made such an idiotic statement as "the field
> can not interact with the charge".
>
> - Randy- Hide quoted text -
gu...@hotmail.com
Work is a bad thing. W= F d , if F = zero not only W = zero but d also
= zero, power/energy consumption is a more appropriate term that
should have been used?
T.M. Sommers
> On Jul 12, 8:06 pm, Randy Poe <poespam-t...@yahoo.com> wrote:
>
>>What they said is no WORK is done. If a planet is
>>in circular orbit around the sun, no work is done
>>there either for the same reasons (motion is perpendicular
>>to the force). Would you say the sun has no effect
>>on the planet? Would you say the sun can not interact
>>with the planet?
>>
>>You are now fixated on the equivalence that "no work"
>>equals "no interaction" and you are flat out wrong.
>
> Unfortunately you are correct. A net displacement of zero (such as
> delta_x = +5 then delta_x =-5) = zero work even if takes a million
> years to do it, mind you energy is lost/depleted.
>
> Work = force x displacement.
The force and displacement are both vectors, and the work is
their vector dot product, not cross product. You must always
consider the vector nature of force, velocity, etc.
> ....in this case the force applied is zero
> and thus no displacement and no net displacement or work will occur.
> The point is that there is no FORCE at all. The Force is proportional
> to the angle, at angle zero the force is 100% at cos 90 the FORCE IS
> ZERO.
No, there is a force (obviously the sun's gravity does not
magically disappear), but the displacement is at right angles to
the force, so the dot product is zero, and there is no work.
T.M. Sommers
> On Jul 12, 5:47 pm, "T.M. Sommers" <t...@nj.net> wrote:
>>gu...@hotmail.com wrote:
>>
>>>Model: A wire moving perpendicularly through a N&S magnetic field
>>
>>>Quote:"It is ****ERRONEOUS**** to think that the magnetic force does
>>>work on the charges to produce an EMF. We recall that the magnetic
>>>force is PERPENDICULAR to the velocity and hence to the displacement
>>>of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
>>>and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
>>>*****MAGNETIC FORCE********."
>>
>>>If one cannot figure out that the reverse is also true and that a
>>>magnetic force also does not affect a perpendicularly moving charge,
>>>absolutely no interaction occurs.
>>
>>Your quotation does not say there is no interaction, it says no
>>work is done. Work is the integral of the vector dot product of
>>the force and the differential displacment. Since the force and
>>the displacement (which is parallel to the velocity) are always
>>perpendicular, the dot product is always zero, and no work is
>>done. If work were done, the speed of the charge would have to
>>change, and it does not.
>
Not if you want a paycheck.
> W= F d ,
Not exactly. F and d are vectors, and W is the dot product.
> if F = zero not only W = zero
Yes.
> but d also
> = zero,
No. Things can move without a force acting on them. Remember
Newton's first law?
> power/energy consumption is a more appropriate term that
> should have been used?
Power is energy per unit time.
When work is done, kinetic energy does change. In the moving
charge in a magnetic field situation under discussion, the
charge's speed does not change, which is another indication that
no work is done.
Eric Gisse
Open up a real electrodynamics text, so you don't sound like a moron
when attempting to discuss a complicated subject which has been
"taught" to you from a freshman text that is meant to introduce people
to physics rather than give a comprehensive understanding of a
specific topic.
As far as me starting school again....at least I started school. You
were posting stupid shit when I *started* my university education, and
here you are still posting stupid shit as I am finishing it.
Randy Poe
No, we are saying the book is correct and your
interpretations of the book's quote are wildly
inaccurate.
- Randy
Randy Poe
In this case we are talking about a nonzero displacement.
> Work = force x displacement.....in this case the force applied is zero
> and thus no displacement and no net displacement or work will occur.
No, work is the dot product of force and displacement.
The reason there is no work is that the two vectors
are perpendicular, and their dot product is zero.
> The point is that there is no FORCE at all.
No, that is wrong. That is not the point. The point which
the book is explicitly making is that there is very
definitely a force vector, and there is very definitely
a displacement vector, but the work (dot product of these
two vectors) is zero.
- Randy
gu...@hotmail.com
Well son, append what information you have = none, zero, zilch that
"you" think is missing and needed that is not quoted from the book.
Since you are so insistant, otherwise stop complaining.
Randy Poe
Work is not a bad thing. It's how kinetic energy of things
changes.
Gravity acts on a falling body. It changes height. Gravity
does work. The body changes KE.
Gravity acts on a body in circular orbit. It does not
change height. Gravity does not do work on this body.
The body's KE does not change. But the direction most
certainly changes. Gravity has a very definite effect
on that body.
Here is part of the quote you are mangling:
"We recall that the magnetic force is
PERPENDICULAR to the velocity and hence
to the displacement "
That doesn't say the force is zero. That doesn't
say the velocity is zero. That says the two
things are perpendicular. Do you understand that
"perpendicular" does not mean "zero"?
- Randy
gu...@hotmail.com
>
> - Show quoted text -
Wildly?
Androcles
"Randy Poe" <poespa...@yahoo.com> wrote in message
news:1184286043.8...@o61g2000hsh.googlegroups.com...
: You have no clue what is meant by "work", do you?
:
: - Randy
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img157.gif
That's 'W', not 'E'.
since v = 0, W = mc^2 (1-1)
What a prat.
gu...@hotmail.com
You can have zero work (net displacement = zero) but have depleted
energy.
W1 = F (+5), W2 = F(-5) , Wt = W1 + W2 = zero, energy deplete not zero
> Gravity acts on a falling body. It changes height. Gravity
> does work. The body changes KE.
>
> Gravity acts on a body in circular orbit. It does not
> change height. Gravity does not do work on this body.
> The body's KE does not change.
Just a thought:
If the mass or velocity changes then KE will change even if gravity or
magnetic field is constant.
A charge is a wave and current(intensity) varies with time(Hz), thus
mass of charge(s) will also vary with time.
> But the direction most
> certainly changes. Gravity has a very definite effect
> on that body.
>
> Here is part of the quote you are mangling:
> "We recall that the magnetic force is
> PERPENDICULAR to the velocity and hence
> to the displacement "
>
> That doesn't say the force is zero. That doesn't
> say the velocity is zero. That says the two
> things are perpendicular. Do you understand that
> "perpendicular" does not mean "zero"?
>
?? ok but """NET""" EFFECT of force on it is zero.
ReQuote:"It is ****ERRONEOUS**** to think that the magnetic force does
work on the charges to produce an EMF.
Also means:"It is ****ERRONEOUS**** to think that the magnetic force
has a ***NET EFFECT*** on the charges to produce an EMF.
-------------
If you move a wire back and forth(force not perpendicular but lateral)
thus net displacement = zero and thus work = zero and yet an EMF is
created...it seems to contradict the REQUOTE statement above?
gu...@hotmail.com
d as in net displacement (as opposed to distance)
>
> > power/energy consumption is a more appropriate term that
> > should have been used?
>
> Power is energy per unit time.
>
> When work is done, kinetic energy does change. In the moving
> charge in a magnetic field situation under discussion, the
> charge's speed does not change, which is another indication that
> no work is done.
>
If you wish to live and breed by the physics books statement I just
posted:
If the wire is parallel with the magnetic field cos (0) = 1 thus work
is no longer zero and yet no EMF will be produced. (statement says no
EMF because work = zero, work = zero because cos(90) = 0.
> --
> Thomas M. Sommers -- t...@nj.net -- AB2SB- Hide quoted text -
gu...@hotmail.com
d as in net displacement (as opposed to distance)
>
> > power/energy consumption is a more appropriate term that
> > should have been used?
>
> Power is energy per unit time.
>
> When work is done, kinetic energy does change. In the moving
> charge in a magnetic field situation under discussion, the
> charge's speed does not change, which is another indication that
> no work is done.
>
If you wish to live and breed by the physics books statement I just
posted:
If the wire is parallel with the magnetic field cos (0) = 1 thus work
is no longer zero and yet no EMF will be produced. (statement says no
EMF because work = zero, work = zero because cos(90) = 0.
> --
gu...@hotmail.com
the d I'm talking about is net displacement (as opposed to distance)
>
> > power/energy consumption is a more appropriate term that
> > should have been used?
>
> Power is energy per unit time.
>
> When work is done, kinetic energy does change. In the moving
> charge in a magnetic field situation under discussion, the
> charge's speed does not change, which is another indication that
> no work is done.
>
If you wish to live and breed by the physics books statement I posted:
If the wire is parallel with the magnetic field cos (0) = 1 thus work
is no longer zero and yet I believe that no EMF will be produced.
(statement says no EMF because work = zero, work = zero because
cos(90) = 0.
> --
Eric Gisse
[...]
>
> Well son, append what information you have = none, zero, zilch that
> "you" think is missing and needed that is not quoted from the book.
The book is probably fine. You are just too fucking stupid to
understand what is in it.
gu...@hotmail.com
There is TWO displacements, the moving charge's displacement = #1 and
is non zero.
#2 is the displacement as a result of the magnetic force is zero.
Work = force times displacement
Work does not equal displacement without a force..... even if it be
continuous, in such a case I believe W = zero.
> > Work = force x displacement.....in this case the force applied is zero
> > and thus no displacement and no net displacement or work will occur.
>
> No, work is the dot product of force and displacement.
>
> The reason there is no work is that the two vectors
> are perpendicular, and their dot product is zero.
>
If I was to stop/reduce/increase the perpendicular velocity of that
charge, work and displacement and dot product would be non-zero.
> > The point is that there is no FORCE at all.
>
> No, that is wrong. That is not the point. The point which
> the book is explicitly making is that there is very
> definitely a force vector, and there is very definitely
> a displacement vector, but the work (dot product of these
> two vectors) is zero.
>
You are correct, there is definetely a force vector but in short the
force does not interfere with the motion of the charge along it's axis
of motion. At any other angle it does interfere.
gu...@hotmail.com
> On Jul 13, 6:05 am, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
> [...]
>
>
>
> > Well son, append what information you have = none, zero, zilch that
> > "you" think is missing and needed that is not quoted from the book.
>
> The book is probably fine. You are just too fucking stupid to
> understand what is in it.
>
See after "reflection" instead of quick complaints/insults now your
saying the book is fine.
>
>
>
>
> > Since you are so insistant, otherwise stop complaining.- Hide quoted text -
Eric Gisse
>
> > When work is done, kinetic energy does change. In the moving
> > charge in a magnetic field situation under discussion, the
> > charge's speed does not change, which is another indication that
> > no work is done.
>
> If you wish to live and breed by the physics books statement I posted:
>
> If the wire is parallel with the magnetic field cos (0) = 1 thus work
> is no longer zero and yet I believe that no EMF will be produced.
> (statement says no EMF because work = zero, work = zero because
> cos(90) = 0.
Pay attention, dipshit. You might learn something.
Work is int(F.dl) - so W = q*int(v x B.dl). The velocity of a particle
always points in the direction of the infinitesimal displacement, and
since v x B is _perpendicular_ to v by definition [now would be a good
time to learn what a cross product is, moron], W = 0 for the force
applied by a magnetic field. Since you don't know what a goddamn
crossproduct is, I guess I have to explicitly explain something extra:
the dot product of two vectors that are perpendicular is zero.
Or *far* more simply put, since the force is always perpendicular to
the direction of travel, no work is done.
Now for the other half of your confusion: EMF.
EMF = - d\phi/dt where \phi = int(B.da). The electromotive force is
created by changing magnetic flux - that is int(B.da). Which has jack-
fuck all to do with work associated with a force.
Now for the example which you continually misunderstand: the infinite
wire.
Use Ampere's law - int(B.dl) = u_0*I
Since you don't know anything about vector calculus, I will make the
quantum leap of assumptions and assume you don't know what a line
integral is. Since I don't feel like teaching you two years of
electromagnetic theory and vector calculus [ever], you will just have
to accept the result or go fuck yourself.
|B|*2piR = u_0*I
B = [u_0 * I / 2*pi*R] * \phi^ [Consult, as I have told you many times
before, an electrodynamics text if you don't understand]
Notice there is NO cosine function. The only functional dependence is
in the current, I.
I know and understand this material because I spent about 6 semesters
in total learning calculus, vector algebra, vector calculus, and
electromagnetic theory. I have spent the last 4 years of my life
learning physics, whereas you have spent that time doing...well,
playing with your dick for all I know. But it clearly wasn't physics
related.
Eric Gisse
>
> > When work is done, kinetic energy does change. In the moving
> > charge in a magnetic field situation under discussion, the
> > charge's speed does not change, which is another indication that
> > no work is done.
>
> If you wish to live and breed by the physics books statement I posted:
>
> If the wire is parallel with the magnetic field cos (0) = 1 thus work
> is no longer zero and yet I believe that no EMF will be produced.
> (statement says no EMF because work = zero, work = zero because
> cos(90) = 0.
Pay attention, dipshit. You might learn something.
|B|*2piR = u_0*I
>
Randy Poe
The force on a charge moving in a magnetic field
is F = qv x B. The value of a cross product is zero
if the vectors (v and B in this case) are parallel.
The magnitude of the cross product is proportional to
sin(theta), not cos(theta). That is why you are seeing
the statement that if v and B are parallel, the force
is zero.
On the other hand, work is given by a dot product between
force and displacement. Not a cross product. A dot product
varies as the cosine. Not the sine. A dot product
is zero when the vectors are perpendicular. Not parallel.
Can you see that the force and the work are two
different things, governed by different rules?
- Randy
Eric Gisse
[snip stupidity]
> ReQuote:"It is ****ERRONEOUS**** to think that the magnetic force does
> work on the charges to produce an EMF.
That's because it doesn't.
> Also means:"It is ****ERRONEOUS**** to think that the magnetic force
> has a ***NET EFFECT*** on the charges to produce an EMF.
Magnetic force has nothing to do with EMF, fuckwad.
Crack open an actual electrodynamics text. One that makes you use
vector calculus.
[snip stupidity]
Randy Poe
> There is TWO displacements, the moving charge's displacement = #1 and
> is non zero.
>
> #2 is the displacement as a result of the magnetic force is zero.
Stop making up physics.
>
> Work = force times displacement
No, work is the dot product of force and displacement.
- Randy
Eric Gisse
[...]
>
> Work = force times displacement
WRONG.
W = int(F.dl)
How many times do you need to be corrected before you learn what is
taught in introductory physics?
[snip stupidity]
Eric Gisse
> On Jul 13, 11:26 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 13, 6:05 am, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
> > [...]
>
> > > Well son, append what information you have = none, zero, zilch that
> > > "you" think is missing and needed that is not quoted from the book.
>
> > The book is probably fine. You are just too fucking stupid to
> > understand what is in it.
>
> See after "reflection" instead of quick complaints/insults now your
> saying the book is fine.
The book is fine for introducing freshman to physics but completely
shitty for the purpose you are using it for.
It is real sad that you can't even understand what is taught in an
introductory physics course.
Eric Gisse
[...]
>
> Can you see that the force and the work are two
> different things, governed by different rules?
>
> - Randy
No. He cannot.
The google archive has him spewing garbage about physics for four
years. Four fucking years of talking about physics and he still
doesn't know what a cross product is.
gu...@hotmail.com
> On Jul 13, 7:12 am, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
>
> [snip stupidity]
>
> > ReQuote:"It is ****ERRONEOUS**** to think that the magnetic force does
> > work on the charges to produce an EMF.
>
> That's because it doesn't.
>
> > Also means:"It is ****ERRONEOUS**** to think that the magnetic force
> > has a ***NET EFFECT*** on the charges to produce an EMF.
>
> Magnetic force has nothing to do with EMF, fuckwad.
>
Now I was going to really insult you but out of respect and because
Randy is correct and you do study/read a lot.
Mind you your reading capacity to understand short paragraphs is
lacking and therefore you turn to insults and it doesn't mean I don't
make mistakes myself.
How can you say:
> Magnetic force has nothing to do with EMF, fuckwad.
When EMF ******MEANS****** Electro MAGNETIC FORCE
gu...@hotmail.com
> You have no clue what is meant by "work", do you?
>
> - Randy
It's a sickening subject Randy and often confusing.
gu...@hotmail.com
Actually I just ment multiplied for the word "times"
Randy Poe
> On Jul 12, 8:20 pm, Randy Poe <poespam-t...@yahoo.com> wrote:
>
> > You have no clue what is meant by "work", do you?
>
>
It's a transfer of energy. What's so sickening about
that?
- Randy
T.M. Sommers
> On Jul 13, 11:58 am, Randy Poe <poespam-t...@yahoo.com> wrote:
>>On Jul 13, 11:35 am, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
>>
>>>There is TWO displacements, the moving charge's displacement = #1 and
>>>is non zero.
>>
>>>#2 is the displacement as a result of the magnetic force is zero.
>>
>>Stop making up physics.
>>
>>>Work = force times displacement
>>
>>No, work is the dot product of force and displacement.
>
Which is why you were corrected. Work and displacement are
vectors, and work is their dot product.
gu...@hotmail.com
>
> - Show quoted text -
Didn't know this was a course in Calculus or Algebra, how do you
calculate the work without multiplying.
gu...@hotmail.com
I don't know, I'm no expert on work. You displace something +5 units
and then return it -5 units to its exact same initial position and NO
WORK is done. Tell it to the poor guy who broke his back pushing it or
the truck that used up the fuel (power) to do this thing which we
ain't allowed to call work = confusing and easily prone to errors
which is what I evendently just did.
T.M. Sommers
>
> Didn't know this was a course in Calculus or Algebra, how do you
> calculate the work without multiplying.
You take the dot product. More specifically, you integrate the
Randy Poe
> On Jul 14, 8:12 pm, Randy Poe <poespam-t...@yahoo.com> wrote:
>
> > On Jul 14, 4:08 pm, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
>
> > > On Jul 12, 8:20 pm, Randy Poe <poespam-t...@yahoo.com> wrote:
>
> > > > You have no clue what is meant by "work", do you?
>
> > > It's a sickening subject Randy and often confusing.
>
> > It's a transfer of energy. What's so sickening about
> > that?
>
> > - Randy
>
> I don't know, I'm no expert on work. You displace something +5 units
> and then return it -5 units to its exact same initial position and NO
> WORK is done. Tell it to the poor guy who broke his back pushing it or
> the truck that used up the fuel (power) to do this thing which we
> ain't allowed to call work
It isn't work on the object. It doesn't transfer any energy
to the object. At the end of this period the object has
no more kinetic energy, and no more potential energy. Since
the object has not changed in energy, no energy was transferred
to it, it makes sense to say no work was done on it.
On the other hand, the guy pushing it against friction
did do work. Work = F . d where the "." means dot
product. Every meter he pushes against F Newtons of force,
he expends at least Fd Joules of energy. I'm not going
to tell that guy he didn't do work, since he did. He
just didn't do it on the object. Every joule he expended
goes into friction and is wasted as heat. It doesn't change
the energy state of the object at all.
> = confusing and easily prone to errors
> which is what I evendently just did.
You would be less prone to errors if you took a course
of some kind rather than trying to make up physics
yourself.
And I don't just mean read books. That doesn't seem to
be a productive path for you. You tend to get out of
books statements that the authors never made. Indeed,
sometimes you get the exact opposite of what the authors
said. Then you blame it on the authors.
- Randy
gu...@hotmail.com
>
> > I don't know, I'm no expert on work. You displace something +5 units
> > and then return it -5 units to its exact same initial position and NO
> > WORK is done. Tell it to the poor guy who broke his back pushing it or
> > the truck that used up the fuel (power) to do this thing which we
> > ain't allowed to call work
>
> It isn't work on the object. It doesn't transfer any energy
> to the object. At the end of this period the object has
> no more kinetic energy, and no more potential energy. Since
> the object has not changed in energy, no energy was transferred
> to it, it makes sense to say no work was done on it.
>
> On the other hand, the guy pushing it against friction
> did do work.
electromotive mechanics.
Sometimes if the guy also started and stopped at same location as the
box = zero displacement for the guy (not only the box) thus one might
determine the guy (as well as the box) has work = zero thus a similar
effect in the more complex quantum or electromotive mechanics can
EASILY lead to mistakes.
> Work = F . d where the "." means dot
> product. Every meter he pushes against F Newtons of force,
> he expends at least Fd Joules of energy. I'm not going
> to tell that guy he didn't do work, since he did. He
> just didn't do it on the object. Every joule he expended
> goes into friction and is wasted as heat. It doesn't change
> the energy state of the object at all.
>
Yes but how do you know when work will generate a reaction based on
the object returning to it's point of origin.
Example displaces a wire back in forth in a magnetic field, then stop
the wire at it's initial position = the wire has received no work
(it's total displacement = zero)....but one cannot ****ALWAYS****
(maybe this case I understand) tell which work will generate
electricity and which one will not.
In fact you were clarifying to me in the past about perpendicular
force (magnetic force) having achieved no work, this is a latteral
force which also achieved no work (in total??) but did generate
electricity.
I passed Physics with memory of their systematic logic, but not my
full understanding of it. They never go into such deep/intricate
detail.
(THINGS CAN GET EVEN MORE COMPLEX...current varies with time thus so
does it's total mass (# of charges) you have a perpendicular Magnetic
Force will generate a difference of Kinetic Energy(since the mass/# of
charges varies with time)...thus Work does not equal zero arising from
this perpendicular magnetic force?)
The Ghost In The Machine
<t...@nj.net>
wrote
on Sun, 15 Jul 2007 11:20:51 -0400
<469a3b52$0$25595$470e...@news.pa.net>:
> gu...@hotmail.com wrote:
>>
>> Didn't know this was a course in Calculus or Algebra, how do you
>> calculate the work without multiplying.
>
> You take the dot product. More specifically, you integrate the
> dot product.
>
The dot product is a sum of scalar products, at least as
conventionally expressed. :-)
--
#191, ewi...@earthlink.net
"Your mother was a hamster and your father smelt of
elderberries!" - Monty Python and the Holy Grail
--
Posted via a free Usenet account from http://www.teranews.com
gu...@hotmail.com
<ew...@sirius.tg00suus7038.net> wrote:
> In sci.physics.relativity, T.M. Sommers
> <t...@nj.net>
> wrote
> on Sun, 15 Jul 2007 11:20:51 -0400
>
> > gu...@hotmail.com wrote:
>
> >> Didn't know this was a course in Calculus or Algebra, how do you
> >> calculate the work without multiplying.
>
> > You take the dot product. More specifically, you integrate the
> > dot product.
>
> The dot product is a sum of scalar products, at least as
> conventionally expressed. :-)
>
> --
> #191, ewi...@earthlink.net
> "Your mother was a hamster and your father smelt of
> elderberries!" - Monty Python and the Holy Grail
>
> --
> Posted via a free Usenet account fromhttp://www.teranews.com
I simply 3x3=9 and nothing more. the others can dot as much as they
want.
Randy Poe
Since work is the integral of F.d over the entire
path, you know whether work was done by seeing whether
this integral is zero or nonzero.
That would be if you wanted to use the actual physical
principle instead of making physics up.
> Example displaces a wire back in forth in a magnetic field, then stop
> the wire at it's initial position = the wire has received no work
> (it's total displacement = zero)....but one cannot ****ALWAYS****
> (maybe this case I understand) tell which work will generate
> electricity and which one will not.
Yes, you can. You integrate F.d and see whether the
result is zero or nonzero.
> In fact you were clarifying to me in the past about perpendicular
> force (magnetic force) having achieved no work, this is a latteral
> force which also achieved no work (in total??) but did generate
> electricity.
>
> I passed Physics with memory of their systematic logic, but not my
> full understanding of it. They never go into such deep/intricate
> detail.
>
> (THINGS CAN GET EVEN MORE COMPLEX...current varies with time thus so
> does it's total mass (# of charges) you have a perpendicular Magnetic
> Force will generate a difference of Kinetic Energy(since the mass/# of
> charges varies with time)...thus Work does not equal zero arising from
> this perpendicular magnetic force?)
If the integral of F.d is nonzero, then work is done.
- Randy
gu...@hotmail.com
>
> - Show quoted text -
Not good with intergrals. Could work = 0, even if integral of F.d is
nonzero?
Don't know how a moving wire (in a magnetic field) can generate
electricity if it's work = 0 (meaning it's final resting point is it's
original starting point) where as Power = W/t ?
T.M. Sommers
> In sci.physics.relativity, T.M. Sommers
> <t...@nj.net>
> wrote
> on Sun, 15 Jul 2007 11:20:51 -0400
> <469a3b52$0$25595$470e...@news.pa.net>:
>>gu...@hotmail.com wrote:
>>
>>>Didn't know this was a course in Calculus or Algebra, how do you
>>>calculate the work without multiplying.
>>
>>You take the dot product. More specifically, you integrate the
>>dot product.
>
> The dot product is a sum of scalar products, at least as
> conventionally expressed. :-)
You don't just multiply the magnitudes of the two vectors, though.
T.M. Sommers
> On Jul 15, 1:11 pm, The Ghost In The Machine
> <ew...@sirius.tg00suus7038.net> wrote:
>>In sci.physics.relativity, T.M. Sommers
>><t...@nj.net>
>> wrote
>>on Sun, 15 Jul 2007 11:20:51 -0400
>><469a3b52$0$25595$470ef...@news.pa.net>:
>>>gu...@hotmail.com wrote:
>>
>>>>Didn't know this was a course in Calculus or Algebra, how do you
>>>>calculate the work without multiplying.
>>>
>>>You take the dot product. More specifically, you integrate the
>>>dot product.
>>
>>The dot product is a sum of scalar products, at least as
>>conventionally expressed. :-)
>
> want.
Then you will get the wrong answer. The dot product depends on
the angle between the two vectors.
gu...@hotmail.com
If there be an angle I would at it to the formula, if there it be
perpendicular or parallel I doth not.
gu...@hotmail.com
> The Ghost In The Machine wrote:
>
> > In sci.physics.relativity, T.M. Sommers
> > <t...@nj.net>
> > wrote
> > on Sun, 15 Jul 2007 11:20:51 -0400
> >>gu...@hotmail.com wrote:
>
> >>>Didn't know this was a course in Calculus or Algebra, how do you
> >>>calculate the work without multiplying.
>
> >>You take the dot product. More specifically, you integrate the
> >>dot product.
>
> > The dot product is a sum of scalar products, at least as
> > conventionally expressed. :-)
>
> You don't just multiply the magnitudes of the two vectors, though.
>
> --
> Thomas M. Sommers -- t...@nj.net -- AB2SB
F = qvB they don't bother writting an angle if there isn't one. If
there is one then they write
F= qvB sin(angle).
I have a McGrawHills Physics SCHOOL book = what they teach and not
just an information book and they don't even use the "X" (meaning
they write F=qvB instead of F= qv X B)
T.M. Sommers
> On Jul 15, 9:43 pm, "T.M. Sommers" <t...@nj.net> wrote:
>>The Ghost In The Machine wrote:
>>>In sci.physics.relativity, T.M. Sommers
>>><t...@nj.net>
>>> wrote
>>>on Sun, 15 Jul 2007 11:20:51 -0400
>>><469a3b52$0$25595$470ef...@news.pa.net>:
>>>>gu...@hotmail.com wrote:
>>>
>>>>>Didn't know this was a course in Calculus or Algebra, how do you
>>>>>calculate the work without multiplying.
>>>>
>>>>You take the dot product. More specifically, you integrate the
>>>>dot product.
>>>
>>>The dot product is a sum of scalar products, at least as
>>>conventionally expressed. :-)
>>
>>You don't just multiply the magnitudes of the two vectors, though.
>
There is always an angle, even if it is 0.
> If
> there is one then they write
>
> F= qvB sin(angle).
Which gives the magnitude, but not the direction, of the resultant.
> I have a McGrawHills Physics SCHOOL book = what they teach and not
> just an information book and they don't even use the "X" (meaning
> they write F=qvB instead of F= qv X B)
It sounds like this book does not use calculus or vector algebra.
If so, dump it and get a book that does use calculus and vector
algebra. Too many simplifications have to be made to try to
explain physics without calculus and vector algebra.
The Ghost In The Machine
<gu...@hotmail.com>
wrote
on Mon, 16 Jul 2007 14:22:44 -0700
<1184620964.1...@n2g2000hse.googlegroups.com>:
The dot product can also be expressed
a dot b = length(a) * length(b) * cos(angle(a,b))
In any event, absent considerations such as hitting dust,
a planet in a pure circular orbit does no work (at least
in Newtonian physics; GR predicts gravitational waves,
with the planet very slowly spiraling inward). The planet
is clearly moving but the force is always perpendicular
to its movement vector.
A planet in an elliptical orbit also does no work but the
considerations are more complicated.
cos(pi/2) = cos(90 degrees) = 0.
--
#191, ewi...@earthlink.net
Useless C++ Programming Idea #889123:
std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);
gu...@hotmail.com
<ew...@sirius.tg00suus7038.net> wrote:
> In sci.physics.relativity, gu...@hotmail.com
> <gu...@hotmail.com>
> wrote
> on Mon, 16 Jul 2007 14:22:44 -0700
Yep
> In any event, absent considerations such as hitting dust,
> a planet in a pure circular orbit does no work (at least
> in Newtonian physics; GR predicts gravitational waves,
> with the planet very slowly spiraling inward). The planet
> is clearly moving but the force is always perpendicular
> to its movement vector.
>
Well for alternators, the magnetic force does some type of work that
generates electricity. And this type of work due to the cross-product
I have more than sufficient confidence to believe it's due to the
torque.
Wire and it's charges are perpendicular to the magnetic field (mind
you the quantity of charges(current) varies with time which is a
difference of kinetic energy in relation with the perpendicular
magnetic field. As well a wire laterally moved back and forth through
this magnetic field once it stops has generated zero displacement,
thus in a way zero work, yet this also generates electricity.
> A planet in an elliptical orbit also does no work but the
> considerations are more complicated.
>
> cos(pi/2) = cos(90 degrees) = 0.
>
> --
> #191, ewi...@earthlink.net
> Useless C++ Programming Idea #889123:
> std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);
>
> --
> Posted via a free Usenet account fromhttp://www.teranews.com- Hide quoted text -
The Ghost In The Machine
<1184912312....@g4g2000hsf.googlegroups.com>:
That torque cuts both ways. If you have a small motor, try this
experiment (disconnected from any power source):
[1] Spin the shaft with the wires open.
[2] Spin the shaft with the wires shorted.
Notice any difference?
>
> Wire and it's charges are perpendicular to the magnetic field (mind
> you the quantity of charges(current) varies with time which is a
> difference of kinetic energy in relation with the perpendicular
> magnetic field. As well a wire laterally moved back and forth through
> this magnetic field once it stops has generated zero displacement,
> thus in a way zero work, yet this also generates electricity.
That electricity also cuts both ways. There's a concept
called "backEMF" -- this is a generated voltage in a coil.
There is also the concept of "power factor". A motor
under no load will have current and voltage 90 degrees
out of phase. A motor under full load will have them
in phase.
>
>> A planet in an elliptical orbit also does no work but the
>> considerations are more complicated.
>>
>> cos(pi/2) = cos(90 degrees) = 0.
>>
>> --
>> #191, ewi...@earthlink.net
>> Useless C++ Programming Idea #889123:
>> std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);
>>
>> --
>> Posted via a free Usenet account fromhttp://www.teranews.com- Hide quoted text -
>>
>> - Show quoted text -
>
>
--
#191, ewi...@earthlink.net
Conventional memory has to be one of the most UNconventional
architectures I've seen in a computer system.
Darwin123
> Model: A wire moving perpendicularly through a N&S magnetic field
>
> Quote:"It is ****ERRONEOUS**** to think that the magnetic force does
> force is PERPENDICULAR to the velocity and hence to the displacement
> of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> *****MAGNETIC FORCE********."
>
> If one cannot figure out that the reverse is also true and that a
> magnetic force also does not affect a perpendicularly moving charge,
> absolutely no interaction occurs.
Microscopically, no work is being done by the magnetic force. The
work is being done by whatever nonmagnetic forces are holding the wire
together. It is the elastic forces of the wire that does the work.
Think of the wire as an empty pipe of finite diameter, and the
electron being in the center of this pipe. The pipe is held together
by elastic forces that do not obey Maxwell's equations. Both pipe and
electron are moving in the same direction at the same speed,
perpendicular to the magnetic field. For the first instant, the
magnetic force acts perpendicular to the motion of the electron. The
electron starts to move in circles. However, it does not absorb energy
since the magnetic force is perpendicular to the direction of motion.
No work is done by the magnetic force. But because it changed
direction, the electron and the pipe are now going in different
directions. The electron now hits the edge of the pipe, which applies
a force with a component in the direction of the electrons motion. Now
the pipe does work on the electron.
People keep on forgetting the cohesive (e.g., elastic) forces
when doing the electromagnetic problems. Pure electromagnetic forces
can not maintain an objects shape. One needs nonelectromagnetic
cohesive forces to do that. And cohesive forces do work.
gu...@hotmail.com
> On Jul 12, 4:39 pm, "gu...@hotmail.com" <gu...@hotmail.com> wrote:
>
> > Model: A wire moving perpendicularly through a N&S magnetic field
>
> > Quote:"It is ****ERRONEOUS**** to think that the magnetic force does
> > work on the charges to produce an EMF. We recall that the magnetic
> > force is PERPENDICULAR to the velocity and hence to the displacement
> > of the charge deltaW= F deltaS Cos(angle) where deltaS = displacement
> > and in this case angle = 90 degrees therefore NO WORK IS DONE BY THE
> > *****MAGNETIC FORCE********."
>
> > If one cannot figure out that the reverse is also true and that a
> > magnetic force also does not affect a perpendicularly moving charge,
> > absolutely no interaction occurs.
>
> Microscopically, no work is being done by the magnetic force.
Actually I was just rewritting a statement from a physics book and
commenting on it.
But now after reflection, I believe it is a torque between the
magnetic field of the charge and the magnetic field of the magnet.
F= gm doesn't mean it's the gravity of one mass applied to another
mass (it is the gravity of both) even though the equation uses only
one G constant. Likewise F= qvB doesn't mean it's the magnetic field
of one applied to a moving charge (it is the magnetic field of both).
gu...@hotmail.com
<ew...@sirius.tg00suus7038.net> wrote:
> In sci.physics.relativity, gu...@hotmail.com
> <gu...@hotmail.com>
> wrote
That's a strange behavior?
>
>
>
>
>
>
> >> A planet in an elliptical orbit also does no work but the
> >> considerations are more complicated.
>
> >> cos(pi/2) = cos(90 degrees) = 0.
>
> >> --
> >> #191, ewi...@earthlink.net
> >> Useless C++ Programming Idea #889123:
> >> std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);
>
> >> --
> >> Posted via a free Usenet account fromhttp://www.teranews.com-Hide quoted text -
>
> >> - Show quoted text -
>
> --
> #191, ewi...@earthlink.net
> Conventional memory has to be one of the most UNconventional
> architectures I've seen in a computer system.
>
> --
> Posted via a free Usenet account fromhttp://www.teranews.com- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
The Ghost In The Machine
<1184999785.0...@w3g2000hsg.googlegroups.com>:
Not at all. If a motor is under no load, it needs no power (except to
offset frictional losses). The only other alternatives are the coils to
heat, or the motor to spin out of control until it flies apart.
P = EI in a DC context; it also turns out that the instantaneous power transmitted
is also the product of voltage and current. If one represents voltage as
E * sin(2*pi*t/f) and current as I * sin(2*pi*t/f + phi)
= I * sin(2*pi*t/f) * cos(phi) + I * cos(2*pi*t/f) * sin(phi),
the product of these two is
E * I * (sin^2(2*pi*t/f) * cos(phi) + sin(2*pi*t/f)*cos(2*pi*t/f)*sin(phi) )
= E * I * (
-(1/2) * (1 - 2*sin^2(2*pi*t/f)) * cos(phi)
+ (1/2) * cos(phi)
+ (1/2) * sin(4*pi*t/f) )
= E * I * (
-(1/2) * (cos(4*pi*t/f)) * cos(phi)
+ (1/2) * cos(phi)
+ (1/2) * sin(4*pi*t/f) )
If one integrates this expression over t from 0 to t, the result is
E * I * (t/2) * cos(phi)
since all other terms are periodic with period t/2 and thus drop.