Re: E=F/q=vB Magnetic Force does not work on the charge

In sci.physics.relativity, guskz@xxxxxxxxxxx
<guskz@xxxxxxxxxxx>
wrote
on Thu, 19 Jul 2007 23:18:32 -0700
<1184912312.697832.67220@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
On Jul 18, 11:25 am, The Ghost In The Machine
<ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics.relativity, gu...@xxxxxxxxxxx
<gu...@xxxxxxxxxxx>
wrote
on Mon, 16 Jul 2007 14:22:44 -0700
<1184620964.145801.131...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:





On Jul 15, 9:45 pm, "T.M. Sommers" <t...@xxxxxx> wrote:
gu...@xxxxxxxxxxx wrote:
On Jul 15, 1:11 pm, The Ghost In The Machine
<ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics.relativity, T.M. Sommers
<t...@xxxxxx>
wrote
on Sun, 15 Jul 2007 11:20:51 -0400
<469a3b52$0$25595$470ef...@xxxxxxxxxxx>:
gu...@xxxxxxxxxxx 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. :-)

I simply 3x3=9 and nothing more. the others can dot as much as they
want.

Then you will get the wrong answer. The dot product depends on
the angle between the two vectors.

--
Thomas M. Sommers -- t...@xxxxxx -- AB2SB- Hide quoted text -

- Show quoted text -

If there be an angle I would at it to the formula, if there it be
perpendicular or parallel I doth not.

The dot product can also be expressed

a dot b = length(a) * length(b) * cos(angle(a,b))


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.

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...@xxxxxxxxxxxxx
Useless C++ Programming Idea #889123:
std::vector<...> v; for(int i = 0; i < v.size(); i++) v.erase(v.begin() + i);

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--
#191, ewill3@xxxxxxxxxxxxx
Conventional memory has to be one of the most UNconventional
architectures I've seen in a computer system.

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Posted via a free Usenet account from http://www.teranews.com

.

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