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

On Jul 13, 10:02 am, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
On Jul 13, 4:19 am, "gu...@xxxxxxxxxxx" <gu...@xxxxxxxxxxx> wrote:





On Jul 12, 8:06 pm, Randy Poe <poespam-t...@xxxxxxxxx> wrote:

On Jul 12, 6:03 pm, "gu...@xxxxxxxxxxx" <gu...@xxxxxxxxxxx> wrote:

On Jul 12, 4:48 pm, Randy Poe <poespam-t...@xxxxxxxxx> wrote:

On Jul 12, 4:39 pm, "g...@xxxxxxxxxxx" <g...@xxxxxxxxxxx> 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,

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.

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.

In this case we are talking about a nonzero displacement.


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.

- Randy- Hide quoted text -

- Show quoted text -


.

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