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

On Jul 13, 4:25 am, "gu...@xxxxxxxxxxx" <gu...@xxxxxxxxxxx> wrote:
On Jul 12, 5:47 pm, "T.M. Sommers" <t...@xxxxxx> wrote:



gu...@xxxxxxxxxxx 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.

--
Thomas M. Sommers -- t...@xxxxxx -- AB2SB

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?

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

.

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