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

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, 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)


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


.

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