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Re: [Phys-l] Magnetic force and work



Hi Bob-
I think your point is easily emphasized by working with free-body
diagrams.
Jack


On Mon, 27 Mar 2006, Bob Sciamanda wrote:

I think I need to emphasize the important idea in this exchange withh Tom
Snyder:

The situation in which two different agents exert equal and opposite forces
on each other or on a particle can (but does not always) represent a
transfer of energy from one agent to the other (perhaps through the
intermediary particle).

1) If I push on a car and accelerate it, the set of equal and oppoite forces
represents a transfer of energy from me to the car.

2) I f I push a crate against a frictional force so as to maintain its
constant velocity, the positive work which I do (against equal and opposite
fricional forces) represents a transfer of energy from me (through the crate
motion) to the various thermal energy modes of the crate and the floor.

3) If I (on frictionless ice skates) push off from an infinitely rigid wall,
there is no energy transfer between the wall and me. The work energy theorem
here guarantees that the work done by the wall's push is numerically equal
to my increase in kinetic energy, but it says nothing to identify the source
of my kinetic energy increase.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
http://www.winbeam.com/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Bob Sciamanda" <trebor@winbeam.com>
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Sent: Monday, March 27, 2006 5:20 PM
Subject: Re: [Phys-l] Magnetic force and work


See embedded comments below:

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
http://www.winbeam.com/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Snyder, Tom" <Tom.Snyder@llcc.edu>
To: "Bob Sciamanda" <trebor@winbeam.com>; "Forum for Physics Educators"
<phys-l@carnot.physics.buffalo.edu>
Sent: Monday, March 27, 2006 1:49 PM
Subject: Re: [Phys-l] Magnetic force and work


Here's a very elementary example that I find helpful.

Suppose you have a hollow tube oriented along the y-axis. Inside the
tube is a small marble that can slide without friction. Neglect
gravity. The marble carries a net positive charge and there is a
uniform B field in the negative z direction.

Imagine you apply an external force to the tube so that the tube slides
with a constant speed in the positive x direction with the tube always
remaining oriented parallel to the y-axis. Initially, let the marble be
at the "lower" end of the tube with zero y-component of velocity. The
marble will slide up the tube with a nonzero y-component of acceleration
as the tube slides to the right. When the marble reaches the top of the
tube it will have gained KE.

Consider the work done by the various forces in the reference frame of
the fixed x-y-z coordinate system.

First take the marble alone as the "system" of interest so that the tube
is part of the "environment". There are only two forces that act on the
marble: the magnetic force and the normal force from the wall of the
tube. The magnetic force does zero work on the marble and the normal
force does an amount of work equal to the gain in KE of the marble.

This last sentence does not parse - the marble has gained KE in the y
direction, but the normal force acts only in the x direction. See below.

However, it is possible to interpret this differently. The x-component
of the magnetic force is negative and equal in magnitude to the normal
force. Thus, they may be considered as cancelling one another. This
leaves just the y-component of the magnetic force which may then be
considered as the force doing the work on the marble.

You gloss over the energetic effects of these "equal and opposite forces".
The normal force does positive work on the marble. Simultaneosly the x
component of the magnetic force does negative work on the marble -
receiving
the energy transferred by the normal force and giving it (by means of
positive work) to the y motion of the marble.


Thus, it seems to
me that the question "which force does the work on the marble?" can be
answered in different ways depending on how you wish to resolve the
forces. Nevertheless, it is always true that the work done by the
_total_ magnetic force is zero.



Next consider the tube as the "system" of interest. The marble is now
part of the "environment". Two forces act on the tube: the normal force
(in the negative x-direction) that the marble exerts on the wall of the
tube and the external applied force (in the positive x-direction) that
you apply to the tube to move the tube with constant speed. These two
forces are equal in magnitude and opposite in direction. Together,
these forces do zero net work on the tube and the KE of the tube remains
constant.

Finally, consider the tube and marble together as the "system". The
internal "action-reaction" normal forces between the marble and the wall
together do zero net work. The total magnetic force also does zero
work. The net work done on the system is the work done by the external
force that you applied to slide the system along the x-axis. For this
system, the gain in energy of the system (namely, the gain in KE of the
marble) can be thought of as coming from the work done by you.

But again, if you wanted to, you could consider the x-component of the
magnetic force as cancelling the externally applied force. The
y-component of the magnetic force is then left to do the work.

Yes, but it must be noted that the work done by the y component of the
magnetic force comes from the positive work done by the external force.
Again you gloss over the work done by the "equal and opposite forces".
This
situation represents an important transfer of energy.

To more closely model an induced electric current in a wire moving
through a B field, you can add friction between the tube and marble such
that the marble slides up the tube with a constant "drift" speed rather
than accelerating up the tube. Then the gain in KE of the marble is
replaced by a gain of thermal energy of the system. (In the electric
circuit, the sliding wire needs to be part of a complete circuit so that
the charge carriers have somewhere to go when they reach the end of the
sliding wire.)

Tom
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l



_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l



_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l


--
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General Custer's unremembered message to his men,
just before leading them into the Little Big Horn Valley