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Re: [Phys-l] Magnetic force and work



Let's examine some of the details here - I think that important concepts
are getting lost in omitted steps; see embedded comments.


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.

The first step here is to draw a diagram - draw the velocity vector for
some time t (I'll assume that the marble starts at the origin at t=0), and
resolve it into components (which I'll call vtube (the x component -
constant) and vy (y component - initially zero)). The whole vector will
begin along the pos. x and begin to elevate towards the y axis as the
motion starts.
Next, draw the perpendicular magnetic force component (initially up,
rotates CCW with the v vector). Resolve this into components (called Fx
and Fy). Note that Fy = Fcos(theta), Fx = Fsin(theta), with theta being
the angle of v's elevation above the x axis.
There's one more force left - the normal force. As the x component of
the momentum is constant, the mag. of this force is equal to Fx =
Fsin(theta).



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.

Seems like that would be true, but it's not:

Take a differential work element due to the mag. force: dW magnetic = F
dot ds = F dot (v dt). Since F is always orthogonal to v, it's orthogonal
to ds also, and the dot product vanishes.
dW magnetic = 0.

Take a diff. element of work due to the normal force: dW normal = N dot ds
= (Fsin(theta)) dot (v dt). To evaluate the dot product, we need to find
the projection of v on the normal force vector, which is just vtube = v
cos(theta).

Now we have, substituting in for the mag. force: dW normal = qvB
sin(theta) * v cos(theta) dt = qBv^2 * sin(theta) * cos(theta) dt.

From the velocity triangle, sub in vy/v for sin(theta) and vtube/v for
cos(theta):

dW normal = qB * vtube * vy * dt.

q, B, and vtube are constants. Moving them outside the integral and
integrating both sides from t=0 to t and W = 0 to W, we get

W = qB * vtube * y

So, the KE is increasing linearly with distance the ball travels up the
tube, but it's clearly done _by_ the normal force, even though the added v
is in the y direction. Remember, the added KE is not "in" any particular
direction, as it's a scalar! A convincing interpretation for me is that B
"transfers" velocity from the x- dim. to the y-dim., and the normal force
"replenishes" that x velocity.

Fantastic example, Tom!

Josh Gates
--------------
Noble and Greenough
Dedham MA