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Re: induced emf again



It is instructive to analyze the work done (and not done) by the qVxB force
in this situation:
General notation: [x] specifies that x is a vector.

Let [i], [j] and [k] be the unit x,y,z vectors : to the right, up the page,
and out of the page, respectively.
The rod is forced (by an external agent) to move to the right with a
constant velocity [v] = v[i]. the magnetic field is everywhere [B] = -B[k]
: into the page. Thus, the (positive) carriers in the moving rod have a
resultant velocity [w] = v[i] + u[j] , where u[j] is the carrier drift
velocity which constitutes the current generated along the rod.

The rate at which the magnetic force does work on a charge carrier is:
P = q*( [w]x[B] ) DOT [w] = q*( [w]x[B] ) DOT ( v[i] + u[j] )

Note the first term is negative, the second is positive, the total is zero.
The first term is power taken from the external agent; the second term is
power given to the current carriers.

Performing the DOT product, and letting TH be the angle betwwen [w] and the
y axis [j]:
(Note this also makes TH the angle between the vector [w]x[B] and the
negative x axis.)

P = q* | [w]x[B] | * (-v*Cos(TH) + u*Sin(TH) )

This is ostensibly zero, because v/u = tan(TH), by construction.

The magnetic force does no work, but acts as a "catalyst" to enable the
external agent to drive the current. I am loosely reminded of the role
played by the force of a wall on a skater pushing off from it. The wall
does "no work" but allows internal bodily energy sources to transfer energy
to the skater's bulk motion.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor