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.