Re: Is the magnetic field a conservative one?

Good head-scratcher question!

Let's begin with a definition:

"The amount of work that must be done when a unit test body is moved along 
an arbitrary curve in the force field defined by a vector function F is the 
line integral of the tangential component of F; that is

W = integral (F dot dR)
        C
    
If there is no dissipation of energy through friction or similar effects, 
then, according to the law of conservation of energy, this integral must be 
zero around every closed path C... Fields for which this is the case are 
said to be conservative".  

(From C.R. Wylie, "Advanced Engineering Mathematics", McGraw Hill, 4th ed., 
1975.  All my engineering books are getting to be 25 years old.  )

Note the implication made by the definition: if the closed path integral of 
WORK is zero, the field is conservative.

This principle is usually illustrated in the context of the electrostatic 
field, in which the electric field vector function E defines the force on a 
unit test charge AND that this force is parallel to E.  That is, F=qE  
Faraday's Law tells us that the integral of E around a closed path is zero 
(remember: in electrostatics E does not vary with time and B is 
zero).  Therefore, the integral of force on a test body around a closed 
path is zero, and we recognize E as conservative.  

A consequence of this ubiquitous example (well, it's ubiquitous to those 
who study electromagnetics, at least!) is that the implication gets 
remembered in slightly twisted form: if the closed path integral of the 
field is zero, the field is conservative.

The mutated form works fine for the electric field (note that the electric 
field is NOT conservative in the presence of changing magnetic fields -- 
Faraday's Law in its more general form).

Things get a bit trickier, as you realized, in applying this principle to 
the magnetostatic case: on the one hand, we're told that a magnetic field 
is conservative (it can do no net work), yet Ampere's Law says the closed 
path integral of the field is non-zero (the integral equals the current 
enclosed).

The way out of the dilemma is to check the fine print: what is the force on 
 a "test body"?  (And what is our "test body"?  One practical answer is to 
use a charged particle)  For the magnetic field, F = q(U x B), where "x" 
represent vector cross product.  U is the velocity vector of charge q.
Note that the force is always perpendicular to the direction of motion of 
charge.  If force is always perpendicular to the direction of motion, the 
motion does no work. Therefore, the integral of work around a closed path 
must be zero, and we conclude that the magnetic field is conservative.

Obviously, a magnetic field can do work over finite paths (as can an 
electrostatic field).  For example, the principle behind a DC motor is to 
recognize that current I flowing through a conductor perpendicular to a 
magnetic field creates a force perpendicular to both the magnetic field and 
the conductor (i.e., this is the cross product rule, above).  If we 
constrain the conductor from motion, no work is done (the motor is stalled 
and current is limited only by the conductor's resistance, so current draw 
becomes very high).  

Now let's let the conductor move in the direction of the resultant force.
We can use the principle of superposition for linear systems to examine 
separately what happens by moving a conductor perpendicularly ("broadside") 
through a magnetic field: the force is perpendicular to the direction of 
motion and the magnetic field, and parallel to the conductor.  This is a 
force on the charges in the conductor and this force is in the opposite 
direction of current flow above.  But F = qE, so this force represent an 
induced voltage V opposing current flow.  We conclude that work is being 
accomplished at the rate P = VI.  

Note that the work is being done by the power source applied to the 
conductor, the magnetic field acts as a sort of "catalyst" to allow this to 
happen and does no work, per se.  For example, we can use permament magnets 
to set up the magnetic field in our DC motor, but they do not become "used 
up" and run down like a battery no matter how long the motor is used.  
(Permanent magnets eventually do wear out, but this is an effect of 
entropic-like effects as the individual magnetic domains in the magnetic 
gradually lose their alignment).

The other refinement which should be pointed out as that arrangements are 
made to mechanically switch currents to different conductors as the motor 
turns (this is typically the job of the "commutator")... otherwise no work 
would be done.  Finally, note that in AC machines, clever arrangement of 
the wiring allows the sinusoidal currents to do the switching for us.