MadSci Network: Physics |
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.
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