Archive for June 25th, 2010

Physics Friday 125

June 25, 2010

Part 5: Work Done on Currents

Consider a point charge of charge q moving from position r0 at time t0 to position r1 at later time t1, and let the curve C be the path it takes between these points. Further, let their be an external electric field E(r). What, then, is the total work done on the charge by this field in the time interval from t0 to t1?

The work done by a force field F on a particle traveling on a curve C is the line integral of F on C:
For an electric force, we have F=qE; and with position as a function of time r(t), then the line element is , where v(t) is the velocity. Thus, our work becomes:

Thus, the work done per unit of time, the power transferred to the particle by the field, for a given moment of time is thus

(by the fundamental theorem of calculus).

Now, let us have a current I flowing along the curve C, with external field E. Consider a line element ds of the current. Let the time for a bit of charge in the current to traverse this length be dt. Since the current is the amount of charge flowing per unit time, in this (infinitesimal) time interval, we have a charge of passing through. And since the velocity of this charge is , we have , and so the work done per unit time on this piece of the current is
, and so the total power delivered to the current by the field is
Now, if the electric field corresponds to an electric potential φ, then , and so we see
and so by the gradient theorem (fundamental theorem of calculus for line integrals):
Thus, if we have a drop in potential, a voltage drop, of , then the power delivered to the current by the source of this voltage difference (and the corresponding electric field) is given by the above ; power is current times voltage. This law is a key tool in the analysis of circuits.

Lastly, let us replace our current in a loop with a general current density distribution . Consider within this distribution a small segment ds parallel in direction to the local current density vector J. So further, let us consider a small path C, with a small cross section Δσ perpendicular to this curve. Then the current in this elemental “tube” is , with , and so the power imparted to this element by the field is

But since ds is in the direction of J, then , and so we have
Now, considering the entire current density distribution as broken up into such elements, we see , and so our total power is given by the volume integral

For magnetic fields, the force exerted on a point charge q with velocity v is the Lorentz force
Thus, the magnetic force is always perpendicular to the path of the particle, , and so the magnetic field does no work.