Part 5: Work Done on Currents

Consider a point charge of charge *q* moving from position **r**_{0} at time *t*_{0} to position **r**_{1} at later time *t*_{1}, 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 *t*_{0} to *t*_{1}?

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**=*q***E**; 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 *d***s** 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 *d***s** 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 *d***s** 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.

## Archive for June 25th, 2010

### Physics Friday 125

June 25, 2010
Advertisements