Re: motional emf

[Brian Whatcott <betwys1@SBCGLOBAL.NET>]

At 12:22 AM 6/26/2004, Kenny Stephens, you wrote:
> //
> Two conducting rods are placed parallel to each other. Let's call them
> rails and say the left end of both rails are connected by a resistance, R.
> Another conducting rod of length L is placed across (perpendicular) to the
> rods and can slide freely along the rails. An external agent acts on the
> rod to give it a uniform velocity, v, parallel to the rails (and away from
> the resistance). A uniform magnetic field is applied perpendicular to the
> plane of the problem (let's say into the page). Assume a current flows
> through the circuit (through R, along one rail, up the rod and returns
> along the other rail) such that the charge carries have a drift velocity v_d.
>
> The text says that each charge carrier, q, in the rod has the velocity v
> and since q moves in a magnetic field it experiences a lorentz force F_M=
> qv cross B. The text then states that the work done by this force pushing
> the charges along the rod is F_M * L= qvBL. Since emf is energy per
> charge, the motional emf between the ends of the rod is E= vBL.
>
> Now this bugs the heck out me because magnetic forces are not supposed to
> do work.

//

> Kenny Stephens

The concept is difficult.
The mantra is, "A magnetic field does no work on a charge carrier"
[because the magnetic force is perpendicular to the charge's direction of
motion.]  One can develop an explanation using the static magnetic field
acting on the  magnetic field  due to the current. A preferred model is to
invoke relativistic charge imbalance  [due to the drift velocity of the
charge carriers] which provides an electric force acting in the direction
of movement - and this is work.


Brian Whatcott    Altus OK    Eureka!