Re: motional emf

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

I'm about to go out to cut the lawn, so I'll be brief - but
not incoherent -  I hope.

At some point an external force must accelerate the bar
from rest to the speed v. This gives the electrons a motion
in the direction of v and an associated kinetic energy. The
magnetic field turns the direction of motion (without
imparting energy to the electrons) so that they have a
component of motion perpendicular to v. The Lorentz force
related to this component produces a force opposite to v -
the rod will slow down unless the external force continues
to act. It is the external force that is producing the
energy.

A slightly different way of approaching  this is to think
of the external force as accelerating the electrons from
rest resulting in a kinetic energy 1/2 m v_x^2 (assuming v
is in the x direction). The magnetic force redirects the
motion - the kinetic energy doesn't change - but it now is
given by 1/2 m v_x^2 + 1/2 m v_y^2. The x term is less, but
the y term has increased. In order to maintain the x term,
the external force must do work - the magnetic force again
redirects this to y motion, etc. In a sense, the external
force does work and is the source of the energy, the
magnetic force simply redirects it to the ends of the rod
resulting in the EMF.  The magnetic field does no work in
this process.

This approach doesn't require a change of frame of
reference.

Bob at PC


*********** REPLY SEPARATOR  ***********

On 6/25/2004 at 10:22 PM Kenny Stephens wrote:

> This is my first posting to the list so please be
tolerant.
> I'm currently reviewing some chapters for a "well-known"
cal-based intro
> physics book. It uses a popular explanation (I've seen it
elsewhere) for
> deriving the emf developed by a conductor moving with
uniform velocity
> through a constant magnetic field. Jsut to be thorough,
here's the setup:
> 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. Using this explanation just sets the students up
for confusion
> and puts me in a pickle to try to justify it.
>
> I prefer the explanation of calculating the changing flux,
Phi_M= BLx,
> through the circuit where x is the position of the rod
measured along the
> rails from the resistance. This gives the emf E=
-dPhi_M/dt= BL(dx/dt)=
> BLv.
>
> After all this yacking, my reason for posting is to get a
range of
> opinions of this text's derivation.
>
> Thanks for the time.
> Kenny Stephens
>
>
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