Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.velocity.net/~trebor/
trebor@velocity.net
----- Original Message -----
From: "Kenny Stephens" <kfstephensii@YAHOO.COM>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Saturday, June 26, 2004 1:22 AM
Subject: motional emf
| 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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