Goofed again!
EMF = BVL should replace the erroneous EMF = qVxB !
Also qVxB=qE should replace qVxB=E !
Bob Sciamanda
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Sunday, May 05, 2002 9:52 AM
Subject: Re: induced emf again
| . . .
| As with other voltage sources, Terminal Voltage = EMF - I*Rint. As the
| external load and current vary, the emf is constant at qVxB, the
terminal
| voltage and the E field in the rod are current dependent, exactly as in
| John's example. To see this conceptually, your second model (a series
of
| dls) is perhaps more useful. Of course the Thevenin equivalent
| conceptually separates the emf and Rint in order to make things
| calculationally simple (and correct).
|
| To answer directly: yes, qVxB = E holds only for zero current (or even
| with current if the rod has zero resistance).
|
| Note that in the external circuit, charges accumulate at the terminals
of
| a resistor so as to establish the necessary driving E field, resulting
in
| an electrostatic potential drop of IR in the direction of the current.
| The internal (rod's) resistance experiences this effect in addition to
the
| effect of the emf, so that charges develop on the terminals which
produce
| the electrostatic potential difference equal to the Terminal Voltage =
| EMF - I*Rint. (Here EMF = qVxB)
|
| Bob Sciamanda (W3NLV)
| Physics, Edinboro Univ of PA (em)
| trebor@velocity.net
| http://www.velocity.net/~trebor
| . . .