Re: Self-inductance, a conceptual question

[Bob Sciamanda <trebor@VELOCITY.NET>]

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Ludwik Kowalski" <kowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, April 17, 2002 9:28 AM
Subject: Re: Self-inductance, a conceptual question



> 2) Here is the dilemma. Suppose I measure B at several
> radii of a Helmholtz coil of 200 turns (when I=2 A) and
> calculate the flux. It turns out to be 2*10^-4 Wb. The
> task is to calculate L from the flux. The textbook definition
> of L is FLUX=L*I. So I divide 2*10^-4 by 2 and conclude
> that L=0.1 mH. What is wrong with this reasoning?

L of a circuit cannot be determined from the value of B and the area
alone - this only gives the flux.  L is concerned with the induction of an
emf  in the circuit by this changing flux - thus L "needs to know" how
many times the circuit in question links the changing  flux.  If the
circuit links the flux N times, the emf in each link is dPHI/dt and the
series combination gives an emf of N*dPHI/dt.

I agree that there certainly are semantic difficulties in texbook
treatments of this.  Some will implicitly include a "silent" N in there
PHI or their dPHI/dt .  If you are just calculating the flux (meaning only
B dot area) then you omit the "second" N. But the physics requires that
the calculation of the emf induced by this changing flux "knows" the
number of flux linkages.  Scott tries to clarify in this manner:

"Any closed circuit will be linked by a magnetic flux produced by its own
current . . . This flux will be proportional to the current , so that we
can write PHI = L*i, where L is the constant of proportionality . . . If
there are several loops in the circuit, as in a solenoid, PHI is taken to
mean the total flux linkage or the sum of the fluxes through each loop.
It is of course the quantity whose time derivative gives the emf . . . emf
= -L*di/dt . . ."