Re: Self-inductance, a conceptual question

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

1) Transformer is not explained before the inductor in most
textbooks. Therefore, justifying N^2 in terms of what
happens in a transformer is, pedagogically, not appropriate.
But it was useful to me.

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?

I know that the correct L is 200 times larger than 0.1 mH
but I do not know how to explain this in terms what students
already know when the problem is presented. The definition
does not say anything about coils "linking the flux of each
other." Why does Tipler write FLUX=N*B*A and not
FLUX=B*A? See my original message below:
Ludwik Kowalski
*******************************************

Ludwik asked:
Suppose I have a single coil with I=0.5 A and the flux is
10^-6 Wb. Then L=2*10^-6 H.

Next I add another coil; the same current flows through two
coils in the same direction. The flux doubles, the current is
the same and I expect L to double. For ten coils the flux will
be ten times larger, for the same I, and L is 2*10^-5 H. This
kind of reasoning leads me to think that L should be
proportional to the number of coils N. But the formulas for L
show that it is usually proportional to N^2.

To verify this conceptual dilemma I am looking into Tipler's
way of introducing self inductance, L. It begins with the
definition of L, FLUX=L*I. Then it derives the formula for
L of a long tightly wound solenoid. The B inside is
mu-o*(N/L)*I. To find the flux I would multiply B by A.
But Tipler writes FLUX=N*B*A. This leads to L being
proportional to N^2. But why does N appear in the flux
formula; it is already hidden in B? In other words, why is
FLUX=N*B*A rather than FLUX=B*A?
Ludwik Kowalski
*********************************************
Bob replied:

> Ludwik wrote:
> > . . .
> > Next I add another coil; the same current flows through two
> > coils in the same direction. The flux doubles, the current is
> > the same and I expect L to double . . .
>
> This assumes that the coils are isolated from each other; ie. that neither
> coil links the flux of the other.
>
> > . . .
> > But Tipler writes FLUX=N*B*A. This leads to L being
> > proportional to N^2. But why does N appear in the flux
> > formula; it is already hidden in B?
>
> Consider a transformer with an open-circuited secondary.  The emf induced in
> the secondary is proportional to both Np (since Phi and dPhi/dt are
> proportional to Np) and Ns (since the turns are in series and the emf
> dPhi/dt is induced in each secondary turn).  Thus the secondary emf is
> proportional to the product Np*Ns.  In a self inductance a single coil plays
> the roles of both primary and secondary.  It both generates the flux and it
> "experiences" the emf  dPhi/dt in each turn. The self induced emf (and the
> self inductance) is proportional to N^2.
>
> Bob Sciamanda (W3NLV)
> Physics, Edinboro Univ of PA (em)
> trebor@velocity.net
> http://www.velocity.net/~trebor