Self-inductance, a conceptual question

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

Responding to Chuck Britton JohnD wrote (in part):

> ...The new equation is particularly simple if you know the
> inductance of the loop, because flux = inductance * current.

This reminds me a difficulty with L I had. 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



> Collecting ideas:
>         voltage = flux dot
>         flux = external_applied_flux + loop's_induced_flux
>         loop's_induced_flux = inductance * current
>         current = voltage / resistance
> which is four equations in four unknowns.
> Not very hard to solve.
>
> If this isn't the answer, please clarify the question.