Re: Self-inductance, a conceptual question

[Bob Sciamanda <trebor@VELOCITY.NET>]

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