W9_parallel_resonance.eps

Week 8: Faraday’s Law and Induction 279

R

B

n

Single turn

for field.increasing

Direction of induced E

∆VL

l

(lower V) (higher V)

I (in)

B

Figure 102: An ordinary (ideal) solenoid withNturns each carrying a currentI(t) is drawn above.
The total flux through the solenoid isNtimes the flux through a single turn.

where the direction is determined from the right hand rule, in the figure above to the left through
the solenoid.

This field isuniformwithin aninfinitesolenoid andvanishes outside of itand we will idealize
it as being uniform in this one and vanishing very rapidly at the ends (neglecting “fringing fields”
outside of the volume of the solenoid, basically, much as we did for electric fields outside of the
volume of an indealized parallel plate capacitor). This idealization will bevalid as long asℓ≫R
and the solenoid is tightly wound as noted.

Next, we find the self-induced flux through asingleturn of the solenoid. Again we idealize the
turn as being a circle in a plane instead of a segment of a helix, with areaπR^2 , so that:

φturn =

∫

S

B~·nˆdA

= BπR^2

= μ 0

N

ℓ

IπR^2 (606)

The solenoid hasNturns,eachwith this flux. Yes, they all count, aseachof them contributes
a piece ∆Vturnto the total potential difference as the current changes, so thetotal will beNtimes
that of just one turn:

φtotal=

(

μ 0 N^2 πR^2

ℓ

)

I (607)

Finally, we find the self-inductance by noting thatφtotal=LIso that:

L=

μ 0 N^2 πR^2

ℓ (608)

Note that we generally makeLpositive by convention and figure out any signs using Lenz’s Law
and a bit of common sense, so inductors don’t come with a polarity or sign.