W9_parallel_resonance.eps

Week 8: Faraday’s Law and Induction 281

inductance of a long straight wire and a rectangular loop above. Again, trynotto have to go

back and look, as the picture should remind you of what you need to do, and the integral itself

is pretty trivial.

c) Multiply the flux for a single turn byN, the number of turns in the solenoid (as once again

eachturn contributes to the overall potential difference) to find the total flux.

d) Divide the flux by the current to find the self-inductance of the solenoid.

e) Think a minute. Suppose the currentI(t) in the direction shown in the figure isincreasing.

What is the direction of the induced electric field around a loop? Suppose it is decreasing,

ditto? Either way, of course, the induced voltage across the two wires leading to/from the

solenoid willopposethe change in the current!

f) If desired, find e.g. the voltageVL=−LdIdtor any other quantities of interest.

Example 8.7.3: Coaxial Cable

I

I

a

b

r l

dA = drl

Figure 104: Coaxial cables have a self-inductance measuredper unit length. At high frequencies the
inductance only depends on the outer radius of the inner conductoraand the inner radius of the
outer conductorb. A strip of areadA=ℓdris shown that may be of use in computingL/ℓ, the
self-inductance per unit length.

This sets up another homework problem, as I’m feeling even lazier than before andyouneed to
do the work in order to learn how! In figure 104 a currentI(t) flows e.g. up the (long-straight) inner
conductor and back on the outer one or vice versa. From Ampere’sLaw you can easily find the
magnetic field in between the inner and outer conductors (where it isconfined; at high frequencies
all of the current will be on the surfaces and we can ignore currentdensity and magnetic field inside
the conductors themselves).

With the field in hand, it should be easy to find the flux through the dark shaded strip shown
(with the parameterℓin it, so this is the flux per lengthℓonce theℓis divided out) and integrate
fromatob, an integral that should by now be boringly familiar to you. Divide by the currentand
ℓto find the self-inductance per unit length of the cable.

That isn’t quiteallof the cases where one can compute the self-inductance of something without
needing to do absurdly difficult integrals or deal with even more heavilyapproximated fields, but it
is pretty close.