5. Consider a coaxial cable which is an (infinitely) long wire of radius a, with given current flowing uniformly up it (so J(t) = I(t)/лa² goes uniformly through the wire). At the outside edge (r=a), there is an infinitesimally thin insulator, and just outside that, an infinitesimally thin sheath that returns all the current, basically a surface current going down the opposite direction carrying a total -I(t) back down. Figure out the B field everywhere in space (direction and magnitude) in terms of givens, and then find the self-inductance per length of this cable. Assume I(t) changes slowly in time, so we can use our usual quasi-static ideas. There are several ways to proceed. You might start from =LI, but this turns out to be tricky here, since the current is not confined to a single path! It is best to think about the total magnetic energy, W, stored by the magnetic field, and use our usual relation between energy and current: WL I²/2.

FINANCIAL ACCOUNTING
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Chapter1: Financial Statements And Business Decisions
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5. Consider a coaxial cable which is an (infinitely) long wire of radius a, with given
current flowing uniformly up it (so J(t) = I(t)/лa² goes uniformly through the wire).
At the outside edge (r=a), there is an infinitesimally thin insulator, and just outside
that, an infinitesimally thin sheath that returns all the current, basically a surface
current going down the opposite direction carrying a total -I(t) back down. Figure out
the B field everywhere in space (direction and magnitude) in terms of givens, and then
find the self-inductance per length of this cable. Assume I(t) changes slowly in time, so
we can use our usual quasi-static ideas. There are several ways to proceed. You might start
from =LI, but this turns out to be tricky here, since the current is not confined to a single
path! It is best to think about the total magnetic energy, W, stored by the magnetic field, and
use our usual relation between energy and current: WL I²/2.
Transcribed Image Text:5. Consider a coaxial cable which is an (infinitely) long wire of radius a, with given current flowing uniformly up it (so J(t) = I(t)/лa² goes uniformly through the wire). At the outside edge (r=a), there is an infinitesimally thin insulator, and just outside that, an infinitesimally thin sheath that returns all the current, basically a surface current going down the opposite direction carrying a total -I(t) back down. Figure out the B field everywhere in space (direction and magnitude) in terms of givens, and then find the self-inductance per length of this cable. Assume I(t) changes slowly in time, so we can use our usual quasi-static ideas. There are several ways to proceed. You might start from =LI, but this turns out to be tricky here, since the current is not confined to a single path! It is best to think about the total magnetic energy, W, stored by the magnetic field, and use our usual relation between energy and current: WL I²/2.
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