Question

3. Self Inductance of coax: 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)/Ta2 goes uniformly through the wire) At the outside edge (r-a), there is an infinitesimally thin insulator, and just outside that, arn 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. Notes: Assume I(t) changes slowly in time, so we can use our usual quasi-static ideas Th ere are several ways to proceed. You might start from d-LI, but this turns out to be tricky here, since the current is not confined to a single path! It's best to think about the total magnetic energy, W, stored by the magnetic field, and use our usual relation between ene W-L/2

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Answer 1

Self inductance of 1 car consider a coaxial cable which is an long wire of radius a with given current flowing uniformly up it. for r lessthanareequalto a from ampere's law we know that B times 2pi r = i mu_0 now, i = I(t)/pia^2 pi r^2 [I(t)/pi a^2 goes uniformly through the wire] i = It(r/a)^2 so that B = I(t)mu_0/2pi r lambda (r/a)^2 rightarrow I(t) (mu_0 r)/2pi a^2 now, figure cut the B field everywhere in space in terms of givens, and then self-inductance per length of this cable. rightarrow for r > a B = 0(as i = 0) self-inductance = L EMF v_l = dd/dt, contourintegral = integral B Da contourintegral - integral^a_0I(t)mu_0 r/2pi a^2 2 pi r dr = I(t) mu_0 r/2pi a^2 2pirdr = I(t)mu_0/2pi a^2 2pi a^3/3 = I(t) mu_0 a/3 now take rightarrow v_l = dphi/dt mu_0 = a/3 di(t)/dt = L. di(t)/dt L = a mu_0/3