Question 5.17: A very long coaxial cable consists of a solid inner conducto......
A very long coaxial cable consists of a solid inner conductor, having a radius a, and a cylindrical shell of negligible thickness, having a radius b. Assuming that the current I flows uniformly in the inner conductor, and returns through the outer conductor as a uniform surface current, find the inductance per unit length of the cable.
From symmetry considerations, we anticipate \pmb{B}=B_{\phi }(\rho )\pmb{a}_{\phi } everywhere.
By using Ampere’s circuital law we obtain
B = 0 (ρ > b) (5-106a)
\pmb{B}=\frac{\mu _{o}I}{2 \pi \rho }\pmb{a}_{\phi } (a < ρ < b) (5-106b)
\pmb{B}=\frac{\mu _{o}\rho I}{2 \pi a^{2} }\pmb{a}_{\phi } (0 ≤ ρ ≤ a) (5-106c)
(1) In the region a < ρ < b, the magnetic flux passing through surface S shown in Fig. 5.32(a) is
\Phi = \int_{S}{ \pmb{B} \pmb{\cdot }d \pmb{s}} = \int_{\rho =a}^{\rho =b}{ \frac{\mu _{o} I}{2 \pi \rho}\ell d\rho } = \frac{\mu _{o} I \ell }{2 \pi }\ln \frac{b}{a}Magnetic flux linkage with the inner conductor per unit length of the cable is
\Lambda _{1} = \frac{\Phi }{l} = \frac{\mu _{o} I }{2 \pi }\ln \frac{b}{a} (5-107)
(2) In the region 0 ≤ ρ ≤ a , let us consider a cylindrical shell of radius \rho _{1}, thickness dρ, and length ℓ as shown in Fig. 5.32(b). The magnetic flux confined to this shell is obtained from (5-106c) as
d\Phi = B_{\phi } \ell d\rho = \frac{\mu _{o} \rho _{1} I }{2 \pi a^{2} }\ell d\rho (5-108)
It should be noted that dΦ in Eq. (5-108) is due to the partial current I_{1 }= \rho^{2}_{1}I/a^{2}, which is enclosed by the shell, not due to the total current I. The differential magnetic flux linkage with the cable, per unit length, is therefore given by the product of dΦ /ℓ and I_{1 } /I:
d\Lambda = \frac{d\Phi }{l} \frac{\rho ^{2}_{1}}{a^{2}} = \frac{\mu _{o} \rho ^{3}_{1} I}{2\pi a^{4}}d\rhoThe total magnetic flux linking the whole inner conductor, per unit length, is
\Lambda_{2} =\int{d\Lambda } =\int_{\rho =0}^{\rho =a}{ \frac{\mu _{o} \rho ^{3} I}{2\pi a^{4}}d\rho } =\frac{\mu _{o}I}{8\pi } (5-109)
In the above equation, subscript 1 is dropped from ρ for generalization.
The total magnetic flux linkage with the coaxial cable, per unit length, equals the sum of \Lambda_{1} and \Lambda_{2}. The inductance per unit length of the coaxial cable is therefore
L = \frac{\Lambda_{1} +\Lambda_{2}}{I} =\frac{\mu _{o}}{2\pi}\ln \frac{b}{a} + \frac{\mu _{o}}{8\pi }\\ \quad \quad \quad \quad \quad \equiv L_{ex}+L_{in} [H/m] (5-110)
In Eq. (5-110), L_{ex} is called the external inductance and L_{in} is called the internal inductance.
