Question 5.20: The coaxial cable shown in Fig. 5.32 carries a current I [A]......
The coaxial cable shown in Fig. 5.32 carries a current I [A], which is non-uniform in the inner conductor such that \pmb{J} = 2 I\rho ^{2} / (\pi a^{4})\pmb{a}_{z} [A/m²] in cylindrical coordinates. Determine the internal inductance per unit length from (a) magnetic flux linkage, and (b) magnetic energy.
In the region 0 ≤ ρ ≤ a , applying Ampere’s circuital law to an Amperian path of a radius \rho_{1}( 0 ≤ \rho_{1} ≤ a ), we obtain
2 \pi \rho_{1} H_{\phi }= \int_{\rho =0}^{\rho = \rho _{1}}\int_{\phi =0}^{\phi = 2\pi }{ \frac{2 I\rho^{2} }{ \pi a^{4}}\rho d\rho d\phi} = I\frac{\rho^{4}_{1} }{a^{4}}Omitting 1 in ρ for generalization, we have
H_{\phi }= \frac{\rho^{3}I }{2 \pi a^{4}} (5-130)
(a) Magnetic flux confined to the cylindrical shell of a radius \rho_{1}, thickness dρ, and length ℓ as shown in Fig. 5.32(b) is
d\Phi = \mu _{o} H_{\phi }\ell d \rho = \frac{\mu _{o}\rho^{3}_{1}I }{2 \pi a^{4}} \ell d\rhoThe differential flux dΦ is produced by a partial current I_{1} flowing in a cylinder of radius \rho_{1}. To find I_{1}, we integrate J over the disc of radius \rho_{1} in the cross section of the inner conductor:
I_{1 } = I \frac{\rho^{4}_{1} }{ a^{4}}The magnetic flux linking with the cylindrical shell, per unit length, is given by the product of dΦ / ℓ and I_{1 } / I, i.e.,
d\Lambda = \frac{d\Phi }{\ell} \frac{I_{1 }}{I} = \frac{\mu _{o}\rho^{7}_{1} I}{2\pi a^{8}}d\rhoTotal flux linkage with the whole inner conductor, per unit length, is
\Lambda = \int{ d\Lambda } =\int_{\rho =0}^{\rho =a}{\frac{\mu _{o}\rho^{7} I}{2\pi a^{8}}d\rho} = \frac{\mu _{o} I}{16 \pi }The internal inductance per unit length is therefore
L = \frac{\Lambda}{I} = \frac{\mu _{o} }{16 \pi } [H/m]. (5-131)
(b) The magnetic energy stored per unit length of the inner conductor is obtained by inserting Eq. (5-130) into Eq. (5-126):
\boxed{W_{m}=\frac{\mu }{2} \int_{\nu}{H^{2} dv} =\frac{1}{2\mu }\int_{\nu}{B^{2}dv}} [J] (5-126)
W_{m} = \frac{\mu _{o}}{2 }\int_{\rho =0}^{\rho =a}\int_{\phi =0}^{\phi =2\pi }{H^{2}\rho d\rho d\phi }= \frac{\mu _{o}}{2} 2\pi\left\lgroup \frac{I}{2\pi a^{4}}\right\rgroup^{2} \int_{\rho =0}^{\rho =a}{\rho ^{7}d\rho } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{\mu _{o}I^{2}}{32\pi } [J/m]The internal inductance per unit length is therefore,
L = \frac{2W_{m}}{I^{2}} = \frac{\mu _{o}}{16\pi } [H/m] (5-132)
The result is the same as that in part (a).