Question 5.19: Referring to the coaxial cable as shown in Fig. 5.32, determ......
Referring to the coaxial cable as shown in Fig. 5.32, determine the inductance per unit length of the cable from the magnetic energy stored in the cable.
In the region a < ρ < b, inserting Eq. (5-106b) into Eq. (5-126), the magnetic energy is
\pmb{B}=\frac{\mu _{o}I}{2 \pi \rho }\pmb{a}_{\phi } (a < ρ < b) (5-106b)
\boxed{W_{m}=\frac{\mu}{2}\int_{\nu}H^{2}dv=\frac{1}{2\mu}\int_{\nu}B^{2}d v} [J] (5-126)
W_{m1} = \frac{1}{2\mu _{o} }\int_{\rho =a}^{\rho =b}\int_{\phi =0}^{\phi =2\pi }{ \left[ \frac{\mu _{o} I}{2\pi \rho } \right]^{2}\rho d\rho d\phi } = \frac{\mu _{o} I^{2}}{4\pi } \ln \frac{b}{a} (5-128a)
In the region 0 ≤ ρ ≤ a , inserting Eq. (5-106c) into Eq. (5-126), the magnetic energy is
\pmb{B}=\frac{\mu _{o}\rho I}{2 \pi a^{2} }\pmb{a}_{\phi } (0 ≤ ρ ≤ a) (5-106c)
W_{m2} = \frac{1}{2\mu _{o} }\int_{\rho =0}^{\rho =a}\int_{\phi =0}^{\phi =2\pi }{ \left[ \frac{\mu _{o}\rho I}{2\pi a^{2} } \right]^{2}\rho d\rho d\phi } = \frac{\mu _{o} I^{2}}{16\pi } (5-128b)
The inductance per unit length of the cable is obtained by inserting Eq. (5- 128) into Eq. (5-127):
\boxed{L = \frac{2W_{m}}{I^{2}}} [J] (5-127)
L =\frac{2}{I^{2}} (W_{m1} + W_{m2}) = \frac{\mu _{o}}{2\pi }\ln \frac{b}{a} +\frac{\mu _{o}}{8\pi } [H/m] (5-129)
We have the same result as Eq. (5-110).
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)