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.

5.32
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In the region < ρ < b, inserting Eq. (5-106b) into Eq. (5-126), the magnetic energy is

B=μoI2πρaϕ\pmb{B}=\frac{\mu _{o}I}{2 \pi \rho }\pmb{a}_{\phi }                                       (< ρ < b)                                                       (5-106b)

Wm=μ2νH2dv=12μνB2dv\boxed{W_{m}=\frac{\mu}{2}\int_{\nu}H^{2}dv=\frac{1}{2\mu}\int_{\nu}B^{2}d v}                                            [J]                                   (5-126)

Wm1=12μoρ=aρ=bϕ=0ϕ=2π[μoI2πρ]2ρdρdϕ=μoI24πlnbaW_{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

B=μoρI2πa2aϕ\pmb{B}=\frac{\mu _{o}\rho I}{2 \pi a^{2} }\pmb{a}_{\phi }                                       (0 ≤ ρ ≤ a)                                                       (5-106c)

Wm2=12μoρ=0ρ=aϕ=0ϕ=2π[μoρI2πa2]2ρdρdϕ=μoI216πW_{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):

L=2WmI2\boxed{L = \frac{2W_{m}}{I^{2}}}                                   [J]                              (5-127)

L=2I2(Wm1+Wm2)=μo2πlnba+μo8πL =\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=Λ1+Λ2I=μo2πlnba+μo8πLex+LinL = \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)

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