Question 8.11: Determine the self-inductance of a coaxial cable of inner ra...

The self-inductance of the inductor can be found in two different ways: by taking the four steps given in Section 8.8 or by using eqs. (8.54):

L=\frac{2W_{m}}{I^{2}}

and (8.66):

W_{m}=\frac{1}{2}\int_{v}B\cdot Hdv=\frac{1}{2}\int_{v}\mu H^{2}dv

Method 1: Consider the cross section of the cable as shown in Figure 8.22. We recall from eq. (7.29):

H=\begin{cases} \frac{I\rho}{2\pi a^{2}}a_{\phi}, & 0\leq \rho\leq a \\ \frac{I}{2\pi \rho}a_{\phi}, & a\leq \rho\leq b \\ \frac{I}{2\pi \rho}\left[1-\frac{\rho^{2}-b^{2}}{t^{2}+2bt}\right]a_{\phi} , & b\leq \rho\leq b+t \\0, & \rho\geq b+t \end{cases}

that by applying Ampère’s circuit law, we obtained for region 1(0\leq \rho\leq a)

B_{1}=\frac{\mu I\rho}{2\pi a^{2}}a_{\phi}

and for region 2(a\leq \rho\leq b)

B_{2}=\frac{\mu I}{2\pi\rho}a_{\phi}

We first find the internal inductance L_{in} by considering the flux linkages due to the inner conductor. From Figure 8.22(a), the flux leaving a differential shell of thickness d\rho is

d\Psi_{1}=B_{1}d\rho dz=\frac{\mu I\rho}{2\pi a^{2}}d\rho dz

The flux linkage is d\Psi_{1} multiplied by the ratio of the area within the path enclosing the flux to the total area, that is

d\lambda_{1}=d\Psi_{1}\cdot\frac{I_{enc}}{I}=d\Psi_{1}\cdot\frac{\pi \rho^{2}}{\pi a^{2}}

because I is uniformly distributed over the cross section for dc excitation. Thus, the total flux linkages within the differential flux element are

d\lambda _{1}=\frac{\mu I\rho d\rho dz}{2\pi a^{2}}\cdot\frac{\rho^{2}}{a^{2}}

For length \ell of the cable

\lambda_{1}=\int_{\rho =0}^{a}\int_{z=0}^{\ell}\frac{\mu I\rho^{3} d\rho dz}{2\pi a^{4}}=\frac{\mu I\ell}{8\pi}

L_{in}=\frac{\lambda_{1}}{I}=\frac{\mu\ell}{8\pi}                                                                             (8.11.1)

The internal inductance per unit length, given by

L^{'}_{in}=\frac{L_{in}}{\ell}=\frac{\mu}{8\pi} H/m                                                              (8.11.2)

is independent of the radius of the conductor or wire. Since the inductance does not depend cable to finding the inductance of any infinitely long straight conductor of finite radius. on a, we can make the wire as thin as possible. Thus eqs (8.11.1) and (8.11.2) are also applicable to finding the inductance of any infinitely long straight conductor of finite radius.

We now determine the external inductance L_{ext} by considering the flux linkages between the inner and the outer conductor as in Figure 8.22(b). For a differential shell of thickness d\rho

d\Psi_{2}=B_{2}d\rho dz=\frac{\mu I}{2\pi\rho}d\rho dz

In this case, the total current I is enclosed within the path enclosing the flux. Hence

\lambda_{2}=\Psi _{2}=\int_{\rho =a}^{b}\int_{z=0}^{\ell}\frac{\mu Id\rho dz}{2\pi\rho }=\frac{\mu I\ell}{2\pi}\ln\frac{b}{a}

L_{ext}=\frac{\lambda_{2}}{I}=\frac{\mu\ell}{2\pi}\ln\frac{b}{a}

Thus

L=L_{in}+L_{ext}=\frac{\mu\ell}{2\pi}\left[\frac{1}{4}+\ln\frac{b}{a}\right]

or the inductance per length is

L^{'}=\frac{L}{\ell}=\frac{\mu}{2\pi}\left[\frac{1}{4}+\ln\frac{b}{a}\right] H/m

Method 2: It is easier to use eqs. (8.54) and (8.66), to determine L, that is

W_{m}=\frac{1}{2}LI^{2}   or  L=\frac{2W_{m}}{I^{2}}

where

W_{m}=\frac{1}{2}\int B\cdot Hdv=\int\frac{B^{2}}{2\mu}dv

Hence

L_{in}=\frac{2}{I^{2}}\int_{v}\frac{B_{1}^{2}}{2\mu}dv=\frac{1}{I^{2}\mu}\iiint\frac{\mu ^{2}I^{2}\rho ^{2}}{4\pi ^{2}a^{4}}\rho d\rho d\phi dz

=\frac{\mu}{4\pi^{2}a^{4}}\int_{0}^{\ell}dz\int_{0}^{2\pi}d\phi \int_{0}^{a}\rho^{3}d\rho=\frac{\mu\ell}{8\pi}

L_{ext}=\frac{2}{I^{2}}\int_{v}\frac{B_{2}^{2}}{2\mu}dv=\frac{1}{I^{2}\mu}\iiint\frac{\mu^{2}I^{2}}{4\pi^{2}\rho^{2}}\rho d\rho d\phi dz

=\frac{\mu}{4\pi^{2}}\int_{0}^{\ell}dz\int_{0}^{2\pi}d\phi \int_{a}^{b}\frac{d\rho}{\rho}=\frac{\mu\ell}{2\pi}\ln\frac{b}{a}

and

L=L_{in}+L_{ext}=\frac{\mu\ell}{2\pi}\left[\frac{1}{4}+\ln\frac{b}{a}\right]

as obtained previously.