Question 8.12: Determine the inductance per unit length of a two-wire trans...

We use the two methods of Example 8.11.

Method 1: We determine L_{in} just as we did in Example 8.11. Thus for region 0\leq \rho\leq a, we obtain

\lambda_{1}=\frac{\mu I\ell}{8\pi}

as before. For region a\leq \rho\leq d-a, the flux linkages between the wires are

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

The flux linkages produced by wire 1 are

\lambda_{1}+\lambda_{2}=\frac{\mu I\ell}{8\pi}+\frac{\mu I\ell}{2\pi}\ln\frac{d-a}{a}

By symmetry, the same amount of flux is produced by current -I in wire 2. Hence the total linkages are

\lambda=2(\lambda_{1}+\lambda_{2})=\frac{\mu I\ell}{\pi}\left[\frac{1}{4}+\ln\frac{d-a}{a}\right]=LI

If d\gg a, the self-inductance per unit length is

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

Method 2: From Example 8.11, we have

L_{in}=\frac{\mu\ell}{8\pi}

Now

L_{ext}=\frac{2}{I^{2}}\int\frac{B^{2}dv}{2\mu}=\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_{0}^{d-a}\frac{d\rho}{\rho}=\frac{\mu\ell}{2\pi}\ln\frac{d-a}{a}

Since the two wires are symmetrical,

L=2(L_{in}+L_{ext})=\frac{\mu\ell}{\pi}\left[\frac{1}{4}+\ln\frac{d-a}{a}\right] H

as obtained earlier.