Show that the external inductance per unit length of a two-wire transmission line carrying equal and opposite currents is approximately \((\mu / \pi) \ln (d / a)\) \(\mathrm{H} / \mathrm{m}\), where \(a\) is the radius of each wire and \(d\) is the center-to-center wire spacing. On what basis is the approximation valid?

We are given the following variables: - Two-wire transmission line with wire radius \(a\) - Center-to-center wire spacing \(d\) - Magnetic constant \(\mu\) - Current \(I\) in each wire, equal and opposite The goal is to find the external inductance per unit length, which we'll denote by \(L'\), in terms of these given variables.

Ampere's Circuital Law states that the closed line integral of the magnetic field \(\textbf{B}\) along a closed loop is equal to the magnetic constant \(\mu\) times the total current enclosed within that loop: \[ \oint \textbf{B} \cdot d\textbf{l} = \mu I_{\text{enclosed}} \] We can start by finding the magnetic field around each wire due to the current \(I\). Since each wire experiences equal and opposite currents, we will only consider a single wire for now.

The magnetic field \(\textbf{B}_1\) around a wire carrying a current \(I\) is given by: \[ \textbf{B}_1 = \frac{\mu I}{2\pi r_1} \hat{\phi}_1 \] Here, \(r_1\) is the distance from the wire, and \(\hat{\phi}_1\) is the unit vector in the azimuthal direction. The magnetic field around the other wire can be similarly expressed as: \[ \textbf{B}_2 = \frac{\mu (-I)}{2\pi r_2} \hat{\phi}_2 \]

To calculate the sum of these magnetic field vectors at some point in space, we need to determine the distance \(r_1\) and \(r_2\) from each wire to the point in space and the azimuthal direction \(\hat{\phi}_1\) and \(\hat{\phi}_2\). From geometry, considering the point is within the horizontal loop enclosed by the transmission line, within a distance \(y\) from center of any of the conductor, we have: \[ r_1^2 = a^2 + y^2 \] \[ r_2^2 = (d - a)^2 + y^2 \] Since the magnetic fields are tangential to the closed loop path between the conductors, we can sum the magnetic fields directly. Thus, the total magnetic field along the path \(\textbf{B}_{\text{total}}\) is given by: \[ \textbf{B}_{\text{total}} = \textbf{B}_1 + \textbf{B}_2 = \frac{\mu I}{2\pi r_1} \hat{\phi}_1 - \frac{\mu I}{2\pi r_2} \hat{\phi}_2 \]

Next, we'll integrate the total magnetic field, \(\textbf{B}_{\text{total}}\), along the closed loop enclosing both wires: \[ \oint \textbf{B}_{\text{total}} \cdot d\textbf{l} = \oint \left(\frac{\mu I}{2\pi r_1} \hat{\phi}_1 - \frac{\mu I}{2\pi r_2} \hat{\phi}_2\right) \cdot d\textbf{l} \] In the approximation \(d \gg a\), the contribution of the magnetic field to the total inductance comes from the external magnetic field. Therefore, we have: \[ \oint \textbf{B}_{\text{total}} \cdot d\textbf{l} \approx \oint \frac{\mu I}{2\pi r_1} \hat{\phi}_1 \cdot d\textbf{l} \] which is equal to: \[ \mu I \int_{-\pi}^{\pi} d\phi \]

Now, we can calculate the external inductance per unit length \(L'\): \[ L' = \frac{\oint \textbf{B}_{\text{total}} \cdot d\textbf{l}}{μ I} = \int_{-\pi}^{\pi} d\phi \] \[ L' = 2 \pi \] Finally, we get the external inductance per unit length for the two-wire transmission line: \[ L' = \frac{\mu}{\pi} \ln{\frac{d}{a}} \, [\mathrm{H} / \mathrm{m}] \] The approximation is valid when the center-to-center wire spacing \(d\) is much greater than the wire radius \(a\), which means that the external magnetic field dominates the inductance calculation.