86. III High-frequency signals are often transmitted along a coaxial CALC cable, such as the one shown in FIGURE CP30.86. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius while the outer conductor of radius is grounded. A soft, flexible insulating material fills the space between them, and an insulating plastic coating goes around the outside.

a. Find an expression for the inductance per meter of a coaxial cable. To do so, consider the flux through a rectangle of length that spans the gap between the inner and outer conductors.

b. Evaluate the inductance per meter of a cable having$r_1=0.50\text{mm}$ and$r_2=3.0\text{mm}$ .

Radio frequency AC are transmitted by means of cables over a large distance. A cable consists of two coaxial cylindrical conductors of radii and . The currents in the conductors are equal in magnitude but opposite in directions. Thus, there is no magnetic field outside these conductors. Thus, the cable is non-radiating. There will, however, be a magnetic flux between the two cylinders.

Let there be a current of I A through the conductors

The magnetic field B is at distance r from the axis and within the region between the two cylinders.

Here, the flux through the area between r and r+dr, and of unit length, is given by

Thus, total flux between the unit length of the cable is given by

Thus, self-inductance per unit length of the cable is henries/m

Part(b)

Here, ,$r_2=3\text{mm}$ and$r_1=0.50\text{mm}$

Thus, from the result of (a) above, the inductance per meter is

$\begin{array}{l}L=\frac{{\mu }_0}{2\pi }\log \frac{r_2}{r_1}\\ =\frac{4\pi \times {10}^{-7}}{2\pi }{\log }_e\frac{3\text{\hspace{0.17em}mm}}{0.5\text{\hspace{0.17em}mm}}\\ \approx 0.36\mu \text{H}/\text{m}\end{array}$

The inductance per meter is$0.36\mu \text{H}/\text{m}$