91Ó°ÊÓ

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

$\begin{array}{l}L=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\log \frac{r_2}{r_1}\\ =\frac{4\mathrm{Ï€}×{10}^{−7}}{2\mathrm{Ï€}}{\log }_e\frac{3\text{″¾³¾}}{0.5\text{″¾³¾}}\\ ≈0.36\mathrm{μ}\text{H}/\text{m}\end{array}$

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