14.3 Energy in a Magnetic Field

Self-Inductance of a Coaxial Cable

Figure 14.11 shows two long, concentric cylindrical shells of radii $R_1$ and $R_2.$ As discussed in Capacitance on capacitance, this configuration is a simplified representation of a coaxial cable. The capacitance per unit length of the cable has already been calculated. Now (a) determine the magnetic energy stored per unit length of the coaxial cable and (b) use this result to find the self-inductance per unit length of the cable.

Strategy

The magnetic field both inside and outside the coaxial cable is determined by Ampère’s law. Based on this magnetic field, we can use Equation 14.22 to calculate the energy density of the magnetic field. The magnetic energy is calculated by an integral of the magnetic energy density times the differential volume over the cylindrical shell. After the integration is carried out, we have a closed-form solution for part (a). The self-inductance per unit length is determined based on this result and Equation 14.22.

Solution

1. We determine the magnetic field between the conductors by applying Ampère’s law to the dashed circular path shown in Figure 14.11(b). Because of the cylindrical symmetry, $\overrightarrow{B}$ is constant along the path, and
   $${\displaystyle \oint \overrightarrow{B}·d\overrightarrow{l}=B(2\pi r)}={\mu }_{0}I.$$
   This gives us
   $$B=\frac{{\mu }_{0}I}{2\pi r}.$$
   In the region outside the cable, a similar application of Ampère’s law shows that $B=0$, since no net current crosses the area bounded by a circular path where $r>R_2.$ This argument also holds when $r<R_1;$ that is, in the region within the inner cylinder. All the magnetic energy of the cable is therefore stored between the two conductors. Since the energy density of the magnetic field is
   $$u_{\text{m}}=\frac{B^2}{2{\mu }_0}$$
   the energy stored in a cylindrical shell of inner radius r, outer radius $r+dr,$ and length l (see part (c) of the figure) is
   $$u_{\text{m}}=\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2}.$$
   Thus, the total energy of the magnetic field in a length l of the cable is
   $$U={\displaystyle {\int }_{R_1}^{R_2}dU}={\displaystyle {\int }_{R_1}^{R_2}\frac{{\mu }_0{I}^2}{8{\pi }^2{r}^2}(2\pi rl)dr}=\frac{{\mu }_0{I}^{2}l}{4\pi }\text{ln}\frac{R_2}{R_1},$$
   and the energy per unit length is $({\mu }_0{I}^2\text{/}4\pi )\text{ln}(R_2\text{/}{R}_1)$.
2. From Equation 14.22,
   $$U=\frac{1}{2}L{I}^2,$$
   where L is the self-inductance of a length l of the coaxial cable. Equating the previous two equations, we find that the self-inductance per unit length of the cable is
   $$\frac{L}{l}=\frac{{\mu }_0}{2\pi }\text{ln}\frac{R_2}{R_1}.$$

Significance

The inductance per unit length depends only on the inner and outer radii as seen in the result. To increase the inductance, we could either increase the outer radius $\left(R_2\right)$ or decrease the inner radius $\left(R_1\right)$. In the limit as the two radii become equal, the inductance goes to zero. In this limit, there is no coaxial cable. Also, the magnetic energy per unit length from part (a) is proportional to the square of the current.