Consider the coaxial cable of Problem 30.46. The conductors carry equal currents i in opposite directions. (a) Use Ampere’s law to find the magnetic field at any point in the volume between the conductors. (b) Use the energy density for a magnetic field, Eq. (30.10), to calculate the energy stored in a thin, cylindrical shell between the two conductors. Let the cylindrical shell have inner radius $r$, outer radius $\mathrm{r}+\mathrm{dr}$, and length l.(c) Integrate your result in part (b) over the volume between the two conductors to find the total energy stored in the magnetic field for a length l of the cable. (d) Use your result in part (c) and Eq. (30.9) to calculate the inductance L of a length l of the cable. Compare your result to L calculated in part (d) of Problem 30.46.

The inner cylindrical shell of radius r, the outer cylindrical shell of radius r+dr , and the length is /

Consider inner cylindrical shell of radius r and outer cylindrical shell of radius

r = a

r + dr = b

The circumference of the coaxial cable is, $\mathrm{l}=2\mathrm{\pi r}$where, r is amperian loop of radius r. Using Ampere's law, we have $\oint \mathrm{B}.\mathrm{dl}={\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{enclosed}}$to calculate the magnitude of the magnetic field as

$$\begin{gathered} \mathrm{B}\oint \mathrm{dl}={\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{enclosed}} \\ \mathrm{Bl}={\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{enclosed}} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{l}}_{\mathrm{enclosed}}}{\mathrm{l}} \end{gathered}$$

The entire current is enclosed by the loop is equal to the current ${\mathrm{l}}_{\mathrm{enclosed}}=0$ Substitute l= $2\mathrm{\pi r}$So, $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi r}}$is the the magnitude of the magnetic field.

The inner cylindrical shell of radius r, the outer cylindrical shell of radius r + dr, and the length is l. The change in volume of the cable is,$\mathrm{dV}=2\mathrm{\pi rldr}$andthe magnetic energy density is$\mathrm{U}=\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}$where,Bis the magnetic field magnitude,${\mathrm{\mu }}_0$is the permeability of the cable and the differential form of energy stored in the cable is dU = udV where,u is the magnetic energy density, dV is the change in volume of the cable. Substitute the values we have,

$$\begin{gathered} \mathrm{dU}=\frac{1}{2{\mathrm{\mu }}_0}{\left(\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi r}}\right)}^{2}2\mathrm{\pi rldr} \\ =\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi r}}\mathrm{dr}\right) \end{gathered}$$

Therefore, the expression for the energy stored in a thin cylindrical shell between the two conductors is$\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi r}}\mathrm{dr}$

The inner cylindrical shell of radius r, the outer cylindrical shell of radius r + dr, and the length is l Refer the figure 1: The limit between the inner and outer cylindrical shell is a to b, a is the inner cylindrical shell of radius and b is the outer cylindrical shell of radius. Integrate equation$\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi r}}\mathrm{dr}\right)$, to total energy stored in the magnetic field for a length of the cable.

$$\begin{gathered} \int \mathrm{dU}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi r}}\mathrm{dr}\right) \\ \mathrm{U}=\mathrm{l}\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\frac{\mathrm{dr}}{\mathrm{r}} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}{\left[\mathrm{Inr}\right]}_{\mathrm{a}}^{\mathrm{b}} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}\left[\mathrm{Inb}-\mathrm{Ina}\right] \\ =\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}} \end{gathered}$$

Therefore, the expression for the total energy stored in the magnetic field for a length of the cable is$\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}$

The inner cylindrical shell of radius r, the outer cylindrical shell of radius r + dr, and the length is l. The energy stored in a inductor is,$\mathrm{U}=\frac{1}{2}{\mathrm{Li}}^2$.Rearrange the equation, to find the inductance of a length of the cable is,$\mathrm{L}=\frac{2\mathrm{U}}{{\mathrm{i}}^2}$where,U is the total energy stored in the magnetic field for a length of the cable, i is the current flows in the inductor. Substitute the values we get,

role="math" localid="1664189028616" $$\begin{gathered} \mathrm{L}=\frac{2\left(\frac{{\mathrm{\mu }}_0{\mathrm{i}}^2\mathrm{l}}{4\mathrm{\pi }}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}\right)}{{\mathrm{i}}^2} \\ =\left(\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi }}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}\right) \end{gathered}$$

Therefore, the inductance of the length of the cable is $\left(\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi }}\mathrm{In}\frac{\mathrm{b}}{\mathrm{a}}\right)$and hence proved.