A small solid conductor with radius a is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius b. The inner and outer 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) Write the expression for the flux${\mathbf{d}}_{{\mathbf{\varphi }}_{\mathbf{B}}}$ through a narrow strip of length l parallel to the axis, of width dr, at a distance r from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current i in the central conductor. (d) Show that the inductance of a length l of the cable is$\mathbf{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\tau \tau }}\left[\ln \frac{b}{a}\right]$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length l of the cable.

The radius of the small solid conductor is a, the inner radius of the small thin-walled tube is b, and the inner and outer conductor carry equal current is i in the opposite direction.

Formula to calculate the circumference of the coaxial cable is,$l=2\mathrm{\pi }r$Using Ampere's law:

$\oint B.dl={\mu }_0{l}_{enclosed}$,to calculate the magnitude of the magnetic field,

$\begin{array}{r}\mathrm{B}\oint \mathrm{dl}={\mu }_0{l}_{\text{enclosed}}\\ \mathrm{BI}={\mu }_0{l}_{\text{enclosed}}\\ \mathrm{B}=\frac{{\mu }_0{\mathrm{l}}_{\text{enclosed}}}{\mathrm{l}}\end{array}$

The entire current is enclosed by the loop is equal to the current $l_{enclosed}=0$ 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 radius of the small solid conductor is a, the inner radius of the small thin-walled tube is b, and the inner and outer conductor carry equal current is i in the opposite direction.

The entire cross-sectional area of the cable is, dA = ldr. The magnetic field at any point in the volume between the conductors is $\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi r}}$, to calculate the magnetic flux is ${\varphi }_B=BA$Differentiate the equation we have, ${\mathrm{d\varphi }}_{\mathrm{B}}=\mathrm{BdA}$.On substituting the value we have,${\mathrm{d\varphi }}_{{}_{\mathrm{B}}}\frac{{\mathrm{\mu }}_0\mathrm{l}}{2\mathrm{\pi r}}\mathrm{ldr}$

Therefore, the expression for the magnetic flux through a narrow strip of length parallel to the axis is$\frac{{\mu }_{0}l}{2\pi r}ldr$

The radius of the small solid conductor is a, the inner radius of the small thin-walled tube is b, and the inner and outer conductor carry equal current is i in the opposite direction. Refer the figure 1: The limit between the inner and outer conductors is a to b, a is the radius of the small solid conductor and b is the inner radius of the small thin-walled tube. Integrate equation $\frac{{\mu }_{0}l}{2\pi r}ldr$, to calculate the total flux produced by a current in the central conductor, we have

$\begin{array}{r}\int \mathrm{d}{\varphi }_{\mathrm{B}}={\int }_{\mathrm{a}}^{\mathrm{b}} \frac{{\mu }_0\mathrm{I}}{2\pi \mathrm{r}}\left(\mathrm{Idr}\right)\\ {\varphi }_{\mathrm{B}}={\mathrm{I}}_{\frac{1}{0}}^{2\pi }{\int }_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{dr}}{\mathrm{r}}\\ {\varphi }_{\mathrm{b}}=\frac{{\mu }_0\mathrm{il}}{2\pi }[\mathrm{lnr}{]}_{\mathrm{a}}^{\mathrm{b}}\\ =\frac{{\mu }_0\mathrm{il}}{2\pi }[\mathrm{lnb}-\mathrm{lna}]\\ {\varphi }_{\mathrm{b}}=\frac{{\mu }_0\mathrm{il}}{2\pi }\left[\mathrm{lnbla}\right]\end{array}$

Therefore, the expression for the total flux produced by a current in the central conductor is$\frac{{\mu }_{0}iI}{2\pi }\left[In\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\right]$

The radius of the small solid conductor is a, the inner radius of the small thin-walled tube is b, and the inner and outer conductor carry equal current is i in the opposite direction. The number of turns is equal to 1 for the coaxial cable. Refer equation$\frac{{\mathrm{\mu }}_0\mathrm{iI}}{2\mathrm{\pi }}\left[\mathrm{lnbla}\right]$ and to calculate the inductance of the cable is,$\mathrm{L}=\frac{{\mathrm{N\varphi }}_{\mathrm{B}}}{\mathrm{i}}$ whereN is the number of turns in the cable, i is current in the cable,${\varphi }_B$is the flux due to current through each turn of cable.

$\begin{array}{r}L=\frac{\frac{{\mu }_0\mathrm{il}}{2\pi }\left[\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right]}{\mathrm{i}}\\ =\frac{{\mu }_0\mathrm{I}}{2\pi }\left[\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right]\end{array}$

Therefore, the expression for the inductance of the length of the cable is and hence proved$\frac{{\mu }_0\mathrm{I}}{2\pi }\left[\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right]$

The radius of the small solid conductor is a, the inner radius of the small thin-walled tube is b, and the inner and outer conductor carry equal current is i in the opposite direction and the expression for the inductance of the length of the cable is$\frac{{\mu }_0\mathrm{I}}{2\pi }\left[\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right]$and to calculate the energy stored in a inductor is,$\cup =\frac{1}{2}L{i}^2$ where,L is the inductor, i is the current flows in the inductor.

$\begin{array}{r}U=\frac{1}{2}\left(\frac{{\mu }_0\mid }{2\pi }\left[\ln \left(\frac{b}{a}\right)\right]\right){\dot{i}}^2\\ =\left(\frac{{\mu }_0{i}^2\mid }{4\pi }\left[\ln \left(\frac{b}{a}\right)\right]\right)\end{array}$

Therefore, the expression for the energy stored in the magnetic field for a length of the cable is$\frac{{\mu }_0{i}^2\mid }{4\pi }\left[\ln \left(\frac{b}{a}\right)\right]$