91影视

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,$l=2\mathrm{蟺}r$Using Ampere's law:

$鈭�B.dl={渭}_0{l}_{enclosed}$,to calculate the magnitude of the magnetic field,

$\begin{array}{r}\mathrm{B}鈭�\mathrm{dl}={渭}_0{l}_{\text{enclosed}}\\ \mathrm{BI}={渭}_0{l}_{\text{enclosed}}\\ \mathrm{B}=\frac{{渭}_0{\mathrm{l}}_{\text{enclosed}}}{\mathrm{l}}\end{array}$

The entire current is enclosed by the loop is equal to the current $l_{enclosed}=0$ Substitute l =$2\mathrm{蟺谤}$ So, $\mathrm{B}=\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺谤}}$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 $\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺谤}}$, to calculate the magnetic flux is ${蠒}_B=BA$Differentiate the equation we have, ${\mathrm{诲蠒}}_{\mathrm{B}}=\mathrm{BdA}$.On substituting the value we have,${\mathrm{诲蠒}}_{{}_{\mathrm{B}}}\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺谤}}\mathrm{ldr}$

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

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 $\frac{{渭}_{0}l}{2蟺r}ldr$, to calculate the total flux produced by a current in the central conductor, we have

$\begin{array}{r}鈭�\mathrm{d}{蠒}_{\mathrm{B}}={鈭�}_{\mathrm{a}}^{\mathrm{b}}鈥�\frac{{渭}_0\mathrm{I}}{2蟺\mathrm{r}}\left(\mathrm{Idr}\right)\\ {蠒}_{\mathrm{B}}={\mathrm{I}}_{\frac{1}{0}}^{2蟺}{鈭�}_{\mathrm{a}}^{\mathrm{b}}鈥�\frac{\mathrm{dr}}{\mathrm{r}}\\ {蠒}_{\mathrm{b}}=\frac{{渭}_0\mathrm{il}}{2蟺}[\mathrm{lnr}{]}_{\mathrm{a}}^{\mathrm{b}}\\ =\frac{{渭}_0\mathrm{il}}{2蟺}[\mathrm{lnb}鈭�\mathrm{lna}]\\ {蠒}_{\mathrm{b}}=\frac{{渭}_0\mathrm{il}}{2蟺}\left[\mathrm{lnbla}\right]\end{array}$

Therefore, the expression for the total flux produced by a current in the central conductor is$\frac{{渭}_{0}iI}{2蟺}\left[In\left(\raisebox{1ex}{$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$\frac{{\mathrm{渭}}_0\mathrm{iI}}{2\mathrm{蟺}}\left[\mathrm{lnbla}\right]$ and to calculate the inductance of the cable is,$\mathrm{L}=\frac{{\mathrm{狈蠒}}_{\mathrm{B}}}{\mathrm{i}}$ whereN is the number of turns in the cable, i is current in the cable,${蠒}_B$is the flux due to current through each turn of cable.

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

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

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$\frac{{渭}_0\mathrm{I}}{2蟺}\left[\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\right]$and to calculate the energy stored in a inductor is,$鈭�=\frac{1}{2}L{i}^2$ where,L is the inductor, i is the current flows in the inductor.

$\begin{array}{r}U=\frac{1}{2}\left(\frac{{渭}_0鈭�}{2蟺}\left[\ln \left(\frac{b}{a}\right)\right]\right){\stackrel{藱}{i}}^2\\ =\left(\frac{{渭}_0{i}^2鈭�}{4蟺}\left[\ln \left(\frac{b}{a}\right)\right]\right)\end{array}$

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