The coaxial cable shown in $figureP29.56$consists of a solid inner conductor of radius $R_1$surrounded by a hollow, very thin outer conductor of radius $R_2$. The two carry equal currents I, but in opposite directions. The current density is uniformly distributed over each conductor.

a. Find expressions for three magnetic fields: within the inner conductor, in the space between the conductors, and outside the outer conductor.

Finding the expressions for three magnetic fields.

To get the magnetic field of Ampere's law need to integrate over the circumference of the enclosed area.

$$\begin{gathered} {\int }_0^l\overrightarrow{B}·\overrightarrow{dI}={\mu }_o{I}_{encl} \\ B{\left[l\right]}_0^{2\pi r}={\mu }_o{I}_{encl} \\ B\left(2\pi r\right)={\mu }_o{I}_{encl} \\ B=\frac{{\mu }_o{I}_{encl}}{2\pi r} \end{gathered}$$

The current density inside the enclosed area (r) equals the current density in the whole wire of radius $\left(R_1\right)$

$J_1=\frac{1}{\pi R_1^2}$

$I$is the current of wire.

$J_r=\frac{I_{encl}}{\pi r^2}$

The current$I_{encl}$

$$\begin{gathered} J_1=J_r \\ \frac{I}{\pi R_1^2}=\frac{I_{encl}}{\pi r^2} \\ {I}_{encl}=\frac{r^2}{R_1^2}I \end{gathered}$$

Need to get B with the expression of I into equation (1).

$B=\frac{{\mu }_o{I}_{encl}}{2\pi r}=\frac{{\mu }_o\left({\displaystyle \frac{r^2}{R_1^2}}\right)}{2\pi r}=\boxed{\frac{{\mu }_{o}Ir}{2\pi R_1^2}}$

The magnetic field is inside the wire when $r\le R_{1}whereB\propto r$

The distance $R_2\ge r\ge R_1$and the current is the same for the wire with the radius

The distance$R_2\ge r\ge R_1$and the current is the same for the wire with the radius $R_1$

$$\begin{gathered} {\int }_0^l\overrightarrow{B}·\overrightarrow{dI}={\mu }_{o}I \\ B{\left[l\right]}_0^{2\pi r}={\mu }_{o}I \\ B\left(2\pi r\right)={\mu }_{o}I \\ \boxed{B=\frac{{\mu }_I}{2\pi r}} \end{gathered}$$

$r\ge R_2$

The current flows at Distance $r\ge R_2$the net current in the path is zero. From Ampere's law the magnetic field will be zero

$$\begin{gathered} \oint \overrightarrow{B}·\overrightarrow{dI}={\mu }_o\left(0\right) \\ \boxed{B=0} \end{gathered}$$

In the magnetic field the surface of the inner wire decreases as r increases till reach$R_2$, after this point it becomes zero.