Question

The coaxial cable shown in FIGURE P29.56 consists of a solid inner conductor of radius ${\mathrm{R}}_1$surrounded by a hollow, very thin outer conductor of radius ${\mathrm{R}}_2$. 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.

b. Draw a graph of B versus r from $\mathrm{r}=0$to $\mathrm{r}=2{\mathrm{R}}_2$if ${\mathrm{R}}_1=\frac{1}{3}{\mathrm{R}}_{2.}$

Short answer

a. The magnetic field in

inner conductor =$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{Ir}}{2{\mathrm{\pi R}}_1^2}$

Space between the conductor =$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}$

Outside the conductor =$\mathrm{B}=0$

b. The graph between the B and r is

Step-by-step solution

Part (a) Step 1: Given information

We have given,

inner conductor of radius =${\mathrm{R}}_1$

outer conductor of radius =${\mathrm{R}}_2$

Current =$\mathrm{I}$

we have to find three magnetic fields within the inner conductor, in the space between the conductors, and outside the outer conductor.

Simplify

Using ampere's law for closed loop

$$\begin{gathered} {\int }_0^{\mathrm{L}}\overrightarrow{B}\overrightarrow{dl}={\mu }_0{I}_{enclosed} \\ Then, \\ B{\left[L\right]}_0^{2\pi r}={\mu }_0{I}_{enclosed} \\ B.2\pi r={\mu }_0{I}_{enclosed} \\ B=\frac{{\mu }_0{I}_{enclosed}}{2\pi r}.................................\left(1\right) \end{gathered}$$

The current density inside the radius${\mathrm{R}}_1$is given as

localid="1649504415272" ${\mathrm{J}}_1=\frac{\mathrm{i}}{\mathrm{A}}=\frac{\mathrm{i}}{{\mathrm{\pi R}}_1^2}$

The current density for radial path will be

localid="1649504486700" ${\mathrm{J}}_{\mathrm{r}}=\frac{\mathrm{i}}{\mathrm{A}}=\frac{{\mathrm{i}}_{\mathrm{enclosed}}}{{\mathrm{\pi r}}^2}$

Then enclosed current will be

localid="1649504528820" $$\begin{gathered} {\mathrm{J}}_1={\mathrm{J}}_{\mathrm{r}} \\ \frac{\mathrm{i}}{{\mathrm{\pi R}}_1^2}=\frac{{\mathrm{I}}_{\mathrm{enclosed}}}{{\mathrm{\pi r}}^2} \\ {\mathrm{I}}_{\mathrm{enclosed}}=\frac{{\mathrm{r}}^2}{{\mathrm{R}}_1^2}\times \mathrm{I} \end{gathered}$$

Put this value in equation (1)

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{Ir}}{2{\mathrm{\pi R}}_1^2}$

Simplify

For the magnet field between the two conductor.

Simply it will be arbitrary

$$\begin{gathered} {\int }_0^{\mathrm{L}}\overrightarrow{B}\overrightarrow{dl}={\mu }_{0}I \\ Then, \\ B{\left[L\right]}_0^{2\pi r}={\mu }_{0}I \\ B.2\pi r={\mu }_{0}i \\ B=\frac{{\mu }_{0}I}{2\pi r} \end{gathered}$$

For the a magnetic field outside the current carrying conductor will be zero. since there is not have any current enclosed.

$\mathrm{B}=0.$

Part (b) Step 1: Given information

We have given,

inner conductor of radius =${\mathrm{R}}_1$

outer conductor of radius =${\mathrm{R}}_2$

Current =$\mathrm{I}$

we have to find three magnetic fields within the inner conductor, in the space between the conductors, and outside the outer conductor graphs.

Simplify

The graph between the three region magnetic field is varying like as shown in the figure below.