Question

A coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility ${\chi }_m$. A current$\mathit{I}$ flows down the inner conductor and returns along the outer one; in each case, the current distributes itself uniformly over the surface (Fig. 6.24). Find the magnetic field in the region between the tubes. As a check, calculate the magnetization and the bound currents, and confirm that (together, of course, with the free currents) they generate the correct field.

Figure 6.24

Short answer

The value of magnetic field in the region between the tubes is $\overrightarrow{B}=\frac{{\mu }_0(1+{\chi }_{\text{m}})I}{2\pi s}\widehat{\varphi }$.

The value of magnetization is$\overrightarrow{M}=\frac{{\chi }_{\text{m}}I}{2\pi s}\widehat{\varphi }$ .

The value of surface bound current are${\overrightarrow{J}}_b=0$ and role="math" localid="1657712479578" ${\overrightarrow{K}}_{b,a}=\frac{{\chi }_{\text{m}}I}{2\pi a}\widehat{z}$.

Step-by-step solution

Write the given data from the question.

Consider a coaxial cable consists of two very long cylindrical tubes, separated by linear insulating material of magnetic susceptibility ${\chi }_{\text{m}}$.

Consider a current$I$ flows down the inner conductor and returns along the outer one.

Determine the formula of magnetic field in the region between the tubes, magnetization and bound currents.

Write the formula of magnetic field in the region between the tubes.

$\overrightarrow{\mathbf{B}}=\mu \overrightarrow{\mathbf{H}}$ …… (1)

Here,$\mu$is permeability androle="math" localid="1657712664384" $\stackrel{}{\mathbf{H}}$is axillary field.

Write the formula of magnetization.

$\overrightarrow{\mathbf{M}}\mathbf{=}{\mathbf{\chi }}_m\overrightarrow{\mathbf{H}}$ …… (2)

Here, ${\mathbf{\chi }}_m$ is magnetic susceptibility and $\overrightarrow{\mathbf{H}}$is auxiliary field.

Write the formula of surface bound current.

${\overrightarrow{\mathbf{J}}}_b\mathbf{=}\mathbf{\nabla }\mathbf{\times }\overrightarrow{\mathbf{M}}$ …… (3)

Here,$\overrightarrow{\mathbf{M}}$ is magnetization.

Write the formula of surface bound current.

${\overrightarrow{\mathbf{K}}}_{b,a}\mathbf{=}\overrightarrow{\mathbf{M}}\mathbf{\times }\mathbf{(}\mathbf{-}\widehat{\mathbf{s}}\mathbf{)}$ …… (4)

Here,$\overrightarrow{\mathbf{M}}$ is magnetization and$\widehat{\mathbf{s}}$ is distance from the axis.

Determine the value of magnetic field in the region between the tubes, magnetization and bound currents.

Lay down the z-axis so that it points down the current on the inner cylinder, which should be allowed to flow to the right. We can locate the auxiliary field by doing an Amperian loop $\left(\text{a,s,b}\right)$;

$\begin{array}{c}2\pi s\overrightarrow{H}=I_{f,enc}\\ =I\\ \overrightarrow{H}=\frac{I}{2\pi s}\varphi \end{array}$

The media is linear, so the magnetic field is:

Determine the magnetic field in the region between the tubes.

$\begin{array}{c}\overrightarrow{B}=\mu \overrightarrow{H}\\ =\frac{{\mu }_0(1+{\chi }_m)I}{2\pi s}\widehat{\varphi }\end{array}$

Therefore, the value of magnetic field in the region between the tubes is

$\overrightarrow{B}=\frac{{\mu }_0(1+{\chi }_{\text{m}})I}{2\pi s}\widehat{\varphi }$

Determine the magnetization.

Substitute$\frac{I}{2\pi s}\varphi$ for$\overrightarrow{H}$into equation (2).

$\overrightarrow{M}=\frac{{\chi }_{m}I}{2\pi s}\widehat{\varphi }$

Therefore, the value of magnetization is$\overrightarrow{M}=\frac{{\chi }_{\text{m}}I}{2\pi s}\widehat{\varphi }$.

Determine the surface bound currents:

Substitute $\frac{{\chi }_{\text{m}}I}{2\pi s}\widehat{\varphi }$for $\overrightarrow{M}$into equation (3)

$\begin{array}{c}{\overrightarrow{J}}_b=\frac{1}{s}\frac{\partial }{\partial s}\left(s{M}_{\varphi }\right)\\ =0\end{array}$

Determine the surface bound currents:

Substitute$\frac{{\chi }_{\text{m}}I}{2\pi s}\widehat{\varphi }$ for$\overrightarrow{M}$ into equation (4)

${\overrightarrow{K}}_{b,a}=\frac{{\chi }_{m}I}{2\pi a}\widehat{z}$

Therefore, the value of surface bound current are ${\overrightarrow{J}}_b=0$and ${\overrightarrow{K}}_{b,a}=\frac{{\chi }_{\text{m}}I}{2\pi a}\widehat{z}$.

If we now disregard the media and just claim that the inner cylinder carries the effective current, we will not require the bound current at:$s=b$

$1+K_{b,a}\left(2\pi a\right)=I(1+{\chi }_m)$

Then the magnetic field is just as above.

$\overrightarrow{B}=\frac{{\mu }_0(1+{\chi }_m)I}{2\pi s}\widehat{\varphi }$