Question

A current $\mathit{I}$flows down a long straight wire of radius. If the wire is made of linear material (copper, say, or aluminium) with susceptibility ${\mathbf{X}}_m$, and the current is distributed uniformly, what is the magnetic field a distance$\mathbf{s}$ from the axis? Find all the bound currents. What is the net bound current flowing down the wire?

Short answer

The value of magnetic field a distance$s$ from the axis is$\overrightarrow{B}=\left\{\begin{array}{l}\frac{{\mu }_0(1+{\chi }_m)Is}{2\pi R^2}\varphi ,s<R\\ \frac{{\mu }_{0}I}{2\pi s}\varphi ,s>R\end{array}\right\}$ .

The value of bound currents are${\overrightarrow{J}}_n=\frac{{\chi }_{m}I}{\pi R^2}\widehat{z}$ and ${\overrightarrow{K}}_b=-\frac{{\chi }_{m}I}{2\pi R}\widehat{z}$.

The value of net bound current flowing down the wire is $0$.

Step-by-step solution

Write the given data from the question.

Consider acurrent $I$flows down a long straight wire of radius.

Consider a wire is made of linear material (copper, say, or aluminium) with susceptibility,${\text{X}}_{\text{m}}$and the current is distributed uniformly.

Determine the formula of magnetic field a distance S from the axis, bound currents and net bound current flowing down the wire.

Write the formula of magnetic field a distance from the axis.

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

Here,$\mathit{\mu }$ is permeability and $\overrightarrow{\mathbf{H}}$is axillary field.

Write the formula of surface bound current.

${\overrightarrow{\mathbf{J}}}_b\mathbf{=}\mathbf{\nabla }\mathbf{\times }\overrightarrow{\mathbf{M}}$ …… (2)

Here,role="math" localid="1657713888787" $\overrightarrow{\mathbf{M}}$is magnetization.

Write the formula ofsurface bound current.

${\overrightarrow{\mathbf{K}}}_b\mathbf{=}\overrightarrow{\mathbf{M}}\mathbf{\times }\widehat{\mathbf{s}}$ …… (3)

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

Write the formula of net bound current flowing down the wire.

role="math" localid="1657713980832" ${\mathbf{I}}_b\mathbf{=}\mathbf{2}{\mathbf{\pi RK}}_b\mathbf{+}{\mathbf{\int }}_{\mathbf{S}}{\mathbf{J}}_{\mathbf{b}}\mathbf{dS}$ …… (4)

Here,$\mathbf{R}$ is radius of wire, ${\mathbf{K}}_b$is surface bound current and${\mathbf{J}}_{\mathbf{b}}$ is surface bound current.

Determine the value of magnetic field a distance from the axis, bound currents and net bound current flowing down the wire.

Determine the auxiliary field using an Amperian loop of arbitrary radii:

$\begin{array}{c}2\pi sH={\mu }_0{I}_{f,enc}=\left\{\begin{array}{l}I{(s/R)}^2,s<R\\ I,s>R\end{array}\right\}\\ \overrightarrow{H}=\left\{\begin{array}{l}\frac{{\mu }_0(1+{\chi }_m)Is}{2\pi R^2}\varphi ,s<R\\ \frac{{\mu }_{0}I}{2\pi s}\varphi ,s>R\end{array}\right\}\end{array}$

Determine the magnetic field a distance from the axis.

Substitute $\left\{\begin{array}{l}\frac{{\mu }_0(1+{\chi }_m)Is}{2\pi R^2}\varphi ,s<R\\ \frac{{\mu }_{0}I}{2\pi s}\varphi ,s>R\end{array}\right\}$for$\overrightarrow{H}$into equation (1).

Determine the magnetization.

$\begin{array}{c}\overrightarrow{M}={\chi }_m\overrightarrow{H}\\ =\frac{{\chi }_m{I}_s}{2\pi R^2}\varphi \end{array}$

Determine the bound currents.

Substitute$\frac{{\chi }_m{I}_s}{2\pi R^2}\varphi$for$\overrightarrow{M}$into equation (2).

$\begin{array}{c}{\overrightarrow{J}}_b=\frac{1}{s}\frac{\partial }{\partial s}\left(s{M}_{\varphi }\right)\\ =\frac{{\chi }_{m}I}{\pi R^2}\widehat{z}\end{array}$

Determine the bound currents.

Substitute $\frac{{\chi }_m{I}_s}{2\pi R^2}\varphi$for $\overrightarrow{M}$into equation (3).

${\overrightarrow{K}}_b=-\frac{{\chi }_{m}I}{2\pi R}\widehat{z}$

Determine the net bound current flowing down the wire is:

Substitute $-\frac{{\chi }_{m}I}{2\pi R}\widehat{z}$for${\overrightarrow{K}}_b$ into equation (4).

$\begin{array}{c}{I}_b=-{\chi }_{m}I+\frac{2}{R^2}{\chi }_{m}I{\int }_0^{R}sds\\ =-{\chi }_{m}I+{\chi }_{m}I\\ =0\end{array}$

Therefore, the value of net bound current flowing down the wire is $0$.