91Ó°ÊÓ

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.

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

$\stackrel{\mathbf{→}}{\mathbf{B}}=\mathrm{μ}\stackrel{\mathbf{→}}{\mathbf{H}}$ …… (1)

Here,$\mathit{μ}$ is permeability and $\stackrel{\mathbf{→}}{\mathbf{H}}$is axillary field.

Write the formula of surface bound current.

${\stackrel{\mathbf{→}}{\mathbf{J}}}_b\mathbf{=}\mathbf{∇}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{M}}$ …… (2)

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

Write the formula ofsurface bound current.

${\stackrel{\mathbf{→}}{\mathbf{K}}}_b\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{M}}\mathbf{×}\widehat{\mathbf{s}}$ …… (3)

Here,$\stackrel{\mathbf{→}}{\mathbf{M}}$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" ${\mathbf{I}}_b\mathbf{=}\mathbf{2}{\mathbf{Ï€¸é°­}}_b\mathbf{+}{\mathbf{∫}}_{\mathbf{S}}{\mathbf{J}}_{\mathbf{b}}\mathbf{dS}$ …… (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 auxiliary field using an Amperian loop of arbitrary radii:

$\begin{array}{c}2\mathrm{π}sH={\mathrm{μ}}_0{I}_{f,enc}=\left\{\begin{array}{l}I{(s/R)}^2\text{ },s<R\\ I\text{ },s>R\end{array}\right\}\\ \stackrel{→}{H}=\left\{\begin{array}{l}\frac{{\mathrm{μ}}_0(1+{\mathrm{χ}}_m)Is}{2\mathrm{π}{R}^2}\mathrm{ϕ}\text{ },s<R\\ \frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}\mathrm{ϕ}\text{ },s>R\end{array}\right\}\end{array}$

Determine the magnetic field a distance from the axis.

Substitute $\left\{\begin{array}{l}\frac{{\mathrm{μ}}_0(1+{\mathrm{χ}}_m)Is}{2\mathrm{π}{R}^2}\mathrm{ϕ}\text{ },s<R\\ \frac{{\mathrm{μ}}_{0}I}{2\mathrm{π}s}\mathrm{ϕ}\text{ },s>R\end{array}\right\}$for$\stackrel{→}{H}$into equation (1).

Determine the magnetization.

$\begin{array}{c}\stackrel{→}{M}={\mathrm{χ}}_m\stackrel{→}{H}\\ =\frac{{\mathrm{χ}}_m{I}_s}{2\mathrm{π}{R}^2}\mathrm{ϕ}\end{array}$

Determine the bound currents.

Substitute$\frac{{\mathrm{χ}}_m{I}_s}{2\mathrm{π}{R}^2}\mathrm{ϕ}$for$\stackrel{→}{M}$into equation (2).

$\begin{array}{c}{\stackrel{→}{J}}_b=\frac{1}{s}\frac{∂}{∂s}\left(s{M}_{\mathrm{ϕ}}\right)\\ =\frac{{\mathrm{χ}}_{m}I}{\mathrm{π}{R}^2}\widehat{z}\end{array}$

Determine the bound currents.

Substitute $\frac{{\mathrm{χ}}_m{I}_s}{2\mathrm{π}{R}^2}\mathrm{ϕ}$for $\stackrel{→}{M}$into equation (3).

${\stackrel{→}{K}}_b=−\frac{{\mathrm{χ}}_{m}I}{2\mathrm{π}R}\widehat{z}$

Determine the net bound current flowing down the wire is:

Substitute $−\frac{{\mathrm{χ}}_{m}I}{2\mathrm{π}R}\widehat{z}$for${\stackrel{→}{K}}_b$ into equation (4).

$\begin{array}{c}{I}_b=−{\mathrm{χ}}_{m}I+\frac{2}{R^2}{\mathrm{χ}}_{m}I{∫}_0^{R}sds\\ =−{\mathrm{χ}}_{m}I+{\mathrm{χ}}_{m}I\\ =0\end{array}$

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