91Ó°ÊÓ

Consider an infinitely long cylinder, of radius R, carries a "frozen-in" magnetization, parallel to the axis.

Write the formula ofall bound currents.

${\stackrel{\mathbf{→}}{\mathbf{K}}}_{\mathbf{b}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{M}}\mathbf{×}\hat{\mathbf{s}}$ …… (1)

Here, $\stackrel{\mathbf{→}}{\mathbf{M}}$is frozen in magnetization and $\hat{\mathbf{s}}$is distance from the axis.

Write the formula of all bound currents.

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

Here, $\stackrel{\mathbf{→}}{\mathbf{M}}$is frozen in magnetization.

Write the formula ofmagnetic field they produce.

$\mathit{B}\mathit{I}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}{\mathit{K}}_{\mathbf{b}}\mathit{I}\mathbf{-}{\mathit{μ}}_{\mathbf{0}}\mathit{I}{\mathbf{∫}}_{\mathbf{s}}^{\mathbf{R}}\mathit{k}\mathit{b}\mathit{s}$ …… (3)

Here,${\mathit{μ}}_{\mathbf{0}}$ is permeability, k is constant, $\hat{\mathbf{s}}$ is distance from the axis and I is current passing area.

Write the formula ofmagnetic field for any loop there is no enclosed free current.

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\stackrel{\mathbf{→}}{\mathbf{M}}$ …… (4)

Here, ${\mathit{μ}}_{\mathbf{0}}$is permeability and$\stackrel{\mathbf{→}}{\mathbf{M}}$ is frozen in magnetization.

Determine the bound currents first:

Substitute k for $\stackrel{→}{M}$and $R\widehat{\mathrm{ϕ}}$for $\hat{s}$into equation (1).

${\stackrel{→}{k}}_b=kR\widehat{\mathrm{ϕ}}$

Determine the bound currents second:

Substitute -k for $\stackrel{→}{M}$and $\widehat{\mathrm{ϕ}}$for $∇$into equation (2).

${\stackrel{→}{J}}_b=-k\widehat{\mathrm{ϕ}}$

Although the magnetic field will be in the z-direction, it will be zero outside since I_enc = 0. We have the following inside, utilising an Amperian rectangle that spans the surface:

$\begin{array}{rcl}BI& =& kri-k(R-s)\\ \stackrel{→}{B}& =& {\mathrm{μ}}_{0}ks\hat{z}\end{array}$

Therefore, the value of magnetic field they produce is $\begin{array}{rcl}\stackrel{→}{B}& =& {\mathrm{μ}}_{0}ks\hat{z}\end{array}$.

In the gray-shaded areas in the image below, the total current is seen passing through the RHS.

Figure 1

for any loop there is no enclosed free current, so we have:

H = 0

Determine the magnetic field.

$\begin{array}{rcl}\stackrel{→}{H}& =& \frac{\stackrel{→}{B}}{{\mathrm{μ}}_0}-\stackrel{→}{M}\\ 0& =& \frac{\stackrel{→}{B}}{{\mathrm{μ}}_0}-\stackrel{→}{M}\\ \stackrel{→}{B}& =& {\mathrm{μ}}_0\stackrel{→}{M}\\ & =& {\mathrm{μ}}_{0}ks\hat{z}\end{array}$

Therefore, the value of magnetic field for any loop there is no enclosed free current, is $\begin{array}{rcl}\stackrel{→}{B}& =& {\mathrm{μ}}_{0}ks\hat{z}\end{array}$.