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

$\mathit{M}\mathbf{=}\mathit{k}\mathit{s}\mathbf{}\hat{\mathbf{z}}\mathbf{,}$

Where is a constant and is the distance from the axis; there is no free current anywhere. Find the magnetic field inside and outside the cylinder by two different methods: (a) As in Sect. 6.2, locate all the bound currents, and calculate the field they produce. (b) Use Ampere's law (in the form of Eq. 6.20) to find, and then get from Eq. 6.18. (Notice that the second method is much faster, and avoids any explicit reference to the bound currents.)

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

Write the formula ofall bound currents.

${\overrightarrow{\mathbf{K}}}_{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{M}}\mathbf{\times }\hat{\mathbf{s}}$ …… (1)

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

Write the formula of all bound currents.

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

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

Write the formula ofmagnetic field they produce.

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

Here,${\mathit{\mu }}_{\mathbf{0}}$ 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.

$\overrightarrow{\mathbf{B}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\overrightarrow{\mathbf{M}}$ …… (4)

Here, ${\mathit{\mu }}_{\mathbf{0}}$is permeability and$\overrightarrow{\mathbf{M}}$ is frozen in magnetization.

Determine the bound currents first:

Substitute k for $\overrightarrow{M}$and $R\hat{\varphi }$for $\hat{s}$into equation (1).

${\overrightarrow{k}}_b=kR\hat{\varphi }$

Determine the bound currents second:

Substitute -k for $\overrightarrow{M}$and $\hat{\varphi }$for $\nabla$into equation (2).

${\overrightarrow{J}}_b=-k\hat{\varphi }$

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:

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

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

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.

$\begin{array}{rcl}\overrightarrow{H}& =& \frac{\overrightarrow{B}}{{\mu }_0}-\overrightarrow{M}\\ 0& =& \frac{\overrightarrow{B}}{{\mu }_0}-\overrightarrow{M}\\ \overrightarrow{B}& =& {\mu }_0\overrightarrow{M}\\ & =& {\mu }_{0}ks\hat{z}\end{array}$

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