91Ó°ÊÓ

Suppose the field inside a large piece of magnetic material is B₀.

Consider a small spherical cavity is hollowed out of the material in (Fig. 6.21).

Assume the cavities are small enough so M, B₀, and H₀ are essentially constant.

Write the formula offield at the center of the cavity, in terms of .B₀

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{0}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{s}}$ …… (1)

Here, ${\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{0}}$is field inside a large piece of magnetic material and ${\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{s}}$is field inside of a uniformly magnetized sphere.

Write the formula of at the center of the cavity, in terms of H₀ and M

$\stackrel{\mathbf{→}}{\mathbf{H}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\stackrel{\mathbf{→}}{\mathbf{B}}$ …… (2)

Here, ${\mathit{μ}}_{\mathbf{0}}$is permeability and $\stackrel{\mathbf{→}}{\mathbf{B}}$ is magnetic field at the center of the cavity in terms of B₀.

Write the formula offield for long needle-shaped cavity perpendicular to M.

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{0}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{s}}$ …… (3)

Here, role="math" localid="1657695281686" ${\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathbf{0}}$is field inside a large piece of magnetic material and ${\stackrel{\mathbf{→}}{\mathbf{B}}}_{\mathit{s}}$is field inside of a uniformly magnetized sphere.

Write the formula of H for long needle-shaped cavity perpendicular to M.

role="math" localid="1657695633493" $\stackrel{\mathbf{→}}{\mathbf{H}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\stackrel{\mathbf{→}}{\mathbf{B}}$ …… (4)

Here, ${\mathbf{μ}}_{\mathbf{0}}$is permeability and $\stackrel{\mathbf{→}}{\mathbf{B}}$ is magnetic field at the center of the cavity in terms of B₀.

Write the formula of H for a thin wafer-shaped cavity perpendicular to M.

$\stackrel{\mathbf{→}}{\mathbf{H}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\stackrel{\mathbf{→}}{\mathbf{B}}$ …… (5)

Here, ${\mathit{μ}}_{\mathbf{0}}$is permeability and $\stackrel{\mathbf{→}}{\mathbf{B}}$ is magnetic field at the center of the cavity in terms of B₀.

The field inside of a uniformly magnetized sphere with magnetization $-\stackrel{→}{M}$is:

role="math" localid="1657696622237" ${\stackrel{→}{B}}_s=-\frac{2}{3}{\mathrm{μ}}_0\stackrel{→}{M}$

Determine the field at the center of the cavity, in terms of B₀ and.

Substitute $-\frac{2}{3}{\mathrm{μ}}_0\stackrel{→}{M}$for ${\stackrel{→}{B}}_s$into equation (1).

role="math" localid="1657696633523" $\stackrel{→}{B}={\stackrel{→}{B}}_0-\frac{2}{3}{\mathrm{μ}}_0\stackrel{→}{M}$

Therefore, the values of field at the center of the cavity, in terms of B₀ and M is $\stackrel{→}{B}={\stackrel{→}{B}}_0-\frac{2}{3}{\mathrm{μ}}_0\stackrel{→}{M}$.

Determine the H at the center of the cavity, in terms of H₀ and M.

Substitute ${\stackrel{→}{B}}_0$for $\stackrel{→}{B}$into equation (2).

$\begin{array}{rcl}\stackrel{→}{H}& =& \frac{1}{{\mathrm{μ}}_0}{\stackrel{→}{B}}_0-\frac{2}{3}\stackrel{→}{M}\\ & =& \frac{1}{{\mathrm{μ}}_0}\left({\mathrm{μ}}_0\stackrel{→}{M}+{\mathrm{μ}}_0{\stackrel{→}{H}}_0\right)-\frac{2}{3}\stackrel{→}{M}\\ & =& {\stackrel{→}{H}}_0+\frac{\stackrel{→}{M}}{3}\end{array}$

Therefore, the value of at the center of the cavity, in terms of H₀ and M is $\begin{array}{rcl}\stackrel{→}{H}& =& {\stackrel{→}{H}}_0+\frac{\stackrel{→}{M}}{3}\end{array}$.

The needle cavity's induced field will resemble an endless cylinder with magnetization $-\stackrel{→}{M}$in this scenario. It causes a field that is:

role="math" localid="1657697322029" ${\stackrel{→}{B}}_s=-{\mathrm{μ}}_0\stackrel{→}{M}$

Determine the field for long needle-shaped cavity perpendicular to $\stackrel{→}{M}$.

Substitute $-{\mathrm{μ}}_0\stackrel{→}{M}$for ${\stackrel{→}{B}}_s$into equation (3).

$\stackrel{→}{B}={\stackrel{→}{B}}_0-{\mathrm{μ}}_0\stackrel{→}{M}$

Determine the H for long needle-shaped cavity perpendicular to $\stackrel{→}{M}$.

Substitute ${\stackrel{→}{B}}_0-{\mathrm{μ}}_0\stackrel{→}{M}$for $\stackrel{→}{B}$into equation (4).

$\begin{array}{rcl}\stackrel{→}{H}& =& \frac{1}{{\mathrm{μ}}_0}\left({\stackrel{→}{B}}_0-{\mathrm{μ}}_0\stackrel{→}{M}\right)\\ & =& \frac{1}{{\mathrm{μ}}_0}{\stackrel{→}{B}}_0-\stackrel{→}{M}\\ & =& {\stackrel{→}{H}}_0\end{array}$

The cavity will only generate a little magnetic field if the wafer is very thin, and the field will be:

Determine the field for a thin wafer-shaped cavity.

$\stackrel{→}{B}={\stackrel{→}{B}}_0$

Determine the H for a thin wafer-shaped cavity perpendicular to M.

$\stackrel{→}{H}={\stackrel{→}{H}}_0+\stackrel{→}{M}$

Therefore, the value of H for a thin wafer-shaped cavity perpendicular to M is $\stackrel{→}{H}=\frac{1}{{\mathrm{μ}}_0}$.