91Ó°ÊÓ

The linear magnetic field strength of one material is $B_1$.

The linear magnetic field strength of another material is$B_2$ .

The magnetic permeability of one material is${\mathrm{μ}}_1$

The magnetic permeability of another material is ${\mathrm{μ}}_2$

The relationship between the magnetic field strength and magnetic field intensity is given as follows.

$\mathbf{B}\mathbf{=}\mathbf{μ\pm á}$ ……. (1)

Here $\mathbf{H}$ is the magnetic field intensity.

The interface between one linear magnetic material and another linear magnetic material parallel component of $H$is continuous.

$H_{above}^{âˆ\yen }=H_{below}^{âˆ\yen }$ …… (2)

Here $H_{above}^{âˆ\yen }$is the above the interface and $H_{below}^{âˆ\yen }$is the below interface.

$below$The perpendicular component is continuous.

$B_{above}^{âŠ\yen }=B_{below}^{âŠ\yen }$ ……. (3)

Here $B_{above}^{âŠ\yen }$is the above the interface and $B_{below}^{âŠ\yen }$is below the interface.

The relationship between the magnetic field strength and magnetic field intensity for one material given by,

$H_{above}^{âˆ\yen }=\frac{B_{above}^{âˆ\yen }}{{\mathrm{μ}}_1}$

The relationship between the magnetic field strength and magnetic field intensity for another material given by,

$H_{below}^{âˆ\yen }=\frac{B_{below}^{âˆ\yen }}{{\mathrm{μ}}_2}$

Recall equation (2),

$H_{above}^{âˆ\yen }=H_{below}^{âˆ\yen }$

Substitute $\frac{B_{above}^{âˆ\yen }}{{\mathrm{μ}}_1}$for $H_{above}^{âˆ\yen }$and $\frac{B_{below}^{âˆ\yen }}{{\mathrm{μ}}_2}$for $H_{below}^{âˆ\yen }$into above equation.

$\frac{B_{above}^{âˆ\yen }}{{\mathrm{μ}}_1}=\frac{B_{below}^{âˆ\yen }}{{\mathrm{μ}}_2}$ ……. (4)

Divide the equation (4) by equation (3).

$\frac{B_{above}^{âˆ\yen }}{{\mathrm{μ}}_1}×\frac{1}{B_{above}^{âŠ\yen }}=\frac{B_{below}^{âˆ\yen }}{{\mathrm{μ}}_2}×\frac{1}{B_{below}^{âŠ\yen }}$

Substitute $B_2$ for $B_{below}$and $B_1$for $B_{above}$into above equation.

$\frac{B_1^{âˆ\yen }}{{\mathrm{μ}}_1}×\frac{1}{B_1^{âŠ\yen }}=\frac{B_2^{âˆ\yen }}{{\mathrm{μ}}_2}×\frac{1}{B_2^{âŠ\yen }}$

$\frac{B_1^{âˆ\yen }}{B_1^{âŠ\yen }{\mathrm{μ}}_1}×\frac{1}{{\mathrm{μ}}_1}=\frac{B_2^{âˆ\yen }}{B_2^{âŠ\yen }}×\frac{1}{{\mathrm{μ}}_2}$ ……. (5)

The angle${\mathrm{θ}}_1$is given by,

$\tan {\mathrm{θ}}_1=\frac{B_1^{âˆ\yen }}{B_1^{âŠ\yen }}$

The angle${\mathrm{θ}}_2$is given by,

$\tan {\mathrm{θ}}_2=\frac{B_2^{âˆ\yen }}{B_2^{âŠ\yen }}$

Substitute$\tan {\mathrm{θ}}_1$ for $\frac{B_1^{âˆ\yen }}{B_1^{âŠ\yen }}$and $\tan {\mathrm{θ}}_2$ for $\frac{B_2^{âˆ\yen }}{B_2^{âŠ\yen }}$into equation (5).

$\begin{array}{c}\tan {\mathrm{θ}}_1×\frac{1}{{\mathrm{μ}}_1}=\tan {\mathrm{θ}}_2×\frac{1}{{\mathrm{μ}}_2}\\ \frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\frac{\frac{1}{{\mathrm{μ}}_1}}{\frac{1}{{\mathrm{μ}}_2}}\\ \frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\frac{{\mathrm{μ}}_2}{{\mathrm{μ}}_1}\end{array}$

Hence the expression$\frac{\tan {\mathrm{θ}}_2}{\tan {\mathrm{θ}}_1}=\frac{{\mathrm{μ}}_2}{{\mathrm{μ}}_1}$ is obtained.