At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that $\mathit{t}\mathit{a}\mathit{n}{\mathit{\theta }}_{\mathbf{2}}\mathbf{/}\mathit{t}\mathit{a}\mathit{n}{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathit{\mu }}_{\mathbf{2}}\mathbf{/}{\mathit{\mu }}_{\mathbf{1}}$ assuming there is no free current at the boundary. Compare Eq. 4.68.

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${\mu }_1$

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

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

$\mathbf{B}\mathbf{=}\mathbf{\mu H}$ ……. (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.

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

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

$below$The perpendicular component is continuous.

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

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

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

$H_{above}^{\parallel }=\frac{B_{above}^{\parallel }}{{\mu }_1}$

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

$H_{below}^{\parallel }=\frac{B_{below}^{\parallel }}{{\mu }_2}$

Recall equation (2),

$H_{above}^{\parallel }=H_{below}^{\parallel }$

Substitute $\frac{B_{above}^{\parallel }}{{\mu }_1}$for $H_{above}^{\parallel }$and $\frac{B_{below}^{\parallel }}{{\mu }_2}$for $H_{below}^{\parallel }$into above equation.

$\frac{B_{above}^{\parallel }}{{\mu }_1}=\frac{B_{below}^{\parallel }}{{\mu }_2}$ ……. (4)

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

$\frac{B_{above}^{\parallel }}{{\mu }_1}\times \frac{1}{B_{above}^{\perp }}=\frac{B_{below}^{\parallel }}{{\mu }_2}\times \frac{1}{B_{below}^{\perp }}$

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

$\frac{B_1^{\parallel }}{{\mu }_1}\times \frac{1}{B_1^{\perp }}=\frac{B_2^{\parallel }}{{\mu }_2}\times \frac{1}{B_2^{\perp }}$

$\frac{B_1^{\parallel }}{B_1^{\perp }{\mu }_1}\times \frac{1}{{\mu }_1}=\frac{B_2^{\parallel }}{B_2^{\perp }}\times \frac{1}{{\mu }_2}$ ……. (5)

The angle${\theta }_1$is given by,

$\tan {\theta }_1=\frac{B_1^{\parallel }}{B_1^{\perp }}$

The angle${\theta }_2$is given by,

$\tan {\theta }_2=\frac{B_2^{\parallel }}{B_2^{\perp }}$

Substitute$\tan {\theta }_1$ for $\frac{B_1^{\parallel }}{B_1^{\perp }}$and $\tan {\theta }_2$ for $\frac{B_2^{\parallel }}{B_2^{\perp }}$into equation (5).

$\begin{array}{c}\tan {\theta }_1\times \frac{1}{{\mu }_1}=\tan {\theta }_2\times \frac{1}{{\mu }_2}\\ \frac{\tan {\theta }_2}{\tan {\theta }_1}=\frac{\frac{1}{{\mu }_1}}{\frac{1}{{\mu }_2}}\\ \frac{\tan {\theta }_2}{\tan {\theta }_1}=\frac{{\mu }_2}{{\mu }_1}\end{array}$

Hence the expression$\frac{\tan {\theta }_2}{\tan {\theta }_1}=\frac{{\mu }_2}{{\mu }_1}$ is obtained.