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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 6.28

Recall from section 6.3.3 "Boundary conditions" that the perpendicular components of $\mathbf{B}$ at the interface are continuous:

$$\begin{array}{}\text{(Eq. 6.26)}& B_{\text{above}}^{\perp }-B_{\text{below}}^{\perp }=0\end{array}$$

And the parallel components of $\mathbf{H}$ are continuous in the absence of surface current:

$$\begin{array}{rl}{\mathbf{H}}_{\text{above}}^{\parallel }-{\mathbf{H}}_{\text{below}}^{\parallel }& =\cancel{{\mathbf{K}}_f}\times \hat{n}\\ & =\mathbf{0}\end{array}$$

In other words, we have:

$$\begin{array}{rl}{B}_1^{\perp }& =B_2^{\perp }\\ {\mathbf{H}}_1^{\parallel }& ={\mathbf{H}}_2^{\parallel }\end{array}$$

Now, recall that

$$\mathbf{B}=\mu \mathbf{H}$$

(where $\mu ={\mu }_0(1+{\chi }_m)$) So

$$\begin{array}{rl}{\mathbf{H}}_1^{\parallel }& ={\mathbf{H}}_2^{\parallel }\\ \Rightarrow \frac{1}{{\mu }_1}{\mathbf{B}}_1^{\parallel }& =\frac{1}{{\mu }_2}{\mathbf{B}}_2^{\parallel }\end{array}$$

By (geometric) definition of tangent, we have:

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

And

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

So

$$\begin{array}{rl}\frac{\tan {\theta }_2}{\tan {\theta }_1}& =\frac{B_2^{\parallel }}{B_2^{\perp }}\frac{1}{\frac{B_1^{\parallel }}{B_1^{\perp }}}\\ & =\frac{B_2^{\parallel }}{B_2^{\perp }}\frac{B_1^{\perp }}{B_1^{\parallel }}\\ \text{(By boundary condition)}& & =\frac{B_2^{\parallel }}{\cancel{B_2^{\perp }}}\frac{\cancel{B_1^{\perp }}}{B_1^{\parallel }}& =\frac{B_2^{\parallel }}{B_1^{\parallel }}\\ & =\frac{{\mu }_2}{{\mu }_1}\end{array}$$

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