At the interface between one linear magnetic material and another, the magnetic field lines bend (Fig. 6.32). Show that \tan \theta_{2} / \tan \theta_{1}=\mu_{2} / \mu_{1}, assuming there is no free current at the boundary. Compare Eq. 4.68.
\tan \theta_{2} / \tan \theta_{1}=\epsilon_{2} / \epsilon_{1} (4.68)
At the interface, the perpendicular component of B is continuous (Eq. 6.26), and the parallel component of H is continuous (Eq. 6.25 with K _{f}=0 ). So B_{1}^{\perp}=B_{2}^{\perp}, H _{1}^{\|}= H _{2}^{\|} . \text {But } B =\mu H \text { (Eq. 6.31), so } \frac{1}{\mu_{1}} B _{1}^{\|}=\frac{1}{\mu_{2}} B _{2}^{\|} .
B_{\text {above }}^{\perp}-B_{\text {below }}^{\perp}=0 (6.26)
H _{\text {above }}^{\|}- H _{\text {below }}^{\prime \prime}= K _{f} \times \hat{ n } (6.25)
B =\mu H (6.31)
Now \tan \theta_{1}=B_{1}^{\|} / B_{1}^{\perp}, \text { and } \tan \theta_{2}=B_{2}^{\|} / B_{2}^{\perp} , so
\frac{\tan \theta_{2}}{\tan \theta_{1}}=\frac{B_{2}^{\|}}{B_{2}^{\perp}} \frac{B_{1}^{\perp}}{B_{1}^{\|}}=\frac{B_{2}^{\|}}{B_{1}^{\|}}=\frac{\mu_{2}}{\mu_{1}}(the same form, though for di↵erent reasons, as Eq. 4.68).
\tan \theta_{2} / \tan \theta_{1}=\epsilon_{2} / \epsilon_{1} (4.68)