Question 8.9: The xy-plane serves as the interface between two different m...

In Example 8.8, K=0, so eq. (8.46):

H_{1t}=H_{2t}   or  \frac{B_{1t}}{\mu_{1}}=\frac{B_{2t}}{\mu_{2}}

was appropriate. In this example, however, k\neq 0, and we must resort to eq. (8.45):

(H_{1}-H_{2})\times a_{n12}=K

in addition to eq. (8.41):

B_{1n}=B_{2n}   or  \mu_{1}H_{1n}=\mu_{2}H_{2n}

Consider the problem as illustrated in Figure 8.18. Let B_{1}=(B_{x},B_{y},B_{z})   in   mWb/m^{2}.

B_{1n}=B_{2n}=8a_{z}\rightarrow B_{z}=8                                                  (8.9.1)

But

H_{2}=\frac{B_{2}}{\mu_{2}}=\frac{1}{4\mu_{o}}(5a_{x}+8a_{z})mA/m                                         (8.9.2)

and

H_{1}=\frac{B_{1}}{\mu_{1}}=\frac{1}{6\mu_{o}}(B_{x}a_{x}+B_{y}a_{y}+B_{z}a_{z})mA/m                               (8.9.3)

Having found the normal components, we can find the tangential components by using

(H_{1}-H_{2})\times a_{n12}=K

or

H_{1}\times a_{n12}=H_{2}\times a_{n12}+K                                         (8.9.4)

Substituting eqs. (8.9.2) and (8.9.3) into eq. (8.9.4) gives

\frac{1}{6\mu_{o}}(B_{x}a_{x}+B_{y}a_{y}+B_{z}a_{z})\times a_{z}=\frac{1}{4\mu_{o}}(5a_{x}+8a_{z})\times a_{z}+\frac{1}{\mu_{o}}a_{y}

Equating components yields

B_{y}=0,       \frac{-B_{x}}{6}=\frac{-5}{4}+1   or  B_{x}=\frac{6}{4}=1.5                                  (8.9.5)

From eqs. (8.9.1) and (8.9.5), we have

B_{1}=1.5a_{x}+8a_{z} mWb/m^{2}

H_{1}=\frac{B_{1}}{\mu_{1}}=\frac{1}{\mu_{o}}(0.25a_{x}+1.33a_{z})mA/m

and

H_{2}=\frac{1}{\mu_{o}}(1.25a_{x}+2a_{z})mA/m

Note that H_{1x} is 1/\mu_{o}mA/m less than H_{2x} because of the current sheet and also that B_{1n}=B_{2n}.