Question

consider the boundary z = 0, $vec{h}_2$ in z < 0, $vec{h}$ in z > 0
$mu_1 = 4mu_0$
$mu_2 = 3mu_0$
$vec{h}_2 = <7,8,9>$
$vec{h} =?$
$vec{k}=2hat{a}_x a/m$
$hat{a}_z\timesvec{k}=vec{h}_{2t}-vec{h}_t$
$vec{h}_t=vec{h}_{2t}-hat{a}_z\timesvec{k}$
$b_{2n}-b_n=mu_2 h_{2n}=mu_1 h_{1n}$
$h_{1n}=\frac{mu_2}{mu_1}h_{2n}=15hat{a}_z$

Explanation:

Step1: Recall boundary conditions

We know that at the boundary $z = 0$, the normal component of the magnetic - flux density is continuous i.e., $B_{1n}=B_{2n}$ or $\mu_1H_{1n}=\mu_2H_{2n}$, and the tangential component of the magnetic field intensity has a relationship with the surface current density $\vec{K}$ given by $\hat{a}_z\times\vec{K}=\vec{H}_{2t}-\vec{H}_{1t}$. Given $\mu_1 = 4\mu_0$, $\mu_2 = 3\mu_0$, $\vec{H}_2=\langle7,8,9
angle$, and $\vec{K} = 2\hat{a}_x$.

Step2: Calculate the normal component of $\vec{H}$

From $\mu_1H_{1n}=\mu_2H_{2n}$, we can find $H_{1n}=\frac{\mu_2}{\mu_1}H_{2n}$. Since the normal direction is along $\hat{a}_z$, and assuming the normal component of $\vec{H}_2$ is $H_{2z} = 9$ (from $\vec{H}_2=\langle7,8,9
angle$), then $H_{1n}=\frac{3\mu_0}{4\mu_0}\times9=\frac{27}{4}$.

Step3: Calculate the tangential component of $\vec{H}$

We have $\hat{a}_z\times\vec{K}=\vec{H}_{2t}-\vec{H}_{1t}$. Since $\vec{K} = 2\hat{a}_x$, $\hat{a}_z\times\vec{K}=2\hat{a}_y$. Let $\vec{H}_{1t}=\langle H_{1x},H_{1y},0
angle$ and $\vec{H}_{2t}=\langle7,8,0
angle$. Then $2\hat{a}_y=\langle7 - H_{1x},8 - H_{1y},0
angle$. So $7 - H_{1x}=0$ gives $H_{1x}=7$ and $8 - H_{1y}=2$ gives $H_{1y}=6$.

Step4: Combine to find $\vec{H}$

The magnetic field intensity $\vec{H}$ in the $z>0$ region has components: $H_x = 7$, $H_y = 6$, $H_z=\frac{27}{4}$. So $\vec{H}=\langle7,6,\frac{27}{4}
angle$.

Answer:

The value of $\vec{H}$ needs to be calculated step - by - step as follows.