Question 8.8: Given that H1 = -2 ax + 6 ay + 4 az A/m in region y - x - 2 ...

Since y-x-2=0 is a plane,y-x\leq 2 or y\leq x+2 is region 1 in Figure 8.17. A point in this region may be used to confirm this. For example, the origin (0,0) is in this region because 0-0-2\lt 0. If we let the surface of the plane be described by f(x,y)=y-x-2 unit vector normal to the plane is given by

a_{n}=\frac{\nabla f}{|\nabla f|}=\frac{a_{y}-a_{x}}{\sqrt{2}}

(a)

M_{1}=\chi_{m1}H_{1}=(\mu_{r1}-1)H_{1}=(5-1)(-2,6,4)=-8a_{x}+24a_{y}+16a_{z}A/m

B_{1}=\mu_{1}H_{1}=\mu_{o}\mu_{r1}H_{1}=4\pi\times 10^{-7}(5)(-2,6,4)=-12.57a_{x}+37.7a_{y}+25.13a_{z}\mu Wb/m^{2}

(b)

H_{1n}=(H_{1}\cdot a_{n})a_{n}=\left[(-2,6,4)\cdot \frac{(-1,1,0)}{\sqrt{2}}\right]\frac{(-1,1,0)}{\sqrt{2}}=-4a_{x}+4a_{y}

But

H_{1}=H_{1n}+H_{1t}

Hence

H_{1t}=H_{1}-H_{1n}=(-2,6,4)-(-4,4,0)=2a_{x}+2a_{y}+4a_{z}

Using the boundary conditions, we have

H_{2t}=H_{1t}=2a_{x}+2a_{y}+4a_{z}

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

or

H_{2n}=\frac{\mu _{1}}{\mu_{2}}H_{1n}=\frac{5}{2}(-4a_{x}+4a_{y})=-10a_{x}+10a_{y}

Thus

H_{2}=H_{2n}+H_{2t}=-8a_{x}+12a_{y}+4a_{z}A/m

and

B_{2}=\mu_{2}H_{2}=\mu_{o}\mu_{r2}H_{2}=(4\pi \times10^{-7})(2)(-8,12,4)

=-20.11a_{x}+30.16a_{y}+10.05a_{z}\mu Wb/m^{2}