Chapter 8, Problem 34P

(a)

To determine

Calculate the normal component of magnetic field intensity $H_{1n}$ for region 1.

Answer to Problem 34P

The normal component of magnetic field intensity $H_{1n}$ for region 1 is $-6.667a_x+6.667a_y-13.333a_z\text{A}/\text{m}$.

Explanation of Solution

Calculation:

Given that,

$f\left(x,y,z\right)=x-y+2z-5$

Consider the expression for the normal unit vector.

$a_n=\frac{\nabla f}{\left|\nabla f\right|}$        (1)

Here,

f is the plane surface function.

Substitute $x-y+2z-5$ for f in equation (1).

$\begin{array}{c}{a}_n=\frac{\nabla \left(x-y+2z-5\right)}{\left|\nabla \left(x-y+2z-5\right)\right|}\\ =\frac{\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\left(x-y+2z-5\right)}{\left|\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\left(x-y+2z-5\right)\right|}\left\{\because \nabla =\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial x}{a}_y+\frac{\partial }{\partial x}{a}_z\right\}\\ =\frac{a_x-a_y+2a_z}{\left|a_x-a_y+2a_z\right|}\\ =\frac{a_x-a_y+2a_z}{\sqrt{1^2+{\left(-1\right)}^2+{\left(2\right)}^2}}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{a}_n=\frac{a_x-a_y+2a_z}{\sqrt{1+1+4}}\\ =\frac{a_x-a_y+2a_z}{\sqrt{6}}\end{array}$

The normal component of magnetic field intensity $H_{1n}$ for region 1 is calculated as follows,

$H_{1n}=\left(H_1\cdot a_{\text{n}}\right)a_{\text{n}}$

Substitute $40a_x+20a_y-30a_z$ for $H_1$, and $\frac{a_x-a_y+2a_z}{\sqrt{6}}$ for $a_{\text{n}}$ in above equation.

$\begin{array}{c}{H}_{1n}=\left[\left(40a_x+20a_y-30a_z\right)\cdot \left(\frac{a_x-a_y+2a_z}{\sqrt{6}}\right)\right]\left(\frac{a_x-a_y+2a_z}{\sqrt{6}}\right)\\ =\left(\frac{1}{\sqrt{6}}\right)\left(40-20-60\right)\left(\frac{a_x-a_y+2a_z}{\sqrt{6}}\right)\\ =\frac{1}{6}\left(-40a_x+40a_y-80a_z\right)\\ =-6.667a_x+6.667a_y-13.333a_z\text{A}/\text{m}\end{array}$

Conclusion:

Thus, the normal component of magnetic field intensity $H_{1n}$ for region 1 is $-6.667a_x+6.667a_y-13.333a_z\text{A}/\text{m}$.

(b)

To determine

Calculate the tangential component of magnetic field intensity $H_{2t}$ for region 2.

Answer to Problem 34P

The tangential component of magnetic field intensity $H_{2t}$ for region 2 is $46.667a_x+13.333a_y-16.667a_z\text{A}/\text{m}$.

Explanation of Solution

Calculation:

Consider the general expression for the magnetic field intensity $H_1$.

$H_1=H_{1t}+H_{1n}$        (2)

Here,

$H_{1t}$ is the tangential component of magnetic field intensity of region 1, and

$H_{1n}$ is the normal component of magnetic field intensity of region 1.

Rearrange the equation (2) as follows,

$H_{1t}=H_1-H_{1n}$

Substitute $40a_x+20a_y-30a_z$ for $H_1$ and $-6.667a_x+6.667a_y-13.333a_z$ for $H_{1n}$ in above equation.

$\begin{array}{c}{H}_{1t}=\left(40a_x+20a_y-30a_z\right)-\left(-6.667a_x+6.667a_y-13.333a_z\right)\\ =40a_x+20a_y-30a_z+6.667a_x-6.667a_y+13.333a_z\\ =46.667a_x+13.333a_y-16.667a_z\end{array}$

For a magnetic boundary condition,

$H_{2t}=H_{1t}$        (3)

Using equation (3),

$H_{2t}=46.667a_x+13.333a_y-16.667a_z\text{A}/\text{m}$

Conclusion:

Thus, the tangential component of magnetic field intensity $H_{2t}$ for region 2 is $46.667a_x+13.333a_y-16.667a_z\text{A}/\text{m}$.

(c)

To determine

Calculate the magnetic flux density $B_{\text{2}}$ for region 2.

Answer to Problem 34P

The magnetic flux density $B_{\text{2}}$ for region 2 is $276.5a_x+100.5a_y-138.2a_z\mu \text{Wb}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Consider the expression for the magnetic boundary condition.

$\begin{array}{l}{B}_{1n}=B_{2n}\\ {\mu }_1{H}_{1n}={\mu }_2{H}_{2n}\left\{\because B=\mu H\right\}\\ H_{2n}=\frac{{\mu }_1}{{\mu }_2}{H}_{1n}\end{array}$

Substitute $5{\mu }_{\text{o}}$ for ${\mu }_2$, $2{\mu }_{\text{o}}$ for ${\mu }_1$, and $-6.667a_x+6.667a_y-13.333a_z$ for $H_{1n}$ in above equation.

$\begin{array}{c}{H}_{2n}=\frac{2{\mu }_{\text{o}}}{5{\mu }_{\text{o}}}\left(-6.667a_x+6.667a_y-13.333a_z\right)\\ =0.4\left(-6.667a_x+6.667a_y-13.333a_z\right)\\ =-2.667a_x+2.667a_y-5.333a_z\end{array}$

Consider the general expression for the magnetic field intensity $H_2$.

$H_2=H_{2t}+H_{2n}$        (4)

Here,

$H_{2t}$ is the tangential component of magnetic field intensity of region 2, and

$H_{2n}$ is the normal component of magnetic field intensity of region 2.

Substitute $46.667a_x+13.333a_y-16.667a_z$ for $H_{2t}$ and $-2.667a_x+2.667a_y-5.333a_z$ for $H_{2n}$ in equation (4).

$\begin{array}{c}{H}_2=46.667a_x+13.333a_y-16.667a_z-2.667a_x+2.667a_y-5.333a_z\\ =44a_x+16a_y-22a_z\end{array}$

The magnetic flux density for region 2 is,

$B_2={\mu }_2{H}_2$        (5)

Here,

${\mu }_2$ is the permeability of region 2, and

$H_2$ is the magnetic field intensity of region 2.

Substitute $44a_x+16a_y-22a_z$ for $H_2$ and $5{\mu }_{\text{o}}$ for ${\mu }_2$ in equation (5).

$\begin{array}{c}{B}_1=5{\mu }_{\text{o}}\left(44a_x+16a_y-22a_z\right)\\ =220{\mu }_{\text{o}}{a}_x+80{\mu }_{\text{o}}{a}_y-110{\mu }_{\text{o}}{a}_z\end{array}$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$ in above equation.

$\begin{array}{c}{B}_1=220\left(4\pi \times {10}^{-7}\right)a_x+80\left(4\pi \times {10}^{-7}\right)a_y-110\left(4\pi \times {10}^{-7}\right)a_z\\ =2.765\times {10}^{-4}{a}_x+1.005\times {10}^{-4}{a}_y-1.382\times {10}^{-4}{a}_z\\ =\left(276.5a_x+100.5a_y-138.2a_z\right)\times {10}^{-6}\text{Wb}/{\text{m}}^{\text{2}}\\ =276.5a_x+100.5a_y-138.2a_z\mu \text{Wb}/{\text{m}}^{\text{2}}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the magnetic flux density $B_{\text{2}}$ for region 2 is $276.5a_x+100.5a_y-138.2a_z\mu \text{Wb}/{\text{m}}^{\text{2}}$.