Chapter 8, Problem 8.27P

Let đ�œ‡_rj = 2 in region 1, defined by 2x + 3y — 4z >1, while p_r2 = 5 in region 2 where 2x + 3y - 4z < 1. In region 1, H₁ = 50ax - 30a_y + 20a_z A/m. Find (o) HM₁; (b) H_r1 (c)H_r2; (d) H_N2 (e) 0_1, the angle between H₁ and a_N2; (f) 0_2, the angle between H₂ and a_N2].

To determine

(a)

The value of ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}$.

Answer to Problem 8.27P

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\text{A/m}$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\end{array}$

Calculation:

Both regions are separated by the surface $2x+3y-4z=1$ . So, the unit normal vector,

   $\begin{array}{l}{\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_N=\frac{1}{\sqrt{2^2+3^2+4^2}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\\ =\frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\end{array}$

So, the normal component of magnetic field intensity:

   $\begin{array}{l}{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=\left({\text{<mover accent="true"><mn>H</mn> <mo>→</mo></mover>}}_1\cdot {\text{<mover accent="true"><mn>a</mn> <mo>^</mo></mover>}}_N\right){\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_N\\ =\left(\left(50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\right)\cdot \frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\right)\frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\\ =-\frac{70}{29}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\\ =-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\text{A/m}\end{array}$

Conclusion:

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\text{A/m}$.

To determine

(b)

The value of ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}$.

Answer to Problem 8.27P

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\text{A/m}\end{array}$

Calculation:

The tangential component of magnetic field intensity:

   $\begin{array}{l}{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}={\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1-{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=\left(50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\right)-\left(-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\right)\\ =54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}\end{array}$

Conclusion:

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}$.

To determine

(c)

The value of ${\text{H}}_{t2}$.

Answer to Problem 8.27P

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t2}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}\end{array}$

Calculation:

The tangential components of magnetic field intensity are continuous across the surface between the regions. So:

:

   ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t2}={\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t1}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}$

Conclusion:

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t2}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}$.

To determine

(d)

The value of ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}$.

Answer to Problem 8.27P

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}=-1.93{\text{a}}_x-2.90{\text{a}}_y+3.86{\text{a}}_z\text{A/m}$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\end{array}$

Calculation:

The normal components of magnetic flux density are continuous across the surface between the regions. So,:

   $\begin{array}{l}{\text{<mover accent="true"><mn>B</mn><mo>→</mo></mover>}}_{N1}={\text{<mover accent="true"><mn>B</mn><mo>→</mo></mover>}}_{N1}\Rightarrow {\mu }_{r1}{\mu }_0{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}={\mu }_{r2}{\mu }_0{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}\\ \Rightarrow {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}=\frac{{\mu }_{r1}}{{\mu }_{r2}}{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N1}=\frac{2}{5}\left(-4.83{\text{a}}_x-7.24{\text{a}}_y+9.66{\text{a}}_z\right)\\ =-1.93{\text{a}}_x-2.90{\text{a}}_y+3.86{\text{a}}_z\text{A/m}\end{array}$

Conclusion:

The required value is, ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}=-1.93{\text{a}}_x-2.90{\text{a}}_y+3.86{\text{a}}_z\text{A/m}$.

To determine

(e)

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$.

Answer to Problem 8.27P

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$ is, ${\theta }_1=102°$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\end{array}$

Calculation:

The unit normal vector:

   ${\text{a}}_{N21}=\frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)$

The required angle:

   $\begin{array}{l}\cos {\theta }_1=\frac{{\text{<mover accent="true"><mn>H</mn> <mo>→</mo></mover>}}_1}{\left|H_1\right|}\cdot {\text{a}}_{N21}=\frac{\left(50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\right)}{\sqrt{{50}^2+{30}^2+{20}^2}}\cdot \frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\\ =-0.21\\ \Rightarrow {\theta }_1=102°\end{array}$

Conclusion:

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$ is, ${\theta }_1=102°$.

To determine

(f)

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_2$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$.

Answer to Problem 8.27P

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_2$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$ is, ${\theta }_2=95°$.

Explanation of Solution

Given Information:

The region 1 is $2x+3y-4z>1$ , and region 2 is $2x+3y-4z<1$.

   $\begin{array}{l}{\mu }_{r1}=2\\ {\mu }_{r2}=5\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_1=50{\text{a}}_x-30{\text{a}}_y+20{\text{a}}_z\text{A/m}\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t2}=54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\text{A/m}\\ {\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}=-1.93{\text{a}}_x-2.90{\text{a}}_y+3.86{\text{a}}_z\text{A/m}\end{array}$

Calculation:

The unit normal vector:

   ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}=\frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)$

The magnetic field intensity:

   $\begin{array}{l}{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_2={\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{t2}+{\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_{N2}=\left(54.83{\text{a}}_x-22.76{\text{a}}_y+10.34{\text{a}}_z\right)+\left(-1.93{\text{a}}_x-2.90{\text{a}}_y+3.86{\text{a}}_z\right)\\ =52.90{\text{a}}_x-25.66{\text{a}}_y+14.20{\text{a}}_z\text{A/m}\end{array}$

The required angle,

   $\begin{array}{l}\cos {\theta }_2=\frac{{\text{<mover accent="true"><mn>H</mn> <mo>→</mo></mover>}}_2}{\left|H_2\right|}\cdot {\text{a}}_{N21}=\frac{\left(52.90{\text{a}}_x-25.66{\text{a}}_y+14.20{\text{a}}_z\right)}{\sqrt{{50.9}^2+{25.66}^2+{14.20}^2}}\cdot \frac{1}{\sqrt{29}}\left(2{\text{a}}_x+3{\text{a}}_y-4{\text{a}}_z\right)\\ =-0.088\\ \Rightarrow {\theta }_2=95°\end{array}$

Conclusion:

The angle between ${\text{<mover accent="true"><mn>H</mn><mo>→</mo></mover>}}_2$ and ${\text{<mover accent="true"><mn>a</mn><mo>^</mo></mover>}}_{N21}$ is, ${\theta }_2=95°$.