Chapter 9, Problem 9.18P

To determine

(a)

The value of $\beta$.

Answer to Problem 9.18P

The value is $\beta =14\cdot 91{\text{m}}^{-1}$.

Explanation of Solution

Given Information:

Breadth, b = 4 cm

Diameter, d = 8 mm

Magnetic field, $H=5\cos \left({10}^{9}t-\beta z\right)a_y\text{A}/\text{m}$

Magnetic permeability, ${\mu }_r=1$

Dielectric constant of the medium, ${\epsilon }_r=20$.

Calculation:

   $H=5\cos \left({10}^{9}t-\beta z\right)a_y\text{A}/\text{m}$ (1)

Comparing with the general magnetic field expression

$H=H_0\cos \left(\omega t-kz\right)\widehat{y}\text{A}/\text{m}$ (2)

We get,

   $\omega ={10}^9{s}^{-1}$ and the wave number k is defined as,

$k=\frac{\omega }{v}$

Velocity of the electromagnetic wave inside the material/medium is, $v=\frac{1}{\sqrt{{\mu }_r{\epsilon }_r{\mu }_0{\epsilon }_0}}$

Speed of light in vacuum is $C=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=3\times {10}^8\text{m}/\text{s}$

Therefore, we get

   $\begin{array}{l}v=\frac{c}{\sqrt{{\mu }_r{\epsilon }_r}}=\frac{3\times {10}^8}{\sqrt{20}}\\ v=0\cdot 6708\times {10}^8\text{m}/\text{s}\end{array}$

Hence,

   $k=\frac{{10}^9}{0.6708\times {10}^8}=14.91{\text{m}}^{-1}$

Thus, we get

   $\beta =k=\frac{{10}^9}{0\cdot 6708\times {10}^8}=14\cdot 91{\text{m}}^{-1}$

Conclusion:

Thus, the value is $\beta =14\cdot 91{\text{m}}^{-1}$.

To determine

(b)

The displacement current density at z = 0.

Answer to Problem 9.18P

The displacement current density is $74\cdot 55\sin \left({10}^{9}t\right)\text{A}/{\text{m}}^{\text{2}}.$

Explanation of Solution

Given Information:

Breadth, b = 4cm

Diameter, d = 8mm

Magnetic field, $H=5\cos \left({10}^{9}t-\beta z\right)a_y\text{A}/\text{m}$

Magnetic permeability, ${\mu }_r=1$

Dielectric constant of the medium, ${\epsilon }_r=20$

Calculation:

The displacement current density, $\frac{\partial D}{\partial t}$ is calculated using the relation derived using Maxwell's equation,

   $\frac{\partial D}{\partial t}=\frac{\partial H}{\partial z}$ (3)

Therefore,

   $\begin{array}{l}\frac{\partial D}{\partial t}=\frac{\partial \left(5\cos \left({10}^{9}t-{\beta }_z\right)\right)}{\partial z}\\ \frac{\partial D}{\partial t}=5\beta \sin \left({10}^{9}t-\beta z\right)\\ \frac{\partial D}{\partial t}=5\times 14\cdot 91\sin \left({10}^{9}t-\beta z\right)\text{A}/{\text{m}}^{\text{2}}\end{array}$

At z = 0 we get,

   $\frac{\partial D}{\partial t}=5\times 14\cdot 91\sin \left({10}^{9}t\right)=74\cdot 55\sin \left({10}^{9}t\right)\text{A}/{\text{m}}^{\text{2}}$

Conclusion:

The displacement current density is, $74\cdot 55\sin \left({10}^{9}t\right)\text{A}/{\text{m}}^{\text{2}}.$

To determine

(c)

The total displacement current crossing the surface.

Answer to Problem 9.18P

The total displacement current is, $D=0.4\sin \left({10}^{9}t-0.745\right)\sin (0.745)a_x\text{A}\text{.}$

Explanation of Solution

Given Information:

Breadth, b = 4cm

Diameter, d = 8mm

Magnetic field, $H=5\cos \left({10}^{9}t-\beta z\right)a_y\text{A}/\text{m}$

Magnetic permeability, ${\mu }_r=1$

Dielectric constant of the medium, ${\epsilon }_r=20$

Calculation:

Total displacement current, crossing the surface is derived as:

   $\begin{array}{l}D=a_x{\displaystyle {\int }_{x=0}^{0.1}dz{\displaystyle {\int }_{y=0}^{b}dy\left(\frac{\partial D}{\partial t}\right)=}}{a}_x{\displaystyle {\int }_{x=0}^{0.1}dz{\displaystyle {\int }_{y=0}^{b}dy\left[74.55\sin \left({10}^{9}t-14.91z\right)\right]}}\\ D=a_x\left(74.55\times 0.04\right)\left(\frac{\cos \left({10}^{9}t-14.91z\right)-\cos \left({10}^{9}t\right)}{14.91}\right)\\ D=0.2\left[\cos \left({10}^{9}t-14.91z\right)-\cos \left({10}^{9}t\right)\right]a_x\end{array}$

   $D=a_x\left(74.55\times 0.04\right)\left(\frac{\cos \left({10}^{9}t-14.91z\right)-\cos \left({10}^{9}t\right)}{14.91}\right)$

   $D=0.2\left[\cos \left({10}^{9}t-14.91z\right)-\cos \left({10}^{9}t\right)\right]a_x$

Using the identity

   $\cos X-\cos Y=-2\sin \left(\frac{X+Y}{2}\right)\sin \left(\frac{X-Y}{2}\right)$

   $\cos \left({10}^{9}t-14.91z\right)-\cos \left({10}^{9}t\right)=2\sin \left({10}^{9}t-0.745\right)\sin (0.745)$

   $D=0.4\sin \left({10}^{9}t-0.745\right)\sin (0.745)a_{x}A$

Conclusion:

The total displacement current is $D=0.4\sin \left({10}^{9}t-0.745\right)\sin (0.745)a_x\text{A}\text{.}$