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Electrostatics

Chapter 05 Propagation of Uniform Plane Waves in Unbounded Space

Exercise Answers

5.1

In free space, given the electric field $E(z,t)=e_y{10}^3\sin (\omega t-\beta z)$ V/m, find the magnetic field strength $H(z,t)$.

Solution:

Taking the cosine as the basis, rewrite the known electric field expression:

$$E(z,t)=e_y{10}^3\cos (\omega t-\beta z-\frac{\pi }{2})\mathrm{V}/\mathrm{m}$$

This is an electromagnetic wave propagating uniformly in the $+z$ direction, with an initial phase of $-{90}^{\circ }$.

The accompanying magnetic field is:

$$H(z,t)=\frac{1}{{\eta }_0}{e}_z\times E(z,t)=\frac{1}{{\eta }_0}{e}_z\times e_y{10}^3\cos (\omega t-\beta z-\frac{\pi }{2})$$$$=-e_x\frac{{10}^3}{120\pi }\cos (\omega t-\beta z-\frac{\pi }{2})$$$$=-e_{x}2\cdot 65\sin (\omega t-\beta z)\mathrm{A}/\mathrm{m}$$

5.2

In an ideal medium (parameters $\mu ={\mu }_0$, $\epsilon ={\epsilon }_r{\epsilon }_0$, $\sigma =0$) there is a uniform plane wave propagating along the $x$ direction, and its electric field instantaneous value expression is known as

$$E(x,t)=e_{y}377\cos ({10}^{9}t-5x)\mathrm{V}/\mathrm{m}$$

Find:

(1) The relative permittivity of this ideal medium;

(2) The magnetic field $H(x,t)$ accompanying $E(x,t)$;

(3) The average power density of this plane wave.

Notice

The first question is too complex to solve those partial differential equations directly. You need to remember the relationship between the propagation speed of electromagnetic waves and the relative dielectric constant (see Example 5.1.2 for details).

Solution:

(1)

Observing the given electric field expression, it represents a uniform plane wave propagating along the $+x$ direction, with a phase velocity of

$$v_p=\frac{\omega }{k}=\frac{{10}^9}{5}\mathrm{m}/\mathrm{s}=2\times {10}^8\mathrm{m}/\mathrm{s}$$

Also,

$$v_p=\frac{1}{\sqrt{\mu \epsilon }}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_r{\epsilon }_0}}=\frac{1}{\sqrt{{\epsilon }_r}}\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=\frac{1}{\sqrt{{\epsilon }_r}}\times 3\times {10}^8$$

Therefore,

$${\epsilon }_r={\left(\frac{3}{2}\right)}^2=2.25$$

(2)

It can also be directly obtained from the relationship $H=\frac{1}{\eta }{e}_n\times E$ to get $H$

$$H=\frac{1}{\eta }{e}_x\times e_{y}377e^{-j5x}=e_z\frac{\sqrt{{\epsilon }_r}}{{\eta }_0}\times 377e^{-j5x}=e_{z}1.5e^{-j5x}\mathrm{A}/\mathrm{m}$$

(3)

The average Poynting vector is

$$S_{av}=\frac{1}{2}\text{Re}[E\times H^{\ast }]=\frac{1}{2}\text{Re}[e_{y}377e^{-j5x}\times e_{z}1.5e^{j5x}]=e_{x}282.75\mathrm{W}/{\mathrm{m}}^2$$

5.3

In air, a uniform plane wave propagates along the $e_y$ direction with a frequency $f=400$ MHz. When $y=0.5$ m, $t=0.2$ ns, the maximum value of the electric field strength $E$ is $250$ V/m, and the unit vector indicating its direction is $e_{x}0.6-e_{z}0.8$. Find the instantaneous expressions for the electric field $E$ and the magnetic field $H$.(Same as Example 5.1.1)

Solution:

The general expression for the electric field strength of a uniform plane wave propagating along the $e_y$ direction is

$$E(y,t)=E_m\cos (\omega t-ky+\varphi )$$

According to the conditions given in this problem, the parameters in the formula are

$$\omega =2\pi f=8\pi \times {10}^8\mathrm{rad}/\mathrm{s}$$$$k=\omega \sqrt{{\mu }_0{\epsilon }_0}=\frac{\omega }{c}=\frac{8\pi \times {10}^8}{3\times {10}^8}\mathrm{rad}/\mathrm{s}=\frac{8\pi }{3}\mathrm{rad}/\mathrm{m}$$$$E_m=250(e_{x}0.6-e_{z}0.8)\mathrm{V}/\mathrm{m}$$

Since $y=0.5$ m, $t=0.2$ ns, $E$ reaches its maximum value, that is

$$E_m\cos (8\pi \times {10}^8\times 0.2\times {10}^{-9}-\frac{8\pi }{3}\times \frac{1}{2}+\varphi )=E_m$$

Thus it is obtained that

$$\varphi =\frac{4\pi }{3}-\frac{4\pi }{25}=\frac{88\pi }{75}$$

Therefore

$$E=(e_{x}150-e_{z}200)\cos (8\pi \times {10}^{8}t-\frac{8\pi }{3}y+\frac{88\pi }{75})\mathrm{V}/\mathrm{m}$$$$H=\frac{1}{{\eta }_0}{e}_y\times E=-(e_x\frac{5}{3\pi }+e_z\frac{5}{4\pi })\cos (8\pi \times {10}^{8}t-\frac{8\pi }{3}y+\frac{88\pi }{75})\mathrm{A}/\mathrm{m}$$

Contributor

Tuntun Yuchiha

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