Question

) A uniform plane wave of 100 kHz travelling in free space strikes a large block of a
material having 4 , 9 and 0 ε = ε0 µ = µ0 σ = normal to the surface. If the incident
magnetic field vector is given by
10 cos( ) ˆ tesla 6 B = ωt −βy z
−
r
write the complete expressions for the incident, reflected, and transmitted field
vectors.

Answer 1

Answer on Question #79566, Physics / Electromagnetism

A uniform plane wave of $100\mathrm{kHz}$ travelling in free space strikes a large block of a material having $\varepsilon = 4\varepsilon_0$ , $\mu = 9\mu_0$ and $\sigma = 0$ normal to the surface. If the incident magnetic field vector is given by

$$\overrightarrow {\mathbf {B}} = 1 0 ^ {- 6} \cos (\omega t - \beta y) \hat {\mathbf {z}} \text { t e s l a}$$

write the complete expressions for the incident, reflected, and transmitted field vectors.

Solution:

When an electromagnetic wave travelling in one dielectric medium impinges on another dielectric medium with a different intrinsic impedance, part of the incident wave is reflected and part is transmitted.

The incident wave fields:

$$\vec {\mathbf {E}} _ {i} = E _ {0} e ^ {- \gamma_ {1} z} \cdot \hat {\mathbf {x}}$$ $$\vec {\mathbf {H}} _ {i} = \frac {E _ {0}}{\eta_ {1}} e ^ {- \gamma_ {1} z} \cdot \hat {\mathbf {y}}$$

The reflected wave fields:

$$\vec {\mathbf {E}} _ {t} = \Gamma E _ {0} e ^ {- \gamma_ {1} z} \cdot \hat {\mathbf {x}}$$ $$\vec {\mathbf {H}} _ {t} = - \Gamma \frac {E _ {0}}{\eta_ {1}} e ^ {\gamma_ {1} z} \cdot \hat {\mathbf {y}}$$

The transmitted wave fields:

$$\vec {\mathbf {E}} _ {t} = \tau E _ {0} e ^ {- \gamma_ {2} z} \cdot \hat {\mathbf {x}}$$ $$\vec {\mathbf {H}} _ {t} = \tau \frac {E _ {0}}{\eta_ {2}} e ^ {- \gamma_ {2} z} \cdot \hat {\mathbf {y}}$$

where $\Gamma$ is the reflection coefficient, $\tau$ - transmission coefficient, $\eta = \sqrt{\mu / \varepsilon}$ .

(i) The incident wave fields:

$\vec{\mathbf{E}}_i$ has the magnitude

$$E _ {0} = H _ {0} \sqrt {\frac {\mu_ {0}}{\varepsilon_ {0}}} = \frac {B _ {0}}{\sqrt {\mu_ {0} \varepsilon_ {0}}} = \frac {1 0 ^ {- 6}}{\sqrt {\mu_ {0} \varepsilon_ {0}}} = 1 0 ^ {- 6} \times 3 \times 1 0 ^ {8} = 3 \times 1 0 ^ {2}$$ $$\vec {\mathbf {E}} _ {i} = E _ {0} e ^ {- \gamma_ {1} z} \cdot \hat {\mathbf {x}} = 3 \times 1 0 ^ {2} \cos (\omega t - \beta y) (- \hat {\mathbf {x}})$$ $$\beta = \frac {\omega}{c} = \frac {2 \pi \times 1 0 0 \times 1 0 ^ {3}}{3 \times 1 0 ^ {8}} = \frac {\pi}{1 5 0 0}$$

So,

$$\vec {\mathbf {E}} _ {i} = 3 \times 1 0 ^ {2} \cos \left(2 \pi \times 1 0 ^ {5} t - \frac {\pi}{1 5 0 0} y\right) (- \hat {\mathbf {x}}) \mathrm {V / m}$$ $$\begin{array}{l} \vec {\mathbf {H}} _ {i} = \frac {\vec {\mathbf {B}} _ {i}}{\mu_ {0}} = \frac {1 0 ^ {- 6}}{\mu_ {0}} \cos (\omega t - \beta y) \hat {\mathbf {z}} = \frac {1 0 ^ {- 6}}{4 \pi \times 1 0 ^ {- 7}} \cos (\omega t - \beta y) = \\ = \frac {2 . 5}{\pi} \cos \left(2 \pi \times 1 0 ^ {5} t - \frac {\pi}{1 5 0 0} y\right) \hat {\mathbf {z}} \mathrm {A / m} \\ \end{array}$$

(ii) We assume that the incident electric field is reflected with a reflection coefficient $\Gamma$ and transmitted with a transmitted with a transmission coefficient $\tau$ . That implies that if the electric field intensity of the incident, reflected and transmitted waves at the boundary ( $z = 0$ ) are $E_{i0}$ , $E_{r0}$ and $E_{t0}$ respectively, then $E_{r0} = \Gamma E_{i0}$ and $E_{t0} = \tau E_{i0}$ .

$$\tau = 1 + \Gamma$$

and

$$\Gamma = \frac {\eta_ {2} - \eta_ {1}}{\eta_ {2} + \eta_ {1}} =$$ $$\eta_ {1} = \sqrt {\frac {\mu_ {0}}{\varepsilon_ {0}}} = 1 2 0 \pi \Omega$$ $$\eta_ {2} = \sqrt {\frac {\mu_ {2}}{\varepsilon_ {2}}} = \sqrt {\frac {9 \mu_ {0}}{4 \varepsilon_ {0}}} = \frac {3}{2} \times 1 2 0 \pi = 1 8 0 \pi \Omega$$ $$\Gamma = \frac {\eta_ {2} - \eta_ {1}}{\eta_ {2} + \eta_ {1}} = \frac {1 8 0 - 1 2 0}{1 8 0 + 1 2 0} = 0. 2$$ $$\tau = \frac {2 \eta_ {2}}{\eta_ {2} + \eta_ {1}} = \frac {2 * 1 8 0 \pi}{3 0 0 \pi} = 1. 2$$

Using the general properties above, we conclude that $(\beta_r = -\beta$ because wave propagates in opposite direction)

$$\vec {\mathbf {E}} _ {r} = 0. 6 \times 1 0 ^ {2} \cos \left(2 \pi \times 1 0 ^ {5} t + \frac {\pi}{1 5 0 0} y\right) (\hat {\mathbf {x}}) \mathrm {V / m}$$ $$\vec {\mathbf {H}} _ {r} = \frac {1 . 5}{\pi} \cos \left(2 \pi \times 1 0 ^ {5} t + \frac {\pi}{1 5 0 0} y\right) \hat {\mathbf {z}} \mathrm {A / m}$$

(iii)

$$\beta_ {2} = 2 \pi f \sqrt {\mu_ {2} \varepsilon_ {2}} = 2 \pi f \sqrt {3 6 \mu_ {0} \varepsilon_ {0}} = 2 * \pi * 1 0 0 * 1 0 ^ {3} * 6 * \frac {1}{3 * 1 0 ^ {8}} = \frac {\pi}{2 5 0}$$

So,

$$\vec {\mathbf {E}} _ {t} = 3. 6 \times 1 0 ^ {2} \cos \left(2 \pi \times 1 0 ^ {5} t - \frac {\pi}{2 5 0} y\right) (- \hat {\mathbf {x}}) \mathrm {V / m}$$ $$\vec{\mathbf{H}}_t = \frac{3}{\pi} \cos \left(2\pi \times 10^5 t - \frac{\pi}{250} y\right) \hat{\mathbf{z}} \ \mathrm{A/m}$$

Answer:

Incident wave:

$$\vec{\mathbf{E}}_i = 3 \times 10^2 \cos \left(2\pi \times 10^5 t - \frac{\pi}{1500} y\right) (-\hat{\mathbf{x}}) \ \mathrm{V/m}$$ $$\vec{\mathbf{H}}_i = \frac{2.5}{\pi} \cos \left(2\pi \times 10^5 t - \frac{\pi}{1500} y\right) \hat{\mathbf{z}} \ \mathrm{A/m}$$

Reflected wave:

$$\vec{\mathbf{E}}_r = 0.6 \times 10^2 \cos \left(2\pi \times 10^5 t + \frac{\pi}{1500} y\right) (\hat{\mathbf{x}}) \ \mathrm{V/m}$$ $$\vec{\mathbf{H}}_r = \frac{1.5}{\pi} \cos \left(2\pi \times 10^5 t + \frac{\pi}{1500} y\right) \hat{\mathbf{z}} \ \mathrm{A/m}$$

Transmitted wave:

$$\vec{\mathbf{E}}_t = 3.6 \times 10^2 \cos \left(2\pi \times 10^5 t - \frac{\pi}{250} y\right) (-\hat{\mathbf{x}}) \ \mathrm{V/m}$$ $$\vec{\mathbf{H}}_t = \frac{3}{\pi} \cos \left(2\pi \times 10^5 t - \frac{\pi}{250} y\right) \hat{\mathbf{z}} \ \mathrm{A/m}$$

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