Question

A uniform plane wave travels through free space, and its instantaneous electric field intensity vector is expressed as

E=15cos(ωt+10πz+θ) yˆ V/m

Where time is in seconds, position z is in meters. The magnetic field intensity of the wave is
H=0.02 A/m at t = 0 and z = 1.15 m. Determine

(a) The operating frequency and initial (for t = 0) phase θ of the electric field in the plane z = 0.
(b) Expression for the instantaneous magnetic field intensity vector
(c) The complex magnetic field intensity vector

Answer 1

Answer on Question#54541 - Physics - Electrical Engineering

A uniform plane wave travels through free space, and its instantaneous electric field intensity vector is expressed as $\pmb{E} = 15\cos (\omega t + 10\pi z + \theta)\hat{\pmb{y}}\frac{\mathrm{V}}{\mathrm{m}}$

Where time is in seconds, position $z$ is in meters. The magnetic field intensity of the wave is $H = 0.02\frac{\mathrm{A}}{\mathrm{m}}$ at $t = 0$ and $z = 1.15\mathrm{m}$. Determine

(a) The operating frequency and initial (for $t = 0$) phase $\theta$ of the electric field in the plane $z = 0$.

(b) Expression for the instantaneous magnetic field intensity vector

(c) The complex magnetic field intensity vector

Solution:

Since the wavenumber (the multiplier of $z$ in the above expression for electric field) is given by

$$k = \frac {\omega}{c},$$

$(c-$ is the speed of light) we obtain

$$\omega = k c = 10 \pi \mathrm{m}^{-1} \cdot 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}} = 9.42 \times 10^{9} \mathrm{s}^{-1}$$

Therefore the operating frequency is

$$f = \frac {\omega}{2 \pi} = \frac {10 \pi \mathrm{m}^{-1} \cdot 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}}{2 \pi} = 1.5 \mathrm{GHz}$$

According to the Maxwell's equations:

$$\nabla \times \boldsymbol{E} = - \mu_{0} \frac {\partial \boldsymbol{H}}{\partial t}$$

Since magnetic field depends on the time in the same manner as electric field, we get the following

$$\nabla \times \boldsymbol{E}|_{t=0,z=1.15\mathrm{m}} = \left. \omega \mu_{0} \boldsymbol{H} \right|_{t=0,z=1.15\mathrm{m}}$$

Then

$$\begin{aligned} &10 \pi (-15 \sin (\omega t + 10 \pi z + \theta)) |_{t=0,z=1.15\mathrm{m}} \hat{\boldsymbol{x}} \frac{\mathrm{V}}{\mathrm{m}} = \mu_{0} \omega H (t = 0, z = 1.15\mathrm{m}) \hat{\boldsymbol{x}} \\ &-150 \pi \sin (11.5 \pi + \theta) \frac{\mathrm{V}}{\mathrm{m}} = 4 \pi \times 10^{-7} \frac{\mathrm{N}}{\mathrm{A}^{2}} \cdot 10 \pi \mathrm{m}^{-1} \cdot 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}} \cdot 0.02 \frac{\mathrm{A}}{\mathrm{m}} \\ &\sin (11.5 \pi + \theta) = - \frac{4 \pi \times 10^{-7} \frac{\mathrm{N}}{\mathrm{A}^{2}} \cdot 10 \pi \mathrm{m}^{-1} \cdot 3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}} \cdot 0.02 \frac{\mathrm{A}}{\mathrm{m}}}{150 \pi \frac{\mathrm{V}}{\mathrm{m}}} = -0.5 \end{aligned}$$

Therefore

$$11.5\pi + \theta = \frac{3\pi}{2} \pm \frac{\pi}{6} + 2\pi n, \quad n \in \mathbb{Z}$$ $$\theta = -10\pi \pm \frac{\pi}{6} + 2\pi n, \quad n \in \mathbb{Z}$$

To be specific let's put $\theta = \frac{\pi}{6}$. Since value of electric field is proportional to the value of the value of the magnetic field we obtain

$$H = H_0 \cos \left(2\pi \cdot 1.5 \ \mathrm{GHz} \cdot t + 10\pi z + \frac{\pi}{6}\right) \hat{x}$$

To find the amplitude of the magnetic field we'll use the fact that $H(t = 0, z = 1.15\,\mathrm{m}) = 0.02\,\frac{\mathrm{A}}{\mathrm{m}}$:

$$H_0 \cos \left(11.5\pi + \frac{\pi}{6}\right) = 0.02 \frac{\mathrm{A}}{\mathrm{m}}$$ $$H_0 = \frac{0.02 \frac{\mathrm{A}}{\mathrm{m}}}{\cos \left(11.5\pi + \frac{\pi}{6}\right)} = 0.025 \frac{\mathrm{A}}{\mathrm{m}}$$

Therefore

$$H = 0.025 \ \cos \left(2\pi \cdot 1.5 \ \mathrm{GHz} \cdot t + 10\pi z + \frac{\pi}{6}\right) \hat{x} \frac{\mathrm{A}}{\mathrm{m}}$$

Since

$$H = \mathrm{Re} \left[ 0.025 \frac{\mathrm{A}}{\mathrm{m}} \hat{x} \, e^{j \left(2\pi \cdot 1.5 \ \mathrm{GHz} \cdot t + 10\pi z + \frac{\pi}{6}\right)} \right],$$

then the complex magnetic field intensity vector is given by

$$0.025 \frac{\mathrm{A}}{\mathrm{m}} e^{j \frac{\pi}{6}} \hat{x}$$

Answer:

(a) $f = \frac{\omega}{2\pi} = \frac{10\pi\,\mathrm{m}^{-1} \cdot 3 \times 10^{\frac{\pi}{\mathrm{m}}}}{\omega^2 \pi} = 1.5\,\mathrm{GHz}$

$$\theta = -10\pi \pm \frac{\pi}{6} + 2\pi n, \quad n \in \mathbb{Z}$$

(b) $H = 0.025 \cos \left(2\pi \cdot 1.5 \ \mathrm{GHz} \cdot t + 10\pi z + \frac{\pi}{6}\right) \hat{x} \frac{\mathrm{A}}{\mathrm{m}}$

(c) $0.025 \frac{\mathrm{A}}{\mathrm{m}} e^{j \frac{\pi}{6}} \hat{x}$

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