Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 5.6.10

Two snapshots, then $ε$ and $μ$

An electromagnetic wave varies only with $y$ and $t$ , in a source-free lossless region and it is propagating along the $a_{y}$ direction. At $t = 0$ , the electric and magnetic field components of this wave are observed as ( $f = 0.2 MHz$ )

E H = sin (\frac{4 π}{3} 1 0^{- 3} y) a_{x}, = - \frac{1}{120 π} sin (\frac{4 π}{3} 1 0^{- 3} y) a_{z} .

Find the explicit expressions for the fields in terms of $y$ and $t$ .
Determine $ε$ and $μ$ of the medium the electromagnetic wave is propagating in.
Find the wavelength.

Try it yourself first.

Strategy

A wave towards $+ y$ is a function of $y - v_{p} t$ only, so the snapshot becomes the full field when $β y$ is replaced by $β y - ω t$ . Then each of the two curl equations links the fields through one material constant: Faraday's law gives $μ$ , Ampère's law gives $ε$ .

Bookkeeping

$ω = 2 π \times 2 \times 1 0^{5} = 4 π \times 1 0^{5} rad/s$ , and from the snapshot $β = \frac{4 π}{3} \times 1 0^{- 3} rad/m$ . Then $v_{p} = ω / β = 4 π \times 1 0^{5} / (\frac{4 π}{3} \times 1 0^{- 3}) = 3 \times 1 0^{8} m/s$ .

Replace $β y$ by $β y - ω t$ in the snapshot, and use $sin θ = cos (θ - π /2)$ and $cos (- u) = cos u$ :

E (y; t) H (y; t) = sin (\frac{4 π}{3} 1 0^{- 3} y - 4 π 1 0^{5} t) a_{x} = cos (4 π 1 0^{5} t - \frac{4 π}{3} 1 0^{- 3} y + \frac{π}{2}) a_{x}, = - \frac{1}{120 π} sin (\frac{4 π}{3} 1 0^{- 3} y - 4 π 1 0^{5} t) a_{z} = \frac{1}{120 π} cos (4 π 1 0^{5} t - \frac{4 π}{3} 1 0^{- 3} y - \frac{π}{2}) a_{z} .

With $e^{j π /2} = j$ and $e^{- j π /2} = - j$ the phasors are

Answer

E (y) H (y) = j e^{- j \frac{4 π}{3} 1 0^{- 3} y} a_{x} V/m, = - \frac{j}{120 π} e^{- j \frac{4 π}{3} 1 0^{- 3} y} a_{z} A/m

b) In the time domain

Write $θ = ω t - β y$ . From (a), $E_{x} = - sin θ$ and $H_{z} = \frac{1}{120 π} sin θ$ . Faraday's law, $\nabla \times E + μ \partial H / \partial t = 0$ :

\nabla \times E μ \frac{\partial H}{\partial t} = a_{x} 0 E_{x} a_{y} \partial / \partial y 0 a_{z} 00 = - \frac{\partial E _{x}}{\partial y} a_{z} = - β cos θ a_{z}, = μ \frac{ω}{120 π} cos θ a_{z},

⟹ [- β + \frac{μ ω}{120 π}] cos θ a_{z} = 0 μ = \frac{120 π β}{ω} = \frac{120 π}{3 \times 1 0 ^{8}} = 4 π \times 1 0^{- 7} = μ_{0} .

Ampère's law, $\nabla \times H - ε \partial E / \partial t = 0$ :

\nabla \times H ε \frac{\partial E}{\partial t} = a_{x} 00 a_{y} \partial / \partial y 0 a_{z} 0 H_{z} = \frac{\partial H _{z}}{\partial y} a_{x} = - \frac{β}{120 π} cos θ a_{x}, = - ε ω cos θ a_{x},

⟹ [- \frac{β}{120 π} + ε ω] cos θ a_{x} = 0 ε = \frac{β}{120 π ω} = \frac{1}{120 π \times 3 \times 1 0 ^{8}} = \frac{1 0 ^{- 9}}{36 π} = ε_{0} .

b) In phasor form

The same result with less work. Every $\partial / \partial y$ is a factor $- j β$ and every $\partial / \partial t$ a factor $j ω$ . With $E = j e^{- j β y} a_{x}$ and $H = - \frac{j}{120 π} e^{- j β y} a_{z}$ , Faraday's law $\nabla \times E = - j ω μ H$ gives

⟹ - \frac{\partial E _{x}}{\partial y} a_{z} = - β e^{- j β y} a_{z} = - \frac{ω μ}{120 π} e^{- j β y} a_{z} \frac{4 π \times 1 0 ^{5} μ}{120 π} = \frac{4 π}{3} \times 1 0^{- 3}, μ = μ_{0},

and Ampère's law $\nabla \times H = j ω ε E$ gives

⟹ \frac{\partial H _{z}}{\partial y} a_{x} = - \frac{β}{120 π} e^{- j β y} a_{x} = j ω ε (j) e^{- j β y} a_{x} 4 π \times 1 0^{5} ε = \frac{\frac{4 π}{3} \times 1 0 ^{- 3}}{120 π}, ε = ε_{0} .

The equations reduce to algebra on the common factor $e^{- j β y}$ , with no sines and cosines to shift.

\frac{4 π}{3} \times 1 0^{- 3} λ = 2 π ⟹ λ = 1.5 km .

Answer

ε = ε_{0}, μ = μ_{0}, λ = 1.5 km

Check

$λ = c / f = 3 \times 1 0^{8} /2 \times 1 0^{5} = 1.5 km$ . At $t = 0$ both fields of (a) reduce to the given snapshots, and $E \times H = \frac{1}{120 π} sin^{2} θ (- a_{x} \times a_{z}) = \frac{1}{120 π} sin^{2} θ a_{y}$ : the power flows to $+ y$ , as stated.

Remark

One Maxwell equation alone gives only $μ$ , or only $ε$ . Together they say $ω / β = 1/ μ ε = 3 \times 1 0^{8} m/s$ and $E / H = μ / ε = 120 π Ω$ : the medium is free space. A shortcut: $η = μ / ε = E / H$ and $v_{p} = 1/ μ ε$ give at once $μ = η / v_{p}$ and $ε = 1/ (η v_{p})$ . In a quarter period, $T /4 = 1.25 μ s$ , every crest moves $λ /4 = 375 m$ towards $+ y$ , which is $3 \times 1 0^{8} m/s$ (Figure 5.6.22), and the average power density is $P_{a v} = \frac{1}{240 π} a_{y} = 1.33 mW/ m^{2}$ .

(a) The fields along y. — (a) The fields along $y$ .

(b) The triad where E_x>0. — (a) The fields along $y$ .

Problem 5.6.10: Two snapshots, then \varepsilon and \mu