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

Problem 6.1.10: \vec E from \vec H, and an elliptical wave

Problem 6.1.10

$E$ from $H$ , and an elliptical wave

In a lossless medium for which $η = 60 π Ω$ , $μ_{r} = 1$ , and

H = - 0.1 cos (ω t - z) a_{x} + 0.5 sin (ω t - z) a_{y} A/m,

determine the relative permittivity and the angular frequency,
find the corresponding electric field,
determine the polarisation of the wave (type, handedness, axial ratio),
find the time-average power density.

Adapted from: M. N. O. Sadiku, Elements of Electromagnetics, Example 10.3.

Try it yourself first.

Strategy

Both components vary with $ω t - z$ : the wave travels towards $+ z$ with $β = 1 rad/m$ . The impedance gives $ε_{r}$ , $β$ gives $ω$ , and $E = η H \times a_{k}$ , the inverse of $H = \frac{1}{η} a_{k} \times E$ for a transverse field, gives the electric field term by term. The polarisation follows from the tip of $E$ at $z = 0$ .

$η = η_{0} / ε_{r}$ : $ε_{r} = 120 π /60 π = 2$ , $ε_{r} = 4$ . From $β = ω ε_{r} / c$ ,

ω = \frac{β c}{ε _{r}} = \frac{1 \times 3 \times 1 0 ^{8}}{2} = 1.5 \times 1 0^{8} rad/s (f = 23.9 MHz, λ = 2 π / β = 6.28 m) .

With $a_{x} \times a_{z} = - a_{y}$ and $a_{y} \times a_{z} = a_{x}$ ,

- 0.1 cos (\cdot) a_{x} \times a_{z} 0.5 sin (\cdot) a_{y} \times a_{z} = 0.1 cos (\cdot) a_{y}, = 0.5 sin (\cdot) a_{x},

and multiplying by $η = 60 π$ ( $30 π = 94.25$ , $6 π = 18.85$ ):

Answer

E = 94.25 sin (ω t - z) a_{x} + 18.85 cos (ω t - z) a_{y} V/m

Check with Ampère's law

\nabla \times H = - \frac{\partial H _{y}}{\partial z} a_{x} + \frac{\partial H _{x}}{\partial z} a_{y} = 0.5 cos (\cdot) a_{x} - 0.1 sin (\cdot) a_{y} = ε \frac{\partial E}{\partial t} .

Integrating in time and dividing by $ε = 4 ε_{0}$ , $E_{x}$ has the amplitude $0.5/ (ω ε) = 0.5/ (1.5 \times 1 0^{8} \times 4 \times 8.842 \times 1 0^{- 12}) = 94.25 V/m$ , as found.

At $z = 0$ , $E (0, t) = 94.25 sin ω t a_{x} + 18.85 cos ω t a_{y}$ :

$ω t$	$E_{x}$ (V/m)	$E_{y}$ (V/m)
$0$	$0$	$18.85$
$π /2$	$94.25$	$0$
$π$	$0$	$- 18.85$
$3 π /2$	$- 94.25$	$0$

The tip moves from $+ y$ to $+ x$ , a rotation about $a_{y} \times a_{x} = - a_{z} = - a_{k}$ . The components are unequal, in time quadrature and aligned with the axes, so the ellipse has its axes along $x$ and $y$ :

Answer

left-handed elliptical (IEEE), AR = \frac{94.25}{18.85} = 5 (14.0 dB), major axis along x

With the handedness test: $E = (- j 94.25 a_{x} + 18.85 a_{y}) e^{- j z}$ , so $E_{r} = 18.85 a_{y}$ , $E_{i} = - 94.25 a_{x}$ , and $(E_{r} \times (- E_{i})) \cdot a_{z} = 18.85 \times 94.25 (a_{y} \times a_{x}) \cdot a_{z} < 0$ : LH.

The two linear parts are orthogonal and carry power independently:

P_{a v} = \frac{∣ E _{0} ∣ ^{2}}{2 η} a_{z} = \frac{94.2 5 ^{2} + 18.8 5 ^{2}}{2 \times 60 π} a_{z} .

Answer

P_{a v} = 24.5 a_{z} W/ m^{2}

Check

From $H$ : $\frac{η}{2} ∣ H_{0} ∣^{2} = 30 π (0. 1^{2} + 0. 5^{2}) = 24.5 W/ m^{2}$ .

Problem 6.1.10. The traces of \vec E (orange) and \vec H (blue, dashed) at z=0, both scaled to the same size, with the wave coming out of the page. Both tips turn clockwise, which for an observer facing the wave is left-handed. The ellipse of \vec H is that of \vec E turned by 90^\circ, since \vec H=\frac1\eta\vec a_z\times\vec E at every instant. — **Figure 6.1.33.** Problem 6.1.10. The traces of $E$ (orange) and $H$ (blue, dashed) at $z = 0$ , both scaled to the same size, with the wave coming out of the page. Both tips turn clockwise, which for an observer facing the wave is left-handed. The ellipse of $H$ is that of $E$ turned by $9 0^{\circ}$ , since $H = \frac{1}{η} a_{z} \times E$ at every instant.

Remark

$H$ turns with $E$ : from $- x$ ( $ω t = 0$ ) to $+ y$ ( $π /2$ ), also about $- a_{z}$ , also left-handed, with its major axis along $y$ . At every instant $E ⊥ H$ and $E \times H = (∣ E ∣^{2} / η) a_{z}$ . As $∣ E ∣$ swings between $18.85$ and $94.25 V/m$ , the instantaneous power density pulses at $2 ω$ between $1.88$ and $47.1 W/ m^{2}$ around its mean $24.5 W/ m^{2}$ .