Uniform Plane Wave

1. A wave with l = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media. In the liquid, the wavelength is measured as 1.0 cm. What is the wave’s frequency (a) in air? (b) in the liquid? (c) What is the liquid’s relative permittivity?

(a)

(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz

(c)

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2. Given s = 1.0x10-5 S/m , er = 2.0, mr = 50., and f = 10. MHz, find g, a, b, and h.

Inserting these into the expressions for g and h,

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3. Suppose in free space, H(x,t) = 100.cos(2px107t – bx + p/4) az mA/m. Find E(x,t).

Since free space is stated,

and then

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4. A 100 MHz wave in free space propagates in the y direction with an amplitude of 1 V/m. If the electric field vector for this wave has only an az component, find the instantaneous expression for the electric and magnetic fields.

From the given information we have and

or .

Now to find H.

So

or

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5. In a lossless, nonmagnetic material with er = 16, H = 100 cos(wt – 10y) az mA/m. Determine the propagation velocity, the angular frequency, and the instantaneous expression for the electric field intensity.

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6. In a media with properties s = 0.00964 S/m , er = 1.0, mr = 100., and f = 100. MHz, a 1.0 mA/m amplitude magnetic field travels in the +x direction with its field vector in the z direction. Find the instantaneous form of the related electric field intensity.

Finally,

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7. In seawater, a propagating electric field is given by E(z,t) = 20.e-az cos(2px106t – bz + 0.5) ay V/m. Assuming e’’=0, find (a) a and b, and (b) the instantaneous form of H.

For seawater we have er = 72, s = 5, and mr = 1.

So:

or with appropriate significant digits:

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8. For Nickel (s = 1.45 x 107, mr = 600), make a table of a, b, h, up, and d for 1Hz, 1kHz, 1MHz, and 1 GHz.

For Ni we have s = 1.45x107S/m, mr = 600

d = 1/a

Table

f(Hz)= / 1 / 103 / 106 / 109
a(Np/m) / 185 / 5860 / 185x103 / 5.9x106
b(rad/m) / 185 / 5860 / 185x103 / 5.9x106
h / 18ej45ºmW / 570ej45º mW / 18ej45º mW / 0.57ej45º W
d / 5.4mm / 170mm / 5.3mm / 170nm
up(m/s) / 12x106 / 12x106 / 12x106 / 12x106

9. A semi-infinite slab exists for z > 0 with s = 300 S/m, er = 10.2, and mr = 1.0. At the surface (z = 0),

E(0,t) = 1.0 cos(p x 106t) ax V/m.

Find the instantaneous expressions for E and H anywhere in the slab.

The general expression for E is:

Here, s > we (i.e. it is a good conductor), so

So now we have

To find B we’ll work in phasors.

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10. A 600 MHz uniform plane wave incident in the z direction on a thick slab of Teflon (er = 2.1, mr = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field intensity at the surface. Assuming s = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c) Pav.

Teflon: s = 0 so a = 0,

and

(a)

(b)

(c)

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11. A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/m amplitude at the surface. How much power passes through a square meter at the surface? How much power passes through a square meter area 10. mm beneath the surface?

Cu:

Now at 10 mm beneath the surface, we have

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12. Given E(z,t) = 10.cos(wt-bz)ax - 20.cos(wt-bz-45°)ay V/m, find the polarization and handedness.

The field can be rewritten as E(z,t) = 10.cos(wt-bz)ax + 20.cos(wt-bz-45°-180°)ay

or E(z,t) = 10.cos(wt-bz)ax + 20.cos(wt-bz+135°)ay

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13. Given

we say that Ey leads Ex for 0° < f < 180°, and that Ey lags Ex when –180° < f < 0°. Determine the handedness for each of these two cases.

For 0 < f < 180°, we have LHP

For 180° < f < 360°, we have RHP

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14. Suppose a UPW in air carrying an average power density of 100 mW/m2 is normally incident on a nonmagnetic material with er = 11. What is the time-averaged power density of the reflected and transmitted waves?

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15. A UPW in a lossless nonmagnetic er = 16 media (for z < 0) is given by

E(z,t) = 10.cos(wt-b1z)ax + 20.cos(wt-b1z+p/3)ay V/m.

This is incident on a lossless media characterized by mr = 12, er = 6.0 (for z > 0). Find the instantaneous expressions for the reflected and transmitted electric field intensities.

or , so

.

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16. The wave Ei = 10.cos(2p x 108t - b1z) ax V/m is incident from air onto a copper conductor. Find Er, Et and the time-averaged power density transmitted at the surface.

For copper we have

so

We find

So Er = -10.cos(2p x 108t + b1z) ax V/m

and

17. A wave specified by Ei = 100.cos(px107t-b1z)ax V/m is incident from air (at z < 0) to a nonmagnetic media (z > 0, s = 0.050 S/m, er = 9.0). Find Er, Et and SWR. Also find the average power densities for the incident, reflected and transmitted waves.

In this problem we find in medium 2 (z > 0) that we = 0.0025 and s = 0.05. These values are too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate:

.

Then,

,

so

,

so

(check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2)

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18. A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident from air (z < 0) onto a slab of lossless, nonmagnetic material with er = 25 (z > 0). The angle of incidence is 40°. Calculate (a) the angle of transmission, (b) the reflection and transmission coefficients, and (c) the incident, reflected and transmitted fields.

(a) The material parameters in this problem are the same as for P5.48. So, once again we have qt = 7.4°. Also, b1 = 2.09 rad/m and b2 = 10.45 rad/m.

(b)

(c)

Incident:

Reflected:

transmitted:

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