Module 1 Assignment
Problem 1:
An electric field propagating in a lossless non-magnetic media is characterized by
πΈ(π¦, π‘) = 100.0 πππ (4π × 106 π‘ − 0.1257π¦) ππ§
π
π
Find:
A. The wave amplitude, frequency, propagation velocity, wavelength, and relative
permittivity of the media.
B. Write the expression for H(y, t), including the direction (unit vector) and units.
Solution:
π
A. Using the general form πΈ(π¦, π‘) = πΈ0 cos(ππ‘ − π½π¦) ππ§ π
π
i.
Wave amplitude = πΈ0 = 100 π
ii.
Frequency = π = 2π =
π
4π×106
2π
2π
2π
= 2 ππ»π§
iii.
Wavelength = π = π½ = 0.1257 = 49.986π
iv.
Propagation velocity = π£π = π β π = 49.986π β 2 β 106 π = 99.97 β 106 π
v.
Relative permittivity = π£π =
1
πΈΜ
Μ
(π¦, π‘) = =
B. π»
π
100.0
π
π2
π
(3β108 )
√ π π
π
πππ (4π × 106 π‘ − 0.1257π¦) ππ₯ π =
√π0
100.0 β √3
πππ (4π × 106 π‘ − 0.1257π¦) ππ₯
π΄
=
π
π΄
=
π
√377
π΄
π»(π¦, π‘) = 0.796 πππ (4π × 106 π‘ − 0.1257π¦) ππ₯
π
πππ (4π × 106 π‘ − 0.1257π¦) ππ₯
Problem 2:
A magnetic field propagating in free space is given by
π»(π§, π‘) = 20. π ππ(π × 108 π‘ + π½π§) ππ₯
Find f, λ.
Solution:
2
, ππ = 1, ππ = π£2 = (99.97β106 )2 = 9.005
π π
π΄
100.0 β √π
π
π΄
π
π
π×108
π
2π
3β108
i.
Frequency = π = 2π =
ii.
Wavelength = π = π = 50β106 = 6π
= 50 ππ»π§