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§