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Problem 5 (15 points): You may leave your answers in terms of pi, but you must express your
answers in SI units. Recall that 4Tx 10 T m/A and c3.0x 10* m/s. A particular plane polarized
electromagnetic wave, with B a frequency of 100 MHz, is traveling through a vacuum' in the +x
direction, as shown. At t = 0, the electric field due to this wave at x-0 has a magnitude of 300
V/m. a) What is the amplitude of the magnetic field, the angular frequency, and the wavelength?
b) What are the vector functions (there are 2) for the wave described? c) What is the magnitude
of the Poynting vector of this wave? d) If the wave is now incident on a polarizer with a
transmission axis of 45o, what will the new intensity of the wave be?
Solution
(A) E_max = 300 V/m
B_max = E_max / c = (300)/(3x10^8 m/s)
B_max = 1 x 10^-6 T ....Ans (Magnetic field amplitude)
w = 2 pi f = 2x pi x 100 x 10^6
w = 6.28 x 10^8 rad/s .....Ans
lambda = v / f = (3 x 10^8) / (100 x 10^6)
= 3 m .....Ans
(B) E = (300 V/m) cos[ 2pi (x/3 - (100 x 10^6)t)] j
B = (1 x 10^-6 T)cos[ pi (x/3 - (100 x 10^6)t)] k
(C) S = E x B / 2u0
S = 119.4 W/m^2
(D) S = S0 (cos(theta))^2
S = (119.4)(cos45)^2
S = 59.7 W/m^2

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Problem 5 (15 points)- You may leave your answers in terms of pi- but.docx

  • 1. Problem 5 (15 points): You may leave your answers in terms of pi, but you must express your answers in SI units. Recall that 4Tx 10 T m/A and c3.0x 10* m/s. A particular plane polarized electromagnetic wave, with B a frequency of 100 MHz, is traveling through a vacuum' in the +x direction, as shown. At t = 0, the electric field due to this wave at x-0 has a magnitude of 300 V/m. a) What is the amplitude of the magnetic field, the angular frequency, and the wavelength? b) What are the vector functions (there are 2) for the wave described? c) What is the magnitude of the Poynting vector of this wave? d) If the wave is now incident on a polarizer with a transmission axis of 45o, what will the new intensity of the wave be? Solution (A) E_max = 300 V/m B_max = E_max / c = (300)/(3x10^8 m/s) B_max = 1 x 10^-6 T ....Ans (Magnetic field amplitude) w = 2 pi f = 2x pi x 100 x 10^6 w = 6.28 x 10^8 rad/s .....Ans lambda = v / f = (3 x 10^8) / (100 x 10^6) = 3 m .....Ans (B) E = (300 V/m) cos[ 2pi (x/3 - (100 x 10^6)t)] j B = (1 x 10^-6 T)cos[ pi (x/3 - (100 x 10^6)t)] k (C) S = E x B / 2u0 S = 119.4 W/m^2
  • 2. (D) S = S0 (cos(theta))^2 S = (119.4)(cos45)^2 S = 59.7 W/m^2