Amplitude of Electric Field - Optics and Modern Physics - Solved Past Paper, Exams of Physics

Download Amplitude of Electric Field - Optics and Modern Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. A 550 nm harmonic EM-wave whose electric field is in the z-direction is traveling in the y-direction in vacuum. The amplitude of the electric field is 600 V/m. a) Determine both ω and k for this wave. ν = v λ = 3 × 108 550 × 10−9 = 5.45 × 1014 Hz ω = 2πν = 2π(5.45 × 1014) = 3.43 × 1015 rad/s k = 2π λ = 2π 550 × 10−9 = 1.14 × 107 m−1 b) What is the amplitude of the magnetic field in this wave? B0 = E0 c = 600 3 × 108 = 2 µT c) Write an expression for E(t) if it is zero at x = 0 and t = 0. I think this was supposed to read, “Write an expression for E(y,t) ...” but that’s not what actually got onto the exam. So, if we take E(t) at y = 0, we would get E(t) = 600 sin(−3.43 × 1015t) Note that it must actually be sin here since the conditions stipulate that E(0) = 0. d) What is the average power transferred by this wave through a flat surface perpendicular to the direction of propagation that is 1 m2? I = cǫ0E 2 0 2 = ( 3 × 108 ) ( 8.854 × 10−12 ) (600)2 2 = 478 W/m 2 Since the question asks for the power through one square meter, we get P = IA = 478 · 1 = 478 W