--- title: "Normal incidence on perfect conductor" subject: "Electromagnetism" chapter: 4 key: "semester_3/electromagnetism:normal-incidence-conducting-plane-100mhz" source: https://www.bctengineer.com/q/semester_3/electromagnetism/normal-incidence-conducting-plane-100mhz site: BCT Engineering --- # Normal incidence on perfect conductor Electromagnetism · Chapter 4: Wave equation and Wave propagation - Asked in: 2067 Mangsir, 8 marks - Video: [4 2 Reflection by Perfect Conductor Normal Incidence](https://www.youtube.com/watch?v=ptX23iGcbQs) — BVCEC ECE e-Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/normal-incidence-conducting-plane-100mhz - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4 ## Question A z-polarized uniform plane wave with frequency 100 MHz propagates in air in the positive x-direction and impinges normally on a perfectly conducting plane at x = 0. Assuming the amplitude of the electric field vector to be 3 mV/m, determine phasor and instantaneous expressions for (a) incident electric and magnetic field vectors (b) reflected electric and magnetic field vectors. ## Short answer Incident E_i = 3e⁻j2.094x a_z mV/m, H_i = −7.96e⁻j2.094x a_y μA/m. Γ = −1, reflected E_r = −3e⁺j2.094x a_z mV/m, H_r = −7.96e⁺j2.094x a_y μA/m. Instantaneous forms with cos(2π×10⁸t ∓ 2.094x). ## Answer Given: f = 100 MHz in air, wave travels in +x, E along a_z, amplitude E₀ = 3 mV/m, perfect conductor at x = 0. ```text ω = 2π×10⁸ rad/s β = ω/c = 2π×10⁸ / 3×10⁸ = 2.094 rad/m (2π/3) η₀ = 120π = 376.7 Ω E₀/η₀ = 3×10⁻³/376.7 = 7.96 μA/m ``` For a perfect conductor η₂ = 0, so Γ = (η₂ − η₁)/(η₂ + η₁) = −1: all power is reflected and the reflected E is the negative of the incident E at x = 0 (tangential E must vanish). The direction of H follows from H = (1/η) a_k × E. #### (a) Incident wave (travelling in +x) ```text E_i (phasor) = 3×10⁻³ e^(−j2.094x) a_z V/m E_i (instant) = 3×10⁻³ cos(2π×10⁸t − 2.094x) a_z V/m a_x × a_z = −a_y H_i (phasor) = −7.96×10⁻⁶ e^(−j2.094x) a_y A/m H_i (instant) = −7.96×10⁻⁶ cos(2π×10⁸t − 2.094x) a_y A/m ``` #### (b) Reflected wave (travelling in −x), Γ = −1 ```text E_r (phasor) = −3×10⁻³ e^(+j2.094x) a_z V/m E_r (instant) = −3×10⁻³ cos(2π×10⁸t + 2.094x) a_z V/m (−a_x) × (−a_z) = a_x × a_z = −a_y H_r (phasor) = −7.96×10⁻⁶ e^(+j2.094x) a_y A/m H_r (instant) = −7.96×10⁻⁶ cos(2π×10⁸t + 2.094x) a_y A/m ``` Check at x = 0: E_i + E_r = 0 (boundary condition satisfied), and H total = −15.9 μA/m a_y, twice the incident H, which equals the surface current density J_s = n × H. The total field in x < 0 is a standing wave: E = −j6×10⁻³ sin(βx) a_z, H = −15.9×10⁻⁶ cos(βx) a_y (phasors). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Gamma at a perfect conductor and why? A. Gamma = +1 since eta2 = 0 B. Gamma = -1 since eta2 = 0, so tangential E vanishes at the surface C. Gamma = 0 since all power is absorbed D. Gamma = -1 because H is zero at the surface Answer: B. Gamma = (0 - eta1)/(0 + eta1) = -1, so E_r = -E_i at x = 0. (Key point: Reflection coefficient) 2. What is beta for 100 MHz in air? A. w/c = 2 pi x 10^8/(3x10^8) = 2.094 rad/m B. 2 pi x 10^8 rad/m C. 1.05 rad/m D. 6.28 rad/m Answer: A. beta = w/c = 2 pi/3 rad/m. (Key point: Phase constant) 3. For E_i along a_z travelling in +x, what is H_i? A. +(E0/eta0) a_y B. -(E0 eta0) a_y C. -(E0/eta0) a_z D. -(E0/eta0) a_y, magnitude 7.96 uA/m, since a_x x a_z = -a_y Answer: D. H = (1/eta) a_k x E and a_x x a_z = -a_y, with E0/eta0 = 3 mV/m / 376.7 = 7.96 uA/m. (Key point: Incident H) 4. How does total H at x = 0 compare with the incident H? A. It is zero, because E total is zero B. It is half the incident H C. It is twice the incident H, because H_r has the same sign as H_i D. It equals the incident H Answer: C. E_r is negative and its direction of travel is reversed, so H_r = -7.96 uA/m a_y same as H_i. The total is 15.9 uA/m, equal to the surface current density. 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