University of the Philippines Diliman

PHYSICS 72

INSTRUCTIONS: Choose the best answer and shade the corresponding circle on your answer sheet. To change your answer, cross-out and sign your original answer and then shade your new answer. No computational devices allowed (e.g. calculators, mobile phones). Any form of cheating in examinations or any act of dishonesty in relation to studies, such as plagiarism, shall be subject to disciplinary action. Following instructions is part of the exam. Useful formulas: Area Volume Sphere (radius=r) 4πr2 4 3 πr3 Cylinder (radius=r, height=h) 2πr2 + 2πrh πr2h sin(π/3) = cos(π/6) = √ 3/2 cos(π/3) = sin(π/6) = 1/2 sin(π/4) = cos(π/4) = √ 2/2 Useful constants: e −1.60 × 10−19 C me 9.1 × 10−31 kg 0 8.854 × 10−12 C 2 /Nm2 k 8.988 × 109 Nm2 /C2 µ0 4π × 10−7Tm/A 1. Wave ka lang! A laser emits a sinusoidal EM wave that travels in vacuum in the +xdirection. The wavelength and angular frequency are λ0 and ω, respectively, and the E field is parallel to the −y-axis, with maximum magnitude of E0. Which of the following is the correct WAVE EQUATIONS for E and B as functions of position and time? A. E = −kˆE0 sin(2π λ0 x + ωt), B = −ˆj E0 c sin(2π λ0 x − ωt) B. E = +ˆiE0 sin(2π λ0 x + ωt), B = −ˆj E0 c sin(2π λ0 x − ωt) C. E = −ˆiE0 sin(2π λ0 x + ωt), B = −ˆj E0 c sin(2π λ0 x − ωt) D. E = +ˆjE0 sin(2π λ0 x − ωt), B = +kˆ E0 c sin(2π λ0 x − ωt) E. E = −ˆjE0 sin(2π λ0 x − ωt), B = −kˆ E0 c sin(2π λ0 x − ωt) 2. EFamp. The electric field amplitude of an electromagnetic wave that travels in vacuum is 13.5 N/C. What it the magnetic field amplitude of the electromagnetic wave? A. 13.5c T B. 13.5c 2 T C. 13.5/c T D. 13.5/c2 T E. 13.5 T For the next two items, consider the material with permittivity and permeability constants of 80 and 2µ0 respectively. The electromagnetic wave that travels through the said material has frequency of 9.00 × 108 Hz. 3. Spanning. What is the wavelength? A. 1/3 m B. 1/9 m C. 1/4 m D. 1/12 m E. 1/6 m A – 2