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MIT 8 02T - Study Guide

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2014 Practice Problem Set 11 Text: Dourmashkin, Belcher, and Liao; Introduction to E & M MIT 8.02 Course Notes Revised. Week Fourteen Interference Problem Set 10 Due Tuesday May 6 at 9 pm and Diffraction W14D1 M/T May 5/6 Maxwell’s Equations and One Dimensional Wave Equation Reading Course Notes: Sections 13.4, 13.6-13.7 W14D2 W/R May 7/8 Energy Flow and the Poynting Vector Polarization Expt 5 MW; Interference Reading Course Notes: Sections 13.8, 13.10, 14.1-14.3 W14D3 F May 9 PS10 Maxwell’s Equations, Energy Flow and the Poynting Vector Reading Course Notes: Section 13.5, 13.6-7, 13.11, 13.12 Week Fifteen Poynting Vector and Energy Flow; Final Review W15D1 M/T May 12/13 Diffraction; Expt. 6: Interference and Diffraction Reading Course Notes: Sections 14.4-14.11 W15D2 W/R May 14/15 Final Review Monday May 19 9 am-12 noon in Johnson Athletic Center Second Floor2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Problem 1: Electromagnetic Waves and the Poynting Vector We have been studying one particular class of electric and magnetic fields solutions called plane sinusoidal traveling waves. One special example is an electromagnetic wave traveling in the positive x-direction with the speed of light c is described by the functions E(x, y, z,t) = Ey(x,t)ˆj = Ey,0sin2πλ(x − ct)⎛⎝⎜⎞⎠⎟ˆj B(x, y, z,t) = Bz(x,t)ˆk =1cEy,0sin2πλ(x − ct)⎛⎝⎜⎞⎠⎟ˆk Suppose the plane wave front (infinite yz-plane) traveling at speed c in the positive x-direction passes through a rectangular volume of space that has area A perpendicular to the direction of propagation and length cΔt, corresponding to the distance the electromagnetic wave travels in time Δt. Inside this volume of space the electric and magnetic fields store energy and that energy changes in time as the wave passes through the volume. The energy per time per area transported by this electromagnetic wave is called the Poynting vector and is defined as S =1µ0E ×B. The power transmitted by the Poynting vector through a surface is the flux given by the expression P(t) =S(t) ⋅ˆn dasurface∫∫. The flux of the Poynting vector on a surface describes the energy that ‘flows’ across the surface. a) What the Poytning vector associated with this wave? b) What is the time-averaged Poynting vector field on the fixed plane x = 0 over one period? The definition of a time average of a periodic function f (t) over one period is f (t) =1Tf (t) dt0T∫ Recall that c = 1 / µ0ε0.3 c) What is the time-average (over one period) of the energy density stored in the electric and magnetic fields at the point (x, y, z,t). Recall that uelec= (1/ 2)ε0E2 and umag= B2/ 2µ0. How is that related to the magnitude of the time-averaged Poynting vector? d) What is the time-averaged energy stored in the electric and magnetic fields in a rectangular volume of cross-sectional area A and length cΔt, where the length cΔt is aligned along the x-axis and cΔt << λ. The last assumption allows us to assume that S(x,t) is nearly uniform across the box. e) What is the time-averaged rate of change of the total energy stored in the electric and magnetic fields in the rectangular volume of cross-sectional area A and length cΔt? f) What is the time-averaged flow of power through the surface of area A located on the surface given by x = xp, perpendicular to the direction of propagation? How does the power that through the rectangular surface (this flux is negative since it flows into the volume) compare to the time derivative of the energy stored in the fields inside the volume?4 Problem 2: Poynting Vector and Radiation Pressure: A plane electromagnetic wave transports energy in the direction of propagation of the wave. The power per square area is given by the Poynting vector S =E ×Bµ0. The power that flows into the rectangular volume cross-sectional area A and length cΔt appears as rate of change of the energy stored in the fields inside the volume. Ppower=S A =ddtUtotal. The electromagnetic wave also transports momentum, and hence can exert a radiation pressure on a surface due to the absorption and reflection of the momentum. The momentum carried by an electromagnetic wave is related to the energy of the wave according to U = cp If the plane electromagnetic wave is completely absorbed by a surface of cross-sectional area A then the momentum pΔ delivered to the surface in a time tΔ is given by Δp =ΔUc. The force that the wave exerts on the surface is then the rate of change of the momentum in time F = limΔt→ 0ΔpΔt= limΔt→ 01cΔUΔt=1cdUdt. Since the rate of change of energy is related to the power flowing across the surface, the force is F =1cdUdt=1cPpower=1cS A. The radiation pressure Ppressure is then defined to be the force per area that the wave exerts on the surface Ppressureabs≡FA=1cS, perfectly absorbing. When the surface completely reflects the wave, then the change in momentum is twice the absorbing case since the wave completely reverses direction,5 Δp = 2ΔUc. Therefore the radiation pressure of a wave on a perfectly reflecting surface is Ppressureref≡FA= 21cS, perfectly reflecting. Suppose a space station is placed in orbit the same distance away from the sun as the earth but on the opposite side of the sun from the earth in a circumpolar orbit (the orbit plane is perpendicular to the plane of the earth’s orbit around the sun). The average earth-sun distance is re,s= 1.50 × 1011m. An astronaut of mass m = 100 kg on a space walk loses contact with the space station. The spacesuit has a cross sectional area of approximately A = 2 m2. The mass of the sun is ms= 1.99 × 1030kg. The gravitational constant is G = 6.67 × 10−11N ⋅ m2⋅ kg−2. a) What is the radiation pressure of the sun on the astronaut’ spacesuit. Assume it is perfectly reflecting? b) What is the force on the astronaut? c) How does this force compare to the force of gravity? Problem 3: Intensity of the Sun At the upper surface of the earth’s atmosphere, the time-averaged magnitude of the Poynting vector, referred to


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