Chap29 - MAXWELLS EQUATIONS AND ELECTROMAGNETIC WAVES 29 EXERCISES Section 29.2 Ambiguity in Ampres Law 13 INTERPRET In this problem we are asked

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MAXWELL’S EQUATIONS AND ELECTROMAGNETIC WAVES 29 EXERCISES Section 29.2 Ambiguity in Ampère’s Law 13. INTERPRET In this problem we are asked to find the displacement current through a surface. DEVELOP As shown in Equation 29.1, Maxwell’s displacement current is 00 0 () E d d dEA dE IA dt dt dt εε ε Φ == = EVALUATE The above equation gives 2 12 2 2 0 (8.85 10 C /N m )(1 cm )(1.5 V/m s) 1.33 nA d dE dt εμ × = ASSESS Displacement current arises from changing electric flux and has units of amperes (A), just like ordinary current. 14. The electric field is approximately uniform in the capacitor, so 0 ( / ) , and / ED EA V d A I t E φ εφ = = 12 2 0 ( / ) / (8.85 10 F/m)(10 cm) (220 V/ms)/(0.5 cm) 3.89 A. Ad dVdt μ = Section 29.4 Electromagnetic Waves 15. INTERPRET We are given the electric and magnetic fields of an electromagnetic wave and asked to find the direction of propagation. DEVELOP The direction of propagation of the electromagnetic wave is the same as the direction of the cross product GG . EB × EVALUATE When E G is parallel to and ˆ j B G is parallel to the direction of propagation is parallel to ˆ , i , × or ˆ ji k × . =− ASSESS For electromagnetic waves in vacuum, the directions of the electric and magnetic fields, and of wave propagation, form a right-handed coordinate system. 16. (a) The peak amplitude is the magnitude of which is ( Ei j + ) , 2. E Note that ˆ 2, ij n += where is a unit vector 45 ° between the positive x and y axes. (b) When ˆ n E G is parallel to (for ˆ n sin( ) kz t ω positive) points 45 ° into the second quadrant (so that Thus, B G , and is in the direction). z ⊥× + B G is parallel to the unit vector / 2 . −+ 0 Section 29.5 Properties of Electromagnetic Waves 17. INTERPRET This problem is about measuring the distance between the Sun and the Earth using light-minutes. DEVELOP A light-minute (abbreviated as c-min) is approximately equal to 81 1 c-min (3 10 m/s)(60 s) 1.8 10 m = × On the other hand, the mean distance of the Earth from the Sun (an Astronomical Unit) is about 11 1.5 10 m. SE R EVALUATE In units of c-min, can be rewritten as SE R 11 10 1c-min (1.5 10 m) 8.33 c-min SE R = × 29.1
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29.2 Chapter 29 ASSESS The result implies that it takes about 8.33 minutes for the sunlight to reach the Earth. 18. Assuming the satellite is approximately overhead, we can estimate the round-trip travel time by / tr c Δ=Δ = 5 (2 36,000 km)/(3 10 km/s) 0.24 s. ×× = 19. INTERPRET In this problem we want to deduce the airplane’s altitude by measuring the travel time of a radio wave signal it sends out. DEVELOP The speed of light is and the total distance traveled is 8 31 0 m / s c 2. rh Δ = EVALUATE Since (for waves traveling with speed c ), the altitude is 2 c Δ= =Δ t 8 (3 10 m/s)(50 s) 7.5 km 22 ct h μ Δ× == = ASSESS The airplane is flying lower than the typical cruising altitude of 12,000 m (35,000 ft) for commercial jet airplanes.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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Chap29 - MAXWELLS EQUATIONS AND ELECTROMAGNETIC WAVES 29 EXERCISES Section 29.2 Ambiguity in Ampres Law 13 INTERPRET In this problem we are asked

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