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Unformatted text preview: Version 023 – midterm 04 – chiu – (56565) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a region of space, the electric field varies according to E = (0 . 04 N / C) sin[(2200 s − 1 ) t ], where t is in seconds. Find the maximum displacement current through a 0 . 5 m 2 area perpendicular to vector E . The permittivity of free space is 8 . 85 × 10 − 12 C 2 / N · m 2 . 1. 3.894e10 2. 1.2744e09 3. 3.3984e09 4. 9.558e10 5. 4.248e10 6. 1.062e09 7. 1.8585e09 8. 3.186e10 9. 3.54e10 10. 1.6992e09 Correct answer: 3 . 894 × 10 − 10 A. Explanation: Let : E = 0 . 04 N / C , ω = 2200 s − 1 , A = 0 . 5 m 2 , and ǫ = 8 . 85 × 10 − 12 C 2 / N · m 2 . The displacement current is I d = ǫ d φ e dt = ǫ d dt ( E A ) = ǫ E A d dt (sin ω t ) = ǫ E A ω cos ω t , so the maximum displacement current is I d,max = ǫ E A ω = (8 . 85 × 10 − 12 C 2 / N · m 2 ) × (0 . 04 N / C)(0 . 5 m 2 )(2200 s − 1 ) = 3 . 894 × 10 − 10 A . 002 10.0 points A thin tungsten filament of length 0 . 591 m radiates 99 . 5 W of power in the form of elec tromagnetic waves. A perfectly absorbing surface in the form of a hollow cylinder of ra dius 2 . 37 cm and length 0 . 591 m is placed concentrically with the filament. Assume: The radiation is emitted in the radial direction, and neglect end effects. The speed of light is 2 . 99792 × 10 8 m / s. Calculate the radiation pressure acting on the cylinder. 1. 1.30925e07 2. 3.77126e06 3. 2.68264e07 4. 3.11199e06 5. 1.59106e07 6. 8.10576e07 7. 9.28679e07 8. 1.4029e05 9. 6.87166e07 10. 9.01525e07 Correct answer: 3 . 77126 × 10 − 6 N / m 2 . Explanation: Let : r = 2 . 37 cm = 0 . 0237 m , ℓ = 0 . 591 m , c = 2 . 99792 × 10 8 m / s , and P = 3 . 77126 × 10 − 6 N / m 2 . The intensity of the radiation reaching the walls of the cylinder is I = ( S ) = P 2 π r ℓ , so the radiation pressure on the walls is p = I c = P 2 π r ℓ c = 3 . 77126 × 10 − 6 N / m 2 2 π (0 . 0237 m) (0 . 591 m) (2 . 99792 × 10 8 m / s) = 3 . 77126 × 10 − 6 N / m 2 . 003 10.0 points Version 023 – midterm 04 – chiu – (56565) 2 An electron oscillating along the yaxis with angular frequency ω and amplitude A passes through the origin at time t = 0 and is moving in the + y direction. Which formula desribes the time dependence of E rad at a point vectorp = ( x, x/ √ 3 , ) ? (Note: ˆ n is the unit vector along the direc tion of vectora ⊥ an instant after t = 0.) 1. − 1 4 πǫ eω 2 A c 2 x sin( ωt )ˆ n 2. √ 3 8 πǫ eω 2 A c 2 x sin( ωt )ˆ n 3. 1 4 πǫ eω 2 A c 2 x sin( ωt )ˆ n 4. − √ 3 16 πǫ eω 2 A c 2 x sin( ωt )ˆ n 5....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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