midterm 04-solutions - Version 023 midterm 04 chiu(56565...

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Version 023 – midterm 04 – chiu – (56565) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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. 3.894e-10 2. 1.2744e-09 3. 3.3984e-09 4. 9.558e-10 5. 4.248e-10 6. 1.062e-09 7. 1.8585e-09 8. 3.186e-10 9. 3.54e-10 10. 1.6992e-09 Correct answer: 3 . 894 × 10 10 A. Explanation: Let : E 0 = 0 . 04 N / C , ω = 2200 s 1 , A = 0 . 5 m 2 , and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . The displacement current is I d = ǫ 0 d φ e dt = ǫ 0 d dt ( E A ) = ǫ 0 E 0 A d dt (sin ω t ) = ǫ 0 E 0 A ω cos ω t , so the maximum displacement current is I d,max = ǫ 0 E 0 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 10.0points 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. 1.30925e-07 2. 3.77126e-06 3. 2.68264e-07 4. 3.11199e-06 5. 1.59106e-07 6. 8.10576e-07 7. 9.28679e-07 8. 1.4029e-05 9. 6.87166e-07 10. 9.01525e-07 Correct answer: 3 . 77126 × 10 6 N / m 2 . Explanation: Let : r = 2 . 37 cm = 0 . 0237 m , = 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 10.0points
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Version 023 – midterm 04 – chiu – (56565) 2 An electron oscillating along the y -axis 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 , 0 ) ? (Note: ˆ n is the unit vector along the direc- tion of vectora an instant after t = 0.) 1. 1 4 πǫ 0 2 A c 2 x sin( ωt n 2. 3 8 πǫ 0 2 A c 2 x sin( ωt n 3. 1 4 πǫ 0 2 A c 2 x sin( ωt n 4. 3 16 πǫ 0 2 A c 2 x sin( ωt n 5. 3 16 πǫ 0 2 A c 2 x sin( ωt n correct 6. 3 16 πǫ 0 2 A c 2 x sin( ωt n 7. 3 8 πǫ 0 2 A c 2 x sin( ωt n 8. 3 16 πǫ 0 2 A c 2 x sin( ωt n Explanation: First we note that the point vectorp makes a 30 angle with the xz plane. Calling the angle between vectorp and the xz plane θ , we can see this from tan θ = parenleftbigg x 3 parenrightbigg x = 1 3 and the fact that tan(30 ) = 1 / 3. Next we need to determine vectora . By geomet- ric considerations, we find that vectora = | vectora | cos θ ˆ a = 3 2 | vectora | ˆ a The expression for | vectora | we obtain by noting that the position of the electron is given by the formula y ( t ) = A sin( ωt ) Differentiating this expression twice with re-
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