Solutions7 - Physics 208, Assignment 7, Solutions, 3/16/09,...

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Unformatted text preview: Physics 208, Assignment 7, Solutions, 3/16/09, RT Problem 1. V=10 V f =10 kHz L (e) R = 10 ohm q t (ms) C =0.1 F (d) R = 0 q i i V V E E U C U C U L U L U +U C L U +U C L (a) (b) f= 1 = ; 2 2LC L= 1 2 104 C V0 = L 2 1 = 2.5 mH. 10-7 1 1 CV02 = Li2 ; 2 2 max max UL = imax = 10-7 10 = 0.063 A. 2.5 10-3 (c) 1 1 1 2 Li = CV02 = 10-7 102 = 0.5 10-5 J. 2 max 2 2 1 Problem 2. (a) The only non-vanishing electic field component of an EM wave is given by E(z, t) = Em sin(kz - t) y. ^ The "phase" kz - t, is constant, for example vanishing when z = (/k)t = ct. So the wave is moving in the positive z direction. (b) The wave is plane-polarized in the y-plane. i.e. E y. ^ (c) For E B to be parallel to ^, with B E = 0 requires B = Bm x, z ^ where Bm remains to be determined. In a plane-polarized plane wave c|E| = |B|. Therefore we have B(z, t) = - Em sin(kz - t) x. ^ c (d) Given = 550 nm, A priori the sign here should have been but the minus sign has been selected so that E B ^, z 3 108 c = 5.4 101 4 Hz. = 0.55 10-6 f= k= = 2f = 2 5.4 101 4 = 3.4 1015 s-1 . A useful, and broadly adhered to, convention is to always quote f in Hertz and in s-1 . This helps reduce the likelihood of factor 2 errors. (e) This EM wave is green, visible light. Problem 3. Isunlight = 1.37 103 W/m2 . (a) The energy passing through 1 m2 in 8 hours is UEM = IAt = (1.37 103 ) (1) (8 3600) = 3.9 107 J. For comparison, a 100 W light bulb uses 100(83600) = 2.88106 J of energy in the same time. (b) The force exerted by sunlight on 1 sq.m of absorber is Fabs. = 1370 I = 4.6 10-6 N. = c 3 108 2 2 = 1.14 107 m-1 . = 0.55 10-6 The mass of the absorber might be about 1 kg, so its weight would be about 9.8 N, which is greater than the radiation force by more than a million. 2 (c) By Eq. (33-26) Erms = Ic0 = (1370) (3 108 ) (4 10- 7) = 0.72103 V/m, which is not very different from the electric field in the filament of a light bulb. The corresponding magnetic field is Brms = Erms /c = 2.4 T. This is about 1/20'th of the earth's magnetic field. (d) The total power radiated from the sum is intensityarea = 1370 4 (1.5 1011 ) = 3.9 1026 J/s. Problem 4. Take ^ pointing along the wires of the wire grid and x pointing z ^ in the wave direction which is perpendicular to the grid. (a) The grid is effectively an insulator for y-components of electric field so the wave passes through the grid without attenuation. (b) The grid is effectively an conductor for z-components of electric field so the wave is substantially reflected (not absorbed as the problem states.) (c) Conduction electrons in the metal of the grid move in response to the incident electric field. But they cannot move perpendicular to the wires without building up a retarding field that prevents their further motion. But electrons moving parallel to the wires move freely. For a good conductor like aluminum this leads to high reflectivity. For a poor conductor like graphite the resistance of the medium would cause the wave to be mainly absorbed. Problem 5. (a) The electric field of a vertical (z-axis) antenna is vertical so the radiation is strongest horizontally (x, y-plane). (b) There is little or no upward radiation. (c) In the horizontal plane, distant from the antenna, the electric field is vertical, so the magnetic field is horizontal. As the magnetic field varies with time it induces an e.m.f. in a conductive loop. For largest e.m.f. the normal vector to the loop should be oriented oriented horizontal, and at right angles to the vector from the transmitting antenna to the receiver (because that is the direction of the magnetic field.) 3 Problem 6. Incident intensity IA , unpolarized, is given. y (a) 111111 000000 111 000 111111 111 000 C A 000000 B 111111 000000 111 000 111 000 11111111111111 00000000000000 111111 11111111111111111111111111111 1111111111111111 000000 00000000000000000000000000000 0000000000000000 x 111 000 111111 111111 000000 000000 111 000 111111 111111 000000 000000 111 000 111111 111111 000000 000000 111 000 111111 111111 000000 000000 z 111111 000000 1111 0000 111 111111 000000 B _ 0000 B+ 000 C 1111 111 000 A 000000 111111 1111 0000 111 000 1111 0000 111 000 11111111111111 00000000000000 111111 111111111111111111 11111 111111111111 000000 000000000000000000 00000 000000000000 1111 0000 111 000 111111 111111 000000 000000 1111 0000 111 000 111111 111111 000000 000000 1111 0000 111 000 111111 111111 000000 000000 1111 0000 111 000 111111 111111 000000 000000 z 111 000 IC IA x y (c) 1 8 (e) 0 111111111111111111111111111 000000000000000000000000000 45 2 90 (a) In region B the intensity is IA /2, y-polarized. (b) No light gets to region C. (c) Inserting the diagonal polaroid has no effect on the light in region B- . But in region B+ the electric field is tilted by 45 degrees and its electric field is altered by factor 1/ 2, so the intensity is IA /4. (d) In region C the electric field has only a z-component, its electric field is altered by another factor 1/ 2, so the intensity is IA /8. 4 ...
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This note was uploaded on 09/01/2009 for the course PHYS 208 taught by Professor Amadeuri during the Spring '08 term at Cornell University (Engineering School).

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