HW12solutions - Physics 213 Homework 12 Solutions Spring...

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Unformatted text preview: Physics 213 Homework 12 Solutions Spring 2008 32.23. [EM Wave Energy & Fields] I DENTIFY & SET UP: At a distance r from a source, the intensity I is given by I = P av / A , where A = 4 r 2 is the surface area of a sphere. I is in turn related to the Electric field strength by I = 1 2 cE max 2 . Also, c=1/( ) 1/2 . EXECUTE: The total power output of the source is P av = I av A = E max 2 2 c (4 r 2 ) , where I is the intensity. E max = P av c 2 r 2 = (60.0 W)(3.00 10 8 m s) 2 (5.00 m) 2 = 12.0 V m. B max = E max c = 12.0 V m 3.00 10 8 m s = 4.00 10 8 T. EVALUATE: E max and B max are both inversely proportional to the distance from the source. 32.55. IDENTIFY and SET UP: The gravitational force is given by Eq.(12.2). Express the mass of the particle in terms of its density and volume. The radiation pressure is given by Eq.(32.32); relate the power output L of the sun to the intensity at a distance r . The radiation force is the pressure times the cross sectional area of the particle. EXECUTE: (a) Gravitational force: F g = G mM r 2 . Particle mass: m = V = 4 3 R 3 . Thus F g = 4 G MR 3 3 r 2 . (b) For a totally absorbing surface p rad = I c . If L is the power output (P in the previous problem) of the sun, the intensity of the solar radiation a distance r from the sun is I = L 4 r 2 . Thus p rad = L 4 cr 2 . The force F rad that corresponds to p rad is in the direction of propagation of the radiation, so F rad = p rad A , where A = R 2 is the component of area of the particle perpendicular to the radiation direction. Thus F rad = L 4 cr 2 ( R 2 ) = LR 2 4 cr 2 . (c) F g = F rad 4 G MR 3 3 r 2 = LR 2 4 cr 2 4 G M 3 R = L 4 c , and so R = 3 L 16 c G M . R = 3(3.9 10 26 W) 16(2.998 10 8 m/s)(3000 kg/m 3 )(6.673 10 11 N m 2 / kg 2 ) (1.99 10 30 kg) = 1.9 10 7 m = 0.19 m....
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HW12solutions - Physics 213 Homework 12 Solutions Spring...

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