Solutions3 - Physics 208, Assignment 3, Solutions, 2/6/09,...

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Unformatted text preview: Physics 208, Assignment 3, Solutions, 2/6/09, RT Problem 1. Note that down and radially inward are synonymous for this problem. E =- 1 . 2 10 2 (V / m) r , n = 0 . 55 10 9 m 3 , n + = 0 . 62 10 9 m 3 , = conductivity = 2 . 7 10 14 1 m 1 , (a) + ions drift in (in direction of E ), giving current in,- ions drift out, giving current in. (b) Both types of ion give inward current. Current always flows in the direction of electric field (unless there is magnetic field present as well.) (c) Current density is j = E =- 2 . 7 10 14 1 . 2 10 2 r =- 3 . 2 10 12 (A / m 2 ) r (d) Since the drift speeds are said to be equal it is only n = n + n + = 1 . 17 10 9 m 3 that enters. The drift speed is v d = E en = 2 . 7 10 14 1 . 2 10 2 1 . 6 10 19 1 . 17 10 9 = 1 . 7 10 2 m / s . (e) A = area of earth s surface = 4 (6 . 4 10 6 ) 2 = 5 . 1 10 14 m 2 . i = total current = jA = 3 . 2 10 12 5 . 1 10 14 = 1650 A, flowing inward. (f) The current flow from one place (upper atmosphere) to another (earth) caused by excess charge (of whatever sign) on the first place, necessarily discharges the first place. This diminishes the atmo- spheric electric field. It is just like the discharging of a capacitor or a battery. Problem 2. (a) Since V = IR the bulb resistance (when lit) is R = V I = 3 . . 30 = 10 . (b) Power dissipated by bulb P = IV = 3 . . 3 = 0 . 9 W , or, if you prefer, P = V 2 R = 3 . 2 10 = 0 . 9 W . 1 (c) Clearly you have to look up how the conductivity (or resistivity which is 1/conductivity) of tungsten depends on temperature. From the text, Table 26.1, p. 689, the temperature coefficient of resistivity for tungsten is 4 . 5 10 3 in fractional change per degree Kelvin. Using centigrade units ( R hot- R cold ) /R cold T hot- T cold = (10- 1 . 2) / 1 . 2 T hot- 20 C = 4 . 5 10 3 . (1) T hot = 20 + 8 . 8 1 . 2 1 4 . 5 10 3 1600 C ....
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Solutions3 - Physics 208, Assignment 3, Solutions, 2/6/09,...

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