ph1bA_08_hw8s

# ph1bA_08_hw8s - Ph1b(analytic track Solution to Homework 8...

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Ph1b (analytic track): Solution to Homework 8 Chefung Chan, Winter 2008 Problem 4-2 ( 5 points ) Each electron makes 3 × 10 8 m/s 240 m = 1 . 25 × 10 6 round trips per second. The charge passing any point, per second, is I = 1 . 25 × 10 6 × 10 11 × 4 . 8 × 10 10 = 6 × 10 7 esu/s = 0 . 02 amps. Problem 4-4 ( 5 points ) (a) (2.5 points) The resistance of the conductor is just ρL A = 3 × 10 6 · cm × 3 × 10 8 cm 7 × π × ( 0 . 073 2 cm ) 2 = 31 kΩ. (b) (2.5 points) On the return path, the current will spread out. We approximate that the current ﬂows through a cylinder of radius 10 m to 100 m. Then, the resistance of the return path is R return = ρL A = 25 × 3 × 10 8 π (1 , 000) 2 2 kΩ to 25 × 3 × 10 8 π (10 , 000) 2 20 Ω, which is smaller than that of the cable. Problem 4-7 ( 5 points ) Consider making a large resistor of this material (a cable for simplicity) which comprises n layers of silver alternating with n layers of tim. Let thickness of silver = L 0 , conductivity of silver = 7 . 2 σ 0 , thickness of tin = 2 L 0 , conductivity of tin = σ 0 .

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