Chap26_08 - Chapter 26 26.1. During the 4.0 min a 5.0 A...

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Chapter 26 26.1. During the 4.0 min a 5.0 A current is set up in a wire, how many (a) Coulombs and (b) electrons pass through any cross section across the wire’s width Δ q = i Δ t = 5.0 A 240 s = 1200 C Δ q = ne n = Δ q e = 1200 C 1.6 × 10 19 C = 7.5 × 10 21 electrons 26.4 A small but measurable current of 1.2 × 10 10 A exists in a copper wire whose diameter is 2.5 mm . The number of charge carriers per unit volume is 8.49 × 10 28 m 3 . Assuming the current is uniform, calculate the (a) current density and (b) the electron drift speed. (a) W can compute the current density from the definition. J = i A = 1.2 × 10 10 A π (.00125 m ) 2 = 2.44 × 10 5 A / m 2 (b) Now that we know J, we can compute the velocity J = ρ v = (# of particles / m 3 ) ( chg / particle ) = 8.49 × 10 28 m 3 1.6 × 10 19 C = 1.39 × 10 10 C / m 3 v = J = 2.44 × 10 5 A / m 2 1.39 × 10 10 C / m 3 = 1.8 × 10 15 m / s 26.7 A beam contains 2 × 10 8 double charged positive ions per cubic centimeter, all of which are
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Chap26_08 - Chapter 26 26.1. During the 4.0 min a 5.0 A...

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