Chapter 26, Problem 034 The figure shows wire section 1 of diameter D1 5.50R and

Question

Chapter 26, Problem 034 The figure shows wire section 1 of diameter D1 5.50R and wire section 2 of diameter D2 3.50R, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's wldth. The electric potential change V along the length L = 1.70 m shown in section 2 is 13.0 V. The number of charge carriers per unit volume is 8.10 × 1028 m-3, what is the drift speed of the conduction electrons in section 1? Di D2 Number the tolerance is +/-5% Click if you would like to Show Work for this question: Units Open Show Work

Explanation / Answer

E = V / L = 13 * 10-6 / 1.70 = 7.65 * 10-6 V/m

J2 = 7.65 * 10-6/1.69 * 10-8 = 452.6 A/m2

J1 A1 = J2 A2

J1 (5.5 R)2 = J2 ( 3.5 R)2

J1 = 3.52/5.52 * J2  = 3.52/5.52 * 452.6 = 183.3 A/m2

drift velocity vd = J1 / n e

= 183.3 A/m2 / (8.10 * 1028 * 1.6 * 10-19)

= 1.41 * 10-8 m/s

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