The figure shows wire section 1 of diameter D1 = 5.50R and wire section 2 of diameter D2 = 4.0OR, 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 width. The electric potential change V along the length L = 2.10 m shown in section 2 is 11.0 µV. The number of charge carriers per unit volume is 8.70 × 1028 m3. What is the drift speed of the conduction electrons in section 1? (2) (1) Number 1.18473025e-39 Units m/s

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Chapter1: Units, Trigonometry. And Vectors
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The figure shows wire section 1 of diameter D1 = 5.50R and wire section 2 of diameter D2 = 4.00R, 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 width. The electric potential change V along the length L = 2.10 m shown in section 2 is 11.0 µV. The number of
charge carriers per unit volume is 8.70 x 1028m3. What is the drift speed of the conduction electrons in section 1?
D1
Da
(2)
(1)
Number
i
1.18473025e-39
Units
m/s
Transcribed Image Text:The figure shows wire section 1 of diameter D1 = 5.50R and wire section 2 of diameter D2 = 4.00R, 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 width. The electric potential change V along the length L = 2.10 m shown in section 2 is 11.0 µV. The number of charge carriers per unit volume is 8.70 x 1028m3. What is the drift speed of the conduction electrons in section 1? D1 Da (2) (1) Number i 1.18473025e-39 Units m/s
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