Chapter 31

# Chapter 31 - 31.1 Solve The wires cross-sectional area is A...

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31.1. Solve: The wire’s cross-sectional area is () 2 23 6 2 1.0 10 m 3.1415 10 m Ar ππ −− == × = × , and the electron current through this wire is 19 1 2.0 10 s e N i t ==× Δ . Using Table 31.1 for the electron density of iron and Equation 31.3, the drift velocity is ( ) 19 1 5 d 28 3 6 2 7.5 10 8.5 10 m i v nA × ×× m/s = 75 μ m/s Assess: The drift speed of electrons in metals is small.

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31.2. Solve: Using Equation 31.3 and Table 31.1, the electron current is () ( ) 2 28 3 3 5 18 1 d 5.9 10 m 0.5 10 m 5.0 10 m/s 2.3 10 s in A v π −− == × × × = × The time for 1 mole of electrons to pass through a cross section of the wire is 23 5 A 18 1 1 mole 6.02 10 2.62 10 s 3.0 days N t i ×× = × × Assess: The drift speed is small, and Avogadro’s number is large. A time of the order of 3 days is reasonable.
31.3. Solve: Equation 31.2 is ed N nAv t . Using Table 31.1 for the electron density, we get 2 e d 4 DN A nv t π == Δ ( ) () ( ) 16 4 e 28 3 4 6 d 41 .0 10 4 9.3 10 m 0.93 mm 5.8 10 m 8.0 10 m/s 320 10 s N D nv t −− × ⇒= = = × = Δ ×× ×

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31.4. Solve: The number of electrons crossing a cross sectional area of a wire is the electron current i . () 2 3 28 3 4 19 1 1.6 10 m 6.0 10 m 2.0 10 m/s 2.4 10 s 2 d in A v π −− ⎛⎞ × == × × = × ⎜⎟ ⎝⎠ The electron density n for aluminum is taken from Table 31.1. The number of electrons passing through the cross section in one day is N e = i Δ t = (2.4 × 10 19 s –1 )(365 days)(24 hr/day)(3600 s/hr) = 7.6 × 10 26 electrons Assess: The large electron density compensates for the small drift velocity to deliver a huge number of electrons in current.
31.5. Solve: Using Equation 31.2, () ( ) ( ) 14 28 3 e 62 4 6 d 1.44 10 6.0 10 m 4.00 10 m 2.00 10 m/s 3.0 10 s N n Av t −− × == = × Δ ×× × From Table 31.1, the metal is aluminum.

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31.6. Model: Use the conduction model to relate the drift speed to the electric field strength. Solve: From Equation 31.7, the electric field is () ( ) ( ) 31 4 d 19 14 9.11 10 kg 2.0 10 m/s 0.023 N/C 1.60 10 C 5.0 10 s mv E e τ −− ×× == =
31.7. Solve: For 2 1.0 cm 1.0 10 m L == × , the surface area of the wire is A = (2 π r ) L = DL (1.0 × 10 3 m)(1.0 × 10 2 m) = (1.0 × 10 5 m 2 ) The surface charge density of the wire is () 11 9 12 2 52 1000 cm 1 cm 1.60 10 C 5.1 10 C/m Q A η −− ×× = × ×

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31.8. Solve: (a) Each gold atom has one conduction electron. Using Avogadro’s number and n as the number of moles, the number of atoms is ( ) 2 AAA A AA A rL mV N nN N N N MM M ρπ ρ == = = The density of gold is = 19,300 kg/m 3 , the atomic mass is M A = 197 g mol 1 , r = 0.50 × 10 3 m, L = 0.10 m, and N A = 6.02 × 10 23 mol 1 . Substituting these values, we get N = 4.6 × 10 21 electrons. (b) If all the electrons are transferred a charge of (4.6 × 10 21 )( 1.60 × 10 19 C) = 740 C will be delivered. To deliver a charge of 32 nC, however, the electrons within a length l have to be delivered. Thus, () 9 10 12 32 10 C 10 cm 4.3 10 cm 4.3 10 m 740 C l −− −× × = ×
31.9. Model: We will use the model of conduction to relate the electric field strength to the mean free time between collisions.

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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 31 - 31.1 Solve The wires cross-sectional area is A...

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