YF_ISM_25

YF_ISM_25 - CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE 25...

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25-1 C URRENT , R ESISTANCE , AND E LECTROMOTIVE F ORCE 25.1. IDENTIFY: / IQ t = . SET UP: 1.0 h 3600 s = EXECUTE: 4 (3.6 A)(3.0)(3600 s) 3.89 10 C. QI t == = × EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of charge. 25.2. IDENTIFY: / t = . Use d I nqvA = to calculate the drift velocity d . v SET UP: 28 3 5.8 10 m . n 19 1.60 10 C q . EXECUTE: (a) 2 420 C 8.75 10 A. 80(60 s) Q I t = × (b) d . = This gives 2 6 d 28 19 3 2 8.75 10 A 1.78 10 m s. (5.8 10 )(1.60 10 C)( (1.3 10 m) ) I v nqA π −− × = × ×× × EVALUATE: d v is smaller than in Example 25.1, because I is smaller in this problem. 25.3. IDENTIFY: / t = . / JIA = . d J nqv = SET UP: 2 (/ 4 ) AD = , with 3 2.05 10 m D . The charge of an electron has magnitude 19 1.60 10 C. e += × EXECUTE: (a) (5.00 A)(1.00 s) 5.00 C. t = The number of electrons is 19 3.12 10 . Q e (b) 62 23 2 5.00 A 1.51 10 A/m . 4 ) 4 ) ( 2 . 0 51 0 m ) I J D ππ = × × (c) 4 d 28 3 19 1.51 10 A/m 1.11 10 m/s 0.111 mm/s. (8.5 10 m )(1.60 10 C) J v nq × = EVALUATE: (a) If I is the same, / = would decrease and d v would decrease. The number of electrons passing through the light bulb in 1.00 s would not change. 25.4. (a) IDENTIFY: By definition, J = I/A and radius is one-half the diameter. SET UP: Solve for the current: I = JA = J π ( D /2) 2 EXECUTE: I = (1.50 × 10 6 A/m 2 )( π )[(0.00102 m)/2] 2 = 1.23 A EVALUATE: This is a realistic current. (b) IDENTIFY: The current density is J = nqv d SET UP: Solve for the drift velocity: v d = J/nq EXECUTE: Since most laboratory wire is copper, we use the value of n for copper, giving d (1.50 10 A/m ) v /[(8.5 × 10 28 el/m 3 )(1.60 × 19 10 C) = 1.1 × 4 10 m/s = 0.11 mm/s EVALUATE: This is a typical drift velocity for ordinary currents and wires. 25.5. IDENTIFY and SET UP: Use Eq. (25.3) to calculate the drift speed and then use that to find the time to travel the length of the wire. EXECUTE: (a) Calculate the drift speed d : v () 2 2 3 4.85 A 1.469 10 A/m 1.025 10 m II J Ar = = × × ( ) 4 d 28 3 19 1.079 10 m/s 8.5 10 / m 1.602 10 C J v × = × 3 4 d 0.710 m 6.58 10 s 110 min. L t v = × = × 25
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25-2 Chapter 25 (b) d 2 I v rnq π = 2 d rnqL L t vI == t is proportional to 2 r and hence to 2 d where 2 dr = is the wire diameter. () 2 34 4.12 mm 6.58 10 s 2.66 10 s 440 min. 2.05 mm t ⎛⎞ = ×= ⎜⎟ ⎝⎠ (c) EVALUATE: The drift speed is proportional to the current density and therefore it is inversely proportional to the square of the diameter of the wire. Increasing the diameter by some factor decreases the drift speed by the square of that factor. 25.6. IDENTIFY: The number of moles of copper atoms is the mass of 3 1.00 m divided by the atomic mass of copper. There are 23 A 6.023 10 N atoms per mole. SET UP: The atomic mass of copper is 63.55 g mole, and its density is 3 8.96 g cm . Example 25.1 says there are 28 8.5 10 × free electrons per 3 m . EXECUTE: The number of copper atoms in 3 1.00 m is 36 3 3 2 3 28 3 (8.96 g cm )(1.00 10 cm m )(6.023 10 atoms mole) 8.49 10 atoms m . 63.55 g mole ×× EVALUATE: Since there are the same number of free 3 electrons m as there are atoms of 3 copper m , the number of free electrons per copper atom is one.
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_25 - CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE 25...

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