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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. I DENTIFY : / I Q t = . S ET U P : 1.0 h 3600 s = E XECUTE : 4 (3.6 A)(3.0)(3600 s) 3.89 10 C. Q It = = = × E VALUATE : Compared to typical charges of objects in electrostatics, this is a huge amount of charge. 25.2. I DENTIFY : / I Q t = . Use d I n q v A = to calculate the drift velocity d . v S ET U P : 28 3 5.8 10 m . n = × 19 1.60 10 C q = × . E XECUTE : (a) 2 420 C 8.75 10 A. 80(60 s) Q I t = = = × (b) d . I n q v A = 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 × = = = × × × × E VALUATE : d v is smaller than in Example 25.1, because I is smaller in this problem. 25.3. I DENTIFY : / I Q t = . / J I A = . d J n q v = S ET U P : 2 ( / 4) A D π = , with 3 2.05 10 m D = × . The charge of an electron has magnitude 19 1.60 10 C. e + = × E XECUTE : (a) (5.00 A)(1.00 s) 5.00 C. Q It = = = The number of electrons is 19 3.12 10 . Q e = × (b) 6 2 2 3 2 5.00 A 1.51 10 A/m . ( / 4) ( /4)(2.05 10 m) I J D π π = = = × × (c) 6 2 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 n q × = = = × = × × E VALUATE : (a) If I is the same, / J I A = 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) I DENTIFY : By definition, J = I/A and radius is one-half the diameter. S ET U P : Solve for the current: I = JA = J ( D /2) 2 E XECUTE : I = (1.50 × 10 6 A/m 2 )( π )[(0.00102 m)/2] 2 = 1.23 A E VALUATE : This is a realistic current. (b) I DENTIFY : The current density is J = nqv d S ET U P : Solve for the drift velocity: v d = J/nq E XECUTE : Since most laboratory wire is copper, we use the value of n for copper, giving 6 2 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 E VALUATE : This is a typical drift velocity for ordinary currents and wires. 25.5. I DENTIFY and S ET U P : Use Eq. (25.3) to calculate the drift speed and then use that to find the time to travel the length of the wire. E XECUTE : (a) Calculate the drift speed d : v ( ) 6 2 2 2 3 4.85 A 1.469 10 A/m 1.025 10 m I I J A r π π = = = = × × ( )( ) 6 2 4 d 28 3 19 1.469 10 A/m 1.079 10 m/s 8.5 10 / m 1.602 10 C J v n q × = = = × × × 3 4 d 0.710 m 6.58 10 s 110 min. 1.079 10 m/s L t v = = = × = × 25
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25-2 Chapter 25 (b) d 2 I v r n q π = 2 d r n q L L t v I π = = t is proportional to 2 r and hence to 2 d where 2 d r = is the wire diameter. ( ) 2 3 4 4.12 mm 6.58 10 s 2.66 10 s 440 min. 2.05 mm t = × = × = (c) E VALUATE : 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. I DENTIFY : 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. S ET U P : 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 . E XECUTE : The number of copper atoms in 3 1.00 m is 3 6 3 3 23 28 3 (8.96 g cm )(1.00 10 cm m )(6.023 10 atoms mole) 8.49 10 atoms m .
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