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Chapter 25 HW Solutions

Chapter 25 HW Solutions - PHYS 241 25.3 Ch 25 Homework...

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PHYS 241 Ch. 25 Homework Solutions 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. 25.5.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 ( 29 6 2 2 2 3 4.85 A 1.469 10 A/m 1.025 10 m I I J A r π π - = = = = × × ( 29 ( 29 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 - = = = × = × (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. ( 29 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. 25.13. I DENTIFY : E J ρ = , where / . J I A = S ET U P : For tungsten 8 5.25 10 m ρ - = × Ω⋅ and for aluminum 8 2.75 10 m. ρ - = × Ω⋅ E XECUTE : (a) tungsten: 8 3 3 2 (5.25 10 m)(0.820 A)

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