PHY_solution-chapter25

# PHY_solution-chapter25 - 25.1: Q = It = (3.6 A)(3)(3600 s)...

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25.1: C. 10 89 . 3 ) s 3600 )( 3 )( A 6 . 3 ( 4 × = = = It Q 25.2: a) A. 10 75 . 8 by given is Current 2 ) s 60 ( 80 C 420 - × = = = t Q I b) A nqv I d = ) ) m 10 3 . 1 ( π )( C 10 6 . 1 )( 10 8 . 5 ( A 10 75 . 8 2 3 19 28 2 - - - × × × × = = nqA I v d = . s m 10 78 . 1 6 - × 25.3: a) ) ) m 10 05 . 2 )( 4 π )( C 10 6 . 1 )( 10 5 . 8 ( A 85 . 4 2 3 19 28 - - × × × = = nqA I v d s m 10 08 . 1 4 - × = min 110 s 6574 time travel s m 10 08 . 1 m 71 . 0 4 = = = = - × d v d b) If the diameter is now 4.12 mm, the time can be calculated using the formula above or comparing the ratio of the areas, and yields a time of 26542 s =442 min. c) The drift velocity depends on the diameter of the wire as an inverse square relationship. 25.4: The cross-sectional area of the wire is . m 10 333 . 1 ) m 10 06 . 2 ( 2 5 2 3 2 - - × = × = = π πr A The current density is 2 5 2 5 m A 10 00 . 6 m 10 333 . 1 A 00 . 8 × = × = = - A I J Therefore ; have We ne J v d = 3 28 19 5 2 5 m electrons 10 94 . 6 ) electron C 10 60 . 1 )( s m 10 40 . 5 ( m A 10 00 . 6 × = × × × = = - - e v J n d 25.5: constant. is so , d d v J v q n J = , 2 2 1 1 d d v J v J = s m 10 00 . 6 ) 20 . 1 00 . 6 )( s m 10 20 . 1 ( ) ( ) ( 4 4 1 2 1 1 2 1 2 - - × = × = = = I I v J J v v d d d 25.6: The atomic weight of copper is mole, g 55 . 63 and its density is . cm g 96 . 8 3 The number of copper atoms in thus is m 00 . 1 3 mole g 55 . 63 ) mole atoms 10 023 . 6 )( m cm 10 00 . 1 )( cm g 96 . 8 ( 23 3 3 6 3 × × 3 28 m atoms 10 49 . 8 × = Since there are the same number of free 3 m electrons as there are atoms of 3 m copper (see Ex. 25.1), The number of free electrons per copper atom is one. 25.7: Consider 1 3 m of silver. kg 10 5 . 10 so , m kg 10 5 . 10 density 3 3 3 × = × = m and mol 10 734 . 9 so , mol kg 10 868 . 107 4 3 × = = × = - M m n M 3 28 A m atoms 10 86 . 5 × = = nN N 8b6bf979eef1a37e326f7b046b923ac540a7fe43.doc Page 1 of 16

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If there is one free electron per . m electrons free 10 86 . 5 are there , m 3 28 3 × This agrees with the value given in Exercise 25.2. 25.8: a) C 0106 . 0 ) C 10 60 . 1 )( 10 68 . 2 10 92 . 3 ( ) ( 19 16 16 Na Cl = × × + × = + = - e n n Q total . mA 6 . 10 A 0106 . 0 s 00 . 1 C 0106 . 0 = = = = t Q I total b) Current flows, by convention, in the direction of positive charge. Thus, current flows with + Na toward the negative electrode. 25.9: a) C. 329 3 65 . 0 55 ) 65 . 0 55 ( | | 8 0 3 8 0 8 0 2 8 0 = + = - = = t t dt t dt I Q b) The same charge would flow in 10 seconds if there was a constant current of: A. 1 . 41 ) s 8 ( ) C 329 ( = = = t Q I 25.10: a) . A/m 10 81 . 6 2 5 ) m 10 3 . 2 ( A 6 . 3 2 3 × = = = - × A I J b) . m V 012 . 0 ) A/m 10 81 . 6 )( m 10 72 . 1 ( 2 5 8 = × × = = - ρJ E c) Time to travel the wire’s length: hrs! 22 min 1333 s 10 0 . 8 A 6 . 3 ) m 10 3 . 2 )( C 10 6 . 1 )( m 10 5 . 8 )( m 0 . 4 ( 4 2 3 19 3 28 = × = × × × = = = - - I nqA l v l t d 25.11: . 125 . 0 ) m
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## This note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.

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PHY_solution-chapter25 - 25.1: Q = It = (3.6 A)(3)(3600 s)...

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