This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Assignment 4 Solutions Chpt 26 1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 10 3 C. (b) The number of electrons N is given by q = Ne , where e is the magnitude of the charge on an electron. Thus, N = q / e = (1200 C)/(1.60 10 19 C) = 7.5 10 21 . 23. The resistance of conductor A is given by 2 A A r L R = , where r A is the radius of the conductor. If r o is the outside diameter of conductor B and r i is its inside diameter, then its crosssectional area is ( r o 2 r i 2 ), and its resistance is ) ( 2 2 i B r r L R = The ratio is . 3 ) 50 . ( ) 50 . ( ) . 1 ( 2 2 2 2 2 2 = = = mm mm mm r r r R R A i B A . 26. Let be the radius of the kite string and 2.00 mm r = 0.50 mm t = be the thickness of the water layer. The crosssectional area of the layer of water is ....
View
Full
Document
 Spring '08
 Parsons
 Charge, Current

Click to edit the document details