Please solve exercise 2-11-3 " Electron excursion distance" from example 2-19 electron velocity V ms at 60 Hz. nd the electron excursion distance (back and forth) on the wire. Problem 2-114. Electron excursion distance. if the voltage applied' In Example 2-19 Is in Arts I. l x 10' 'rn (runs). 0: about 100 atomic spaces. Example 2-19. Electron velocity. If 10 V is applied between the ends of a I-mm. diameter copper wire 100 m long, find (a) current. (b) field, and (c) electron velocity. T = 290 K. Take electron mobility . = 0.004 m? V-1s" and charge density p 1.4 x 1010 Cm-3. Solution. From (10) and Table 2-1, the resistance of the wire is given by W 100 5.7 X 101 X (TT/4) x 10-6 =220 Therefore, from Ohm's law, the current club V 10 1 = = 4.5 A Ans. (a) 2.2 and field V 10 E = T. 100 = 0.1 Vm Ans. (b) and from (3) and Table 2-1. the electron velocity " = Eu. = 0.1 X 0.004 m s ! = 0.4 mms ! Ans. (c) Alternatively. OE Va 5.7 X 107 x 0.1 p 1.4 X 1010 = 4 X 10 * ms" = 0.4 mm s" Ans. (c)