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31.46. Solve: (a) A current of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is 1.8x10-" C/s 1.6x10" C Since the channel opens only for 1.0 ms, the total number of potassium ions that pass through the channel is (1.125107 s*)(1.0x10> s)=1.13x10* atoms. (b) The current density in the ion channel is =1.125x10" st _I___18pA 18x10 A p= 25x10" Alm? A x(0.30nm/2) (0.1510 m) 31.61. Model: Assume the battery is ideal. Solve: (a) The electric field inside the wire is E = AV,,,/L . Attaching the wire to the battery makes AV,,. = AV ig =1.5 V . Thus, E= 15V =10V/m 015m (b) Using Table 31.2, the current density is hs ; Sook == =O 86108 Amt p 15x10%Om (©) The current in a wire is related to the potential difference by J = AV,,,/R . Thus, pa Aten 15V Lois I 2A The resistance is related to the wire’s geometry by 1.5x10* Q m)(0.15 pa PEE og y= [PE |(Se10* om\(015 m) < = =3.1x107 m = 031mm A oar aR (0.75 Q) Thus, the wire’s diameter is d = 27 = 0.62 mm. 31.67. Model: The copper wire is uniform. Solve: Equation 31.22 relates the current in a wire to the potential difference across it: (17x10 Om)(20 m)(8.0 A) 7 =087V (1.0107 m) t=Aav sav -24- pL 4 The resistivity p of copper is taken from Table 31.2. Assess: While voltage drops in household wires are small compared to the applied voltage, voltage drops in transmission wires between homes and power plants could be quite large. The power company transports energy in a way that minimizes the voltage drop, as we will leam in Chapter 36. 
