1039 Chapter 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) = 1.2×103C. (b) The number of electrons Nis given by q= Ne, where eis the magnitude of the charge on an electron. Thus, N= q/e= (1200 C)/(1.60 ×10–19C) = 7.5 ×1021. 2. Suppose the charge on the sphere increases by Δqin time Δt. Then, in that time its potential increases by 0,4qVrπεΔΔ=where ris the radius of the sphere. This means 04.qrVπεΔ=ΔNow, Δq= (iin– iout) Δt, where iinis the current entering the sphere and ioutis the current leaving. Thus, ()()()()09inoutinout30.10 m1000 V48.9910F/m1.0000020 A1.0000000 A5.610s.rVqtiiiiπε−ΔΔΔ ===−−×−=×3. We adapt the discussion in the text to a moving two-dimensional collection of charges. Using σfor the charge per unit area and wfor the belt width, we can see that the transport of charge is expressed in the relationship i= σvw, which leads to σ==××=×−−−ivw100103050106 710626Am smC m2bgch.. 4. We express the magnitude of the current density vector in SI units by converting the diameter values in mils to inches (by dividing by 1000) and then converting to meters (by multiplying by 0.0254) and finally using 224.iiiJARDππ===
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