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Unformatted text preview: ELECTRIC CURRENT E XERCISES Section 24.1 Electric Current 14. The current is the amount of charge passing a given point in the wire, per unit time, so in one second, (1.5 A)(1s) 1.5 C. q I t = = = The number of electrons in this amount of charge is 19 1.5 C/1.6 10 18 C 9.38 10 . = 15. I NTERPRET This problem is about the number of charges involved in setting up an electric current. D EVELOP For a steady current, the amount of charge crossing a given area in time t can be found by using Equation 24.1a, / . I Q t = E VALUATE A battery rated at 80 A h can supply a net charge of 5 (80 C/s)(3600 s) 2.88 10 C Q I t = = = A SSESS By definition, 1 A 1C/s, = or 1C 1 A s. = So the result makes sense. 16. The charge moving through the membrane each second is 30 nC. Since singlycharged ions carry one elementary charge (about 160 zC), this corresponds to 11 23 30 nC/(160 zC/ion) (1.88 10 ion)/(6.02 10 ion/mol) 0.311 = = pmol. (See Example 24.2(a), Appendix B, and Table 1.1.) Chemical drugtesting instrumentation can detect amounts of substances this low. 17. I NTERPRET In this problem we are asked to find the current density, given the electric current and the cross section. D EVELOP The current density J , is defined as the current per unit area, or / . J I A = The area of the cross section is 2 2 /4. A R d = = E VALUATE The cross section of a wire is uniform, so the density is 6 2 2 3 2 10 A 7.65 10 A/m /4 (1.29 10 m) /4 I I J A d  = = = = A SSESS If the current I is kept fixed, the smaller the crosssectional area A , the greater the current density J . Section 24.2 Conduction Mechanisms 18. Aluminum obeys Ohms law, so 8 2 / (0.085 V/m)/(2.65 10 m) 3.21 MA/m J E E  = = = = (see Table 24.1). 19. I NTERPRET In this problem we are asked to calculate the electric field in a currentcarrying conductor. D EVELOP To find the electric field, we make use of Ohms law (which applies to silver), / , J E = and the definition of current density, which is assumed to be uniform in the wire. E VALUATE Using Table 24.1, we find the electric field to be 8 2 4 2 (1.59 10 m)(7.5 A) 0.168 V/m /4 (9.5 10 m) /4 I E J d  = = = = A SSESS The value is a lot smaller than the electric field we discussed in electrostatic situations. Since silver is such a good conductor, a small field can drive a substantial current. 20. Assuming a uniform current density obeying Ohms law, we find 2 1 4 / / , or 4 / J E I d d I E = = = = 1/2 2[(0.22 m)(350 mA)/ (21 V/m)] 6.83 cm = (see Table 24.1). 24.1 24 24.2 Chapter 24 21. I NTERPRET This problem is about applying Ohms law to find the resistivity of a rod....
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 Spring '09
 Prof.Kim
 Physics, Charge, Current

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