This preview shows pages 1–3. 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: 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 singly-charged 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 drug-testing 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 cross-sectional 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 current-carrying 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....
View Full Document