# chap24 - ELECTRIC CURRENT 24 EXERCISES Section 24.1...

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24.1 ELECTRIC CURRENT 24 EXERCISES 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. qI t Δ=Δ= = The number of electrons in this amount of charge is1.5 19 C/1.6 10 × 18 C9 . 3 81 0 . t 15. INTERPRET This problem is about the number of charges involved in setting up an electric current. DEVELOP For a steady current, the amount of charge crossing a given area in time Δ can be found by using Equation 24.1a, /. I Qt =Δ Δ EVALUATE A battery rated at 80 A h can supply a net charge of 5 (80 C/s)(3600 s) 2.88 10 C QI t = × ASSESS By definition,1A or1C 1C/s, = 1A s. = So the result makes sense. 16. The charge moving through the membrane each second is Since singly-charged ions carry one elementary charge (about160 this corresponds to 30 nC. zC), 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. INTERPRET In this problem we are asked to find the current density, given the electric current and the cross section. DEVELOP The current density J , is defined as the current per unit area, or J IA = The area of the cross section is 22 /4. AR d ππ == EVALUATE The cross section of a wire is uniform, so the density is 62 23 2 10 A 7.65 10 A/m /4 (1.29 10 m) /4 II J Ad = = × × ASSESS 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 Ohm’s law, so (see Table 24.1). 82 / (0.085 V/m)/(2.65 10 m) 3.21 MA/m JE E σρ === × Ω = 19. INTERPRET In this problem we are asked to calculate the electric field in a current-carrying conductor. DEVELOP To find the electric field, we make use of Ohm’s law (which applies to silver), /, J E ρ = and the definition of current density, which is assumed to be uniform in the wire. EVALUATE Using Table 24.1, we find the electric field to be 8 24 2 (1.59 10 m)(7.5 A) 0.168 V/m /4 (9.5 10 m) /4 I EJ d ×Ω = = × ASSESS 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 Ohm’s law, we find 2 1 4 // , o r 4 / J EId d I E ρπ π = = 1/2 2[(0.22 m)(350 mA)/ (21 V/m)] 6.83 cm Ω⋅ = (see Table 24.1).

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24.2 Chapter 24 21. INTERPRET This problem is about applying Ohm’s law to find the resistivity of a rod. DEVELOP If the rod has a uniform current density and obeys Ohm’s law (Equations 24.4a and 4b), then its resistivity is 2 (/ 4 ) / EEE d JI A I π ρ == = EVALUATE Substituting the values given in the problem, we find the resistivity of the rod to be 22 2 6 ( /4) (1.4 V/m) (10 m) /4 2.20 10 m 50 A Ed I ππ == = × Ω ASSESS With reference to Table 24.1, we see that the value is within the range of resistivity of a metallic conductor.
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chap24 - ELECTRIC CURRENT 24 EXERCISES Section 24.1...

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