Class Examples for Chapter 26

# Class Examples for Chapter 26 - Class Examples for Chapter...

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1 Class Examples for Chapter 26 9. We note that the radial width r = 10 m is small enough (compared to r = 1.20 mm) that we can make the approximation 22 Br rdr Br r r   Thus, the enclosed current is 2 Br 2 r = 18.1 A. Performing the integral gives the same answer. 18. (a) i = V / R = 23.0 V/15.0 10 –3 = 1.53 10 3 A. (b) The cross-sectional area is A r D 2 1 4 2 . Thus, the magnitude of the current density vector is J i A i D 4 4 153 10 541 10 2 3 3 2 7 2 . . A 6.00 10 m A / m . c h c h (c) The resistivity is 3 3 2 8 (15.0 10 ) (6.00 10 m) 10.6 10 m. 4(4.00 m) RA L   (d) The material is platinum. 24. (a) Since the material is the same, the resistivity is the same, which implies (by Eq. 26-11) that the electric fields (in the various rods) are directly proportional to their current-densities. Thus, J 1 : J 2 : J 3 are in the ratio 2.5/4/1.5 (see Fig. 26-25). Now the currents in the rods must be the same (they are “in series”) so J 1 A 1 = J 3 A

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Class Examples for Chapter 26 - Class Examples for Chapter...

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