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Unformatted text preview: 1 CHAPTER 15 Solutions for Exercises E15.1 If one grasps the wire with the right hand and with the thumb pointing north, the fingers point west under the wire and curl around to point east above the wire. E15.2 If one places the fingers of the right hand on the periphery of the clock pointing clockwise, the thumb points into the clock face. E15.3 z y x q u u u B u f 14 5 19 10 602 . 1 10 ) 10 602 . 1 ( − − × − = × × − = × = in which u x , u y , and u z are unit vectors along the respective axes. E15.4 N 5 ) 90 sin( 5 . ) 1 ( 10 ) sin( = = = o l θ B i f E15.5 (a) mWb 927 . 3 ) 05 . ( 5 . 2 2 = = = = π π φ r B BA turns mWb 39.27 = = φ λ N (b) V 27 . 39 10 10 27 . 39 3 3 − = × − = = − − dt d e λ More information would be needed to determine the polarity of the voltage by use of Lenz’s law. Thus the minus sign of the result is not meaningful. E15.6 T 10 4 10 2 20 10 4 2 4 2 7 − − − × = × × = = π π π μ r I B E15.7 By Ampère’s law, the integral equals the sum of the currents flowing through the surface bounded by the path. The reference direction for the currents relates to the direction of integration by the righthand rule. Thus, for each part the integral equals the sum of the currents flowing upward. Referring to Figure 15.9 in the book, we have ∫ = ⋅ 1 Path A 10 l d H ∫ = − = ⋅ 2 Path A 10 10 l d H ∫ − = ⋅ 3 Path A 10 l d H 2 E15.8 Refer to Figure 15.9 in the book. Conceptually the lefthand wire produces a field in the region surrounding it given by T 10 2 10 2 10 10 4 2 5 1 7 − − − × = × × = = π π π μ r I B By the righthand rule, the direction of this field is in the direction of Path 1. The field in turn produces a force on the righthand wire given by N 10 2 ) 10 )( 1 ( 10 2 4 5 − × = × = = i B f l By the righthand rule, the direction of the force is such that the wires repel one another....
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This homework help was uploaded on 03/19/2008 for the course EE 331 taught by Professor Preston during the Fall '06 term at University of Texas.
 Fall '06
 Preston

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