60299-Chapter_15

60299-Chapter_15 - CHAPTER 15 Exercises E15.1 If one grasps...

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Unformatted text preview: CHAPTER 15 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) φ mWb 927 . 3 ) 05 . ( 5 . 2 2 = = = = π π 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 right-hand rule. Thus, for each part the integral equals the sum of the currents flowing upward. Refering 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 506 E15.8 Refer to Figure 15.9 in the book. Conceptually the left-hand 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 right-hand rule, the direction of this field is in the direction of Path 1. The field in turn produces a force on the right-hand wire given by N 10 2 ) 10 )( 1 ( 10 2 4 5 − × = × = = i B f l By the right-hand rule, the direction of the force is such that the wires repel one another. E15.9 The magnetic circuit is: The reluctance of the iron is: 4 7 2 iron iron iron 10 4 10 4 5000 10 27 − − − × × × × × = = π μ μ A R r l 3 iron 10 4 . 107 × = R The reluctance of the air gap is: 4 7 2 gap gap gap 10 9 10 4 10 − − − × × × = = π μ A R l 6 gap 10 842 . 8 × = R Then we have mWb 45 . 10 9 5 . 4 gap gap = × × = = − A B φ A 027 . 4 1000 ) 10 45 . )( 10 842 . 8 10 4 . 107 ( ) ( 3 6 3 gap iron = × × + × = + = − N R R i φ E15.10 Refer to Example 15.6 in the book. Neglecting the reluctance of the iron, we have: = c R 6 4 7 2 10 842 . 8 10 9 10 4 10 1 × = × × × × = = − − − π μ a gap a A R l 507 6 4 7 2 10 366 . 6 10 25 . 6 10 4 10 5 . × = × × × × = = − − − π μ b gap b A R l Wb 1 . 113 10 842 . 8 2 500 6 μ φ = × × = = a a R Ni T 1257 ....
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This note was uploaded on 10/08/2009 for the course AERO 215 taught by Professor Tyler during the Spring '09 term at Texas A&M.

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60299-Chapter_15 - CHAPTER 15 Exercises E15.1 If one grasps...

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