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Unformatted text preview: Chapter 3 Problems Solutions 3.1 Based on the cross product of [3.2] with the vectors B and I separated by 30 , the mag nitude of the force is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 sin 30 0.25 2cos 5cos 0.5 F i t B t t t = = l ( 29 2 1.25cos 0.625 0.625cos 2 N F t t = = + The force F acts perpendicular to a plane that combines both B and I with direction de termined as the direction that the thumb of the right hand points when the fingers are crossed from I to B . 3.2 The Hfield is directed CCW at any particular radius r measured perpendicular to the centerline of the conductor. The current enclosed by a circular path of radius r is 2 2 2 2 4 4 r r Ir I I D D = = By use of [3.4], the magnitude of the Hfield at any distance / 2 r D is 2 2 2 4 2 2 2 / r I Ir D Ir H r r D = = = After substituting known values, ( 29 2 2 4 5 6.366 r r H D D = = 3.3 The assumption of / dB dt = used in derivation of [3.10] is no longer valid; thus, ( 29 d B x dB e B v x dt dt = = + l l l ( 29 1 2 2 sin cos e B B t v x B t = + + l l The first term of e is commonly called a speed voltage . The second term is known as a transformer voltage . 3.4 The MATLAB file cckt1.m is changed by use of an editor so that 2 4 0.10 m = = l l . After execution, the screen gives 19 3.5 The solution is identical to Example 3.3 through the point of calculating the core mmf c F . Starting at that point and neglecting air gap fringing terms in [3.17], ( 29 ( 29 ( 29 7 6 3 4 10 0.05 0.05 1.571 10 H 0.002 o g d  = = = P l The required air gap mmf is 6 0.004 2546.1 At 1.571 10 T g g  = = = F P The total coil current follows as 2826.7 2546.1 10.75 A 500 c g I N + + = = = F F Using the coil current of Example 3.3 as the correct value, 10.75 10.54 % Error 100% 1.99% 10.54 = = 3.6 ( a ) The leakage permeance P l of Example 3.4 is valid; however, the air gap permeance has changed in value. By [3.17], ( 29 3 3 0.52 0.308 g o d d = + + + P l l ( 29 ( 29 ( 29 ( 29 7 6 0.05 0.05 4 10 0.52 0.05 0.05 0.308 0.003 1.113 10 H 0.003  = + + + = cr B and cr H of Example 3.4 are unchanged. Only the mean length path 3 m l need be re calculated. 3 2 0.10 0.05 0.003 0.147 m m h = + = + = l l The mmf across the leakage flux path F l and the resulting leakage flux l follow as 2 4 3 2 2 m m cr m g H = + + + F P l l l l 6 0.20 0.20 0.004 4083.5 0.147 5010.9 At 2 2 1.113 10 = + + + = F l 20 ( 29 ( 29 8 5010.9 9.425 10 0.472 mWb  = = = F P l l l The coil flux is then 4 0.472 4.472 mWb T = + = + = l ( b ) The lefthand core flux density is ( 29 ( 29 ( 29 1 0.004472 1.883 T 0.05 0.05 0.95 T cl B dSF = = = l Either enter Fig. 3.12 or use hm27.m to find 10,131 At cl H = . The coil mmf follows by 2 4 1 2 2 m m cl m NI H = + + + F l l l l 0.20 0.20 10,131 0.15 5010.9 8556.7 2 2 NI...
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This document was uploaded on 01/20/2012.
 Spring '09

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