3209-2009-Solutions8

3209-2009-Solutions8 - ECE 3209 Electromagnetic Fields...

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Unformatted text preview: ECE 3209 Electromagnetic Fields University of Virginia Fall 2009 Homework # 8 Solutions 1. Magnetic fields of current distributions. (a) The hairpin turn can be considered as half the superposition of two infinite staight-line currents that border a loop of current: The total field at P is thus, B ( P ) = 1 2 B straight lines + B loop = 1 2 2 I 2 R | {z } B one line + I 4 Z 2 Rd R 2 ~ B ( P ) = I 2 R + I 4 R = I 2 R 1 + 1 2 , into the page (b) For the tightly-wound finite solenoid, we can use the result from the example worked in class where we found the B-field a distance z above the center of a loop of current (radius a ): ~ B ( z ) = I 2 a 2 ( a 2 + z 2 ) 3 2 z For a solenoid with n turns per unit length, and current I in each winding, the current carried by a width dz of the solenoid is I = nI dz . Thus, dB z = nI 2 a 2 dz ( a 2 + ( z- z ) 2 ) 3 2 B z = nI a 2 2 Z dz ( a 2 + ( z- z ) 2 ) 3 2 To integrate, lets make the substitutions x = z- z , dx =- dz and x = a tan u , dx = a sec 2 u : Z dz ( a 2 + ( z- z ) 2 ) 3 2- Z du a 2 sec u = Z cos u a 2 du = sin u a 2 Thus, B z = nI a 2 2 z- z a 2 p a 2 + ( z- z ) 2 = nI 2 cos fl fl fl 2 1 ~ B = z nI 2 cos 2- cos 1 2. B-field at center of regular N-sided polygon. For an N-sided regular polygon, each side subtends an angle of 2 = 2 /N : The B-field at the center will be the superposition of the B-fields for each side of the polygon....
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3209-2009-Solutions8 - ECE 3209 Electromagnetic Fields...

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