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Unformatted text preview: Chapter 10 OddNumbered 10.1. In Fig. 10.4, let B = 0 . 2cos120 t T, and assume that the conductor joining the two ends of the resistor is perfect. It may be assumed that the magnetic field produced by I ( t ) is negligible. Find: a) V ab ( t ): Since B is constant over the loop area, the ux is = (0 . 15) 2 B = 1 . 41 10 2 cos120 t Wb. Now, emf = V ba ( t ) = d /dt = (120 )(1 . 41 10 2 )sin120 t . Then V ab ( t ) = V ba ( t ) = 5 . 33sin120 t V . b) I ( t ) = V ba ( t ) /R = 5 . 33sin(120 t ) / 250 = 21 . 3sin(120 t ) mA 10.3. Given H = 300 a z cos(3 10 8 t y ) A/m in free space, find the emf developed in the general a direction about the closed path having corners at a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic ux will be: = Z 1 Z 1 300 cos(3 10 8 t y ) dxdy = 300 sin(3 10 8 t y )  1 = 300 sin(3 10 8 t 1) sin(3 10 8 t ) Wb Then emf = d dt = 300(3 10 8 )(4 10 7 ) cos(3 10 8 t 1) cos(3 10 8 t ) = 1 . 13 10 5 cos(3 10 8 t 1) cos(3 10 8 t ) V b) corners at (0,0,0), (2 ,0,0), (2 ,2 ,0), (0,2 ,0): In this case, the ux is = 2 300 sin(3 10 8 t y )  2 = 0 The emf is therefore 0 . 10.5. The location of the sliding bar in Fig. 10.5 is given by x = 5 t +2 t 3 , and the separation of the two rails is 20 cm. Let B = 0 . 8 x 2 a z T. Find the voltmeter reading at: a) t = 0 . 4 s: The ux through the loop will be = Z . 2 Z x . 8( x ) 2 dx dy = . 16 3 x 3 = . 16 3 (5 t + 2 t 3 ) 3 Wb Then emf = d dt = . 16 3 (3)(5 t +2 t 3 ) 2 (5+6 t 2 ) = (0 . 16)[5( . 4)+2( . 4) 3 ] 2 [5+6( . 4) 2 ] = 4 . 32 V b) x = 0 . 6 m: Have 0 . 6 = 5 t + 2 t 3 , from which we find t = 0 . 1193. Thus emf = (0 . 16)[5( . 1193) + 2( . 1193) 3 ] 2 [5 + 6( . 1193) 2 ] = 2 . 93 V 110 10.7. The rails in Fig. 10.7 each have a resistance of 2.2 / m. The bar moves to the right at a constant speed of 9 m/s in a uniform magnetic field of 0.8 T. Find I ( t ), 0 < t < 1 s, if the bar is at x = 2 m at t = 0 and a) a 0.3 resistor is present across the left end with the right end opencircuited: The ux in the lefthand closed loop is l = B area = (0 . 8)(0 . 2)(2 + 9 t ) Then, emf l = d l /dt = (0 . 16)(9) = 1 . 44 V. With the bar in motion, the loop resistance is increasing with time, and is given by R l ( t ) = 0 . 3+2[2 . 2(2+9 t )]. The current is now I l ( t ) = emf l R l ( t ) = 1 . 44 9 . 1 + 39 . 6 t A Note that the sign of the current indicates that it is owing in the direction opposite that shown in the figure....
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This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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