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Unformatted text preview: CHAPTER 10 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 eld 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 10.2. In Fig. 10.1, replace the voltmeter with a resistance, R . a) Find the current I that ows as a result of the motion of the sliding bar: The current is found through I = 1 R I E d L = 1 R d m dt Taking the normal to the path integral as a z , the path direction will be counterclockwise when viewed from above (in the a z direction). The minus sign in the equation indicates that the current will therefore ow clockwise , since the magnetic ux is increasing with time. The ux of B is m = Bdvt , and so  I  = 1 R d m dt = Bdv R (clockwise) b) The bar current results in a force exerted on the bar as it moves. Determine this force: F = Z Id L B = Z d Idx a x B a z = Z d Bdv R a x B a z = B 2 d 2 v R a y N c) Determine the mechanical power required to maintain a constant velocity v and show that this power is equal to the power absorbed by R . The mechanical power is P m = F v = ( Bdv ) 2 R W The electrical power is P e = I 2 R = ( Bdv ) 2 R = P m 10.3. Given H = 300 a z cos(3 10 8 t y ) A/m in free space, Fnd 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 ) dx dy = 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 . 10.4. Conductor surfaces are located at = 1cm and = 2cm in free space. The volume 1cm < < 2cm contains the Felds H = (2 / )cos(6 10 8 t 2 z ) A/m and E = (240 / )cos(6 10 8 t 2 z ) V/m. a) Show that these two Felds satisfy Eq. (6), Sec. 10.1: Have E = E z a = 2 (240 ) sin(6 10 8 t 2 z ) a = 480 2 sin(6 10 8 t 2 z ) a Then B t = 2 (6 10 8 ) sin(6 10 8 t 2 z ) a = (8 10 7 )(6 10 8 ) sin(6 10 8 t 2 z ) = 480 2 sin(6 10 8 t 2 z ) a b) Evaluate both integrals in Eq. (4) for the planar surface deFned by b) Evaluate both integrals in Eq....
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 Spring '08
 CITRIN
 Electromagnet

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