20075ee1_1_HW10 - CHAPTER 10 10.1 In Fig 10.4 let B = 0.2...

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CHAPTER 10 10.1. In Fig. 10.4, let B = 0 . 2cos120 πt 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 flux is Φ = π (0 . 15) 2 B = 1 . 41 × 10 2 cos120 πt Wb. Now, emf = V ba ( t ) = d Φ /dt = (120 π )(1 . 41 × 10 2 )sin120 πt . Then V ab ( t ) = V ba ( t ) = 5 . 33sin120 πt V . b) I ( t ) = V ba ( t ) /R = 5 . 33sin(120 πt ) / 250 = 21 . 3sin(120 πt ) mA 10.2. In Fig. 10.1, replace the voltmeter with a resistance, R . a) Find the current I that flows as a result of the motion of the sliding bar: The current is found through I = 1 R E · d L = 1 R d Φ m dt Taking the normal to the path integral as a z , the path direction will be counter-clockwise when viewed from above (in the a z direction). The minus sign in the equation indicates that the current will therefore flow clockwise , since the magnetic flux is increasing with time. The flux of B is Φ m = Bdvt , and so | I | = 1 R d Φ m dt = Bdv R (clockwise) b) The bar current results in a force exerted on the bar as it moves. Determine this force: F = Id L × B = d 0 Idx a x × B a z = d 0 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 = Fv = ( Bdv ) 2 R W The electrical power is P e = I 2 R = ( Bdv ) 2 R = P m
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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 φ direction about the closed path having corners at a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic flux will be: Φ = 1 0 1 0 300 µ 0 cos(3 × 10 8 t y ) dx dy = 300 µ 0 sin(3 × 10 8 t y ) | 1 0 = 300 µ 0 sin(3 × 10 8 t 1) sin(3 × 10 8 t ) Wb Then emf = d Φ dt = 300(3 × 10 8 )(4 π × 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 π ,0,0), (2 π ,2 π ,0), (0,2 π ,0): In this case, the flux is Φ = 2 π × 300 µ 0 sin(3 × 10 8 t y ) | 2 π 0 = 0 The emf is therefore 0 . 10.4. Conductor surfaces are located at ρ = 1cm and ρ = 2cm in free space. The volume 1cm < ρ < 2cm contains the fields H φ = (2 )cos(6 × 10 8 πt 2 πz ) A/m and E ρ = (240 π/ρ )cos(6 × 10 8 πt 2 πz ) V/m.
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