chp20solutions

chp20solutions - Chapter 20 Induced Voltages and Inductance...

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Unformatted text preview: Chapter 20 Induced Voltages and Inductance Problem Solutions 20.1 The magnetic flux through the area enclosed by the loop is ( ) ( ) ( ) 2 2 2 2 cos cos0 0.30 T 0.25 m 5.9 10 T m B BA B r θ π π-   Φ = = ° = = × ⋅   20.5 (a) Every field line that comes up through the area A on one side of the wire goes back down through area A on the other side of the wire. Thus, the net flux through the coil is zero (b) The magnetic field is parallel to the plane of the coil , so 90.0 θ = ° . Therefore, cos cos90.0 B BA BA θ Φ = = ° = 20.8 ( ) ( ) ( ) 2 3 4 3 1.5 T 1.6 10 m cos0 cos 1.0 10 V 0.10 mV 120 10 s B B A t t π θ ε---  - × ° ΔΦ Δ     = = = = × = Δ Δ × 20.11 The magnitude of the induced emf is ( ) cos B B A t t θ ε Δ ΔΦ = = Δ Δ If the normal to the plane of the loop is considered to point in the original direction of the magnetic field, then 0 and 180 i f θ θ = ° = ° . Thus, we find ( ) ( ) ( ) 2 2 0.20 T cos180 0.30 T cos0 0.30 m 9.4 10 V 94 mV 1.5 s π ε- ° - ° = = × = 175 176       CHAPTER 20 20.13 The required induced emf is ( )( ) 0.10 A 8.0 0.80 V IR ε = = Ω = . From cos B B NA t t θ ε ΔΦ Δ   = =   Δ Δ   ( ) ( )( ) 0.80 V 2.7 T s cos 75 0.050 m 0.080 m cos0 B t NA θ ε Δ = = = Δ °     20.16 The magnitude of the average emf is ( ) ( ) ( ) ( ) ( ) 4 2 cos 200 1.1 T 100 10 m cos0 cos180 44 V 0.10 s B NBA N t t θ ε- Δ   ΔΦ   = = Δ Δ × ° - ° = = Therefore, the average induced current is 44 V 8.8 A 5.0 I R ε = = = Ω 20.18 From B v ε = , the required speed is ( )( ) ( )( ) 0.500 A 6.00 1.00 m s 2.50 T 1.20 m IR v B B ε Ω =...
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chp20solutions - Chapter 20 Induced Voltages and Inductance...

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