22_Chapter 20 HomeworkCH20 Induced Voltages and Inductance

22_Chapter 20 HomeworkCH20 Induced Voltages and Inductance...

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Unformatted text preview: 196 CHAPTER 20 20.58 When A and B are 3.00 m apart, the area enclosed by the loop consists of four triangular sections, each having hypotenuse of 3.00 m, altitude of 1.50 m, and base of 2 2 ( 3.00 m ) − (1.50 m ) = 2.60 m The decrease in the enclosed area has been A 3.00 m 3.00 m 3.00 m 3.00 m B 2 1 ∆A = Ai − A f = ( 3.00 m ) − 4 ( 1.50 m ) ( 2.60 m ) = 1.21 m 2 2 The average induced current has been I av = ε av ( ∆Φ B = R R ∆t ) = B ( ∆A ∆t ) R = ( 0.100 T ) ( 1.21 m 2 0.100 s ) 10.0 Ω = 0.121 A As the enclosed area decreases, the flux (directed into the page) through this area also decreases. Thus, the induced current will be directed clockwise around the loop to create additional flux directed into the page through the enclosed area. 20.59 If d is the distance from the lightning bolt to the center of the coil, then ε av = N ( ∆Φ B ) ∆t = N ( ∆B ) A N µ0 ( ∆I ) 2π d A N µ0 ( ∆I ) A = = 2π d ( ∆t ) ∆t ∆t 2 100 ( 4π × 10 −7 T ⋅ m A ) ( 6.02 × 106 A − 0 ) π ( 0.800 m ) = −6 2π ( 200 m ) ( 10.5 × 10 s ) = 1.15 × 10 5 V= 115 kV 20.60 The flux through the surface area of the tent is the same as that through the tent base. Thus, as the tent is flattened, the change is flux is ∆Φ B = B ( ∆Abase ) = B L ( 2L ) − L ( 2 ⋅ L cosθ ) = 2 L2 B ( 1 − cosθ ) The magnitude of the average induced emf is then ε av 2 ∆Φ B 2 L B ( 1 − cosθ ) 2 ( 1.5 m ) ( 0.30 T ) ( 1 − cos 60° ) = = = = 6.8 V 0.10 s ∆t ∆t 2 ...
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This document was uploaded on 10/20/2011.

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