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Homework 10 Solutions

Homework 10 Solutions - Chapter 30 1 The flux B = BA cos...

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1165 Chapter 30 1. The flux Φ B BA = cos θ does not change as the loop is rotated. Faraday’s law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is zero. 2. Using Faraday’s law, the induced emf is ( ) ( ) ( )( )( ) 2 2 2 0.12m 0.800T 0.750m/s 0.452V. B d r d BA d dA dr B B rB dt dt dt dt dt π ε π π Φ = − = − = − = − = − = − = 3. The total induced emf is given by ( ) 2 0 0 0 2 ( ) ( ) 1.5 A (120)(4 T m A)(22000/m) 0.016m 0.025 s 0.16V. B d dB d di di N NA NA ni N nA N n r dt dt dt dt dt ε μ μ μ π Φ = − = − = − = − = − = − × = -7 p 10 p Ohm’s law then yields | |/ 0.016 V /5.3 0.030 A i R ε = = Ω = . 4. (a) We use ε = – d Φ B / dt = – π r 2 dB / dt . For 0 < t < 2.0 s: ( ) 2 2 2 0.5T 0.12m 1.1 10 V. 2.0s dB r dt ε = − = − = − × p p (b) For 2.0 s < t < 4.0 s: ε dB / dt = 0. (c) For 4.0 s < t < 6.0 s: ε = − = − F H G I K J = × p p r dB dt 2 2 2 012 05 6 0 4 0 11 10 . . . . . . m T s s V b g 5. The field (due to the current in the straight wire) is out of the page in the upper half of the circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle.
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