ch20solnsAM

# ch20solnsAM - Problem Solutions 20.1 The magnetic flux...

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175 Problem Solutions 20.1 The magnetic flux through the area enclosed by the loop is Φ B = BA cos θ = B π r 2 ( ) cos0 ° = 0.30 T ( ) 0.25 m ( ) 2 = 5.9 × 10 2 T m 2 20.2 The magnetic flux through the loop is given by Φ B = BA cos where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and is the angle the magnetic field makes with the normal to the plane of the loop. Thus, Φ B = BA cos = 5.00 × 10 5 T ( ) 20.0 cm 2 10 -2 m 1 cm 2 cos = 1.00 × 10 7 T m 2 ( ) cos (a) When B ur is perpendicular to the plane of the loop, = 0 ° and Φ B = 1.00 × 10 7 T m 2 (b) If = 30.0 ° , then Φ B = 1.00 × 10 7 T m 2 ( ) cos30.0 ° = 8.66 × 10 8 T m 2 (c) If = 90.0 ° , then Φ B = 1.00 × 10 7 T m 2 ( ) cos90.0 ° = 0 20.3 The magnetic flux through the loop is given by Φ B = BA cos where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and is the angle the magnetic field makes with the normal to the plane of the loop. Thus, Φ B = BA cos = 0.300 T ( ) 2.00 m ( ) 2 cos50.0 ° = 7.71 × 10 1 T m 2 20.6 The magnetic field generated by the current in the solenoid is B = μ 0 nI = 4 × 10 7 T ( ) 250 0.200 m 15.0 A ( ) = 2.36 × 10 2 T and the flux through each turn on the solenoid is Φ B = BA cos = 2.36 × 10 2 T ( ) 4.00 × 10 2 m ( ) 2 4 cos0 ° = 2.96 × 10 5 T m 2

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176 CHAPTER 20 20.8 ε = ΔΦ B Δ t = Δ B ( ) A cos θ Δ t = 1.5 T 0 ( ) π 1.6 × 10 3 m ( ) 2 cos0 ° 120 × 10 3 s = 1.0 × 10 4 V = 0.10 mV 20.10 = ΔΦ B Δ t = B Δ A ( ) cos Δ t = 0.15 T ( ) 0.12 m ( ) 2 0 cos0 ° 0.20 s = 3.4 × 10 2 V = 34 mV 20.13 The required induced emf is = IR = 0.10 A ( ) 8.0 Ω ( ) = 0.80 V . From = ΔΦ B Δ t = Δ B Δ t NA cos Δ B Δ t = NA cos = 0.80 V 75 ( ) 0.050 m ( ) 0.080 m ( ) cos0 ° = 20.14 The initial magnetic field inside the solenoid is B = μ 0 nI = 4 × 10 7 T ( ) 100 0.200 m 3.00 A ( ) = 1.88 × 10 3 T (a) Φ B = BA cos = 1.88 × 10 3 T ( ) 1.00 × 10 2 m ( ) 2 cos0 ° = 1.88 × 10 7 T m 2 (b) When the current is zero, the flux through the loop is Φ B = 0 and the average induced emf has been = ΔΦ B Δ t = 1.88 × 10 7 T m 2 0 3.00 s = 6.28 × 10 8 V
Induced Voltages and Inductance 177 20.17 If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it makes with the normal to the plane of the coil is θ = 62.0 ° . Thus, ε = N ΔΦ B ( ) Δ t = NB Δ A ( ) cos Δ t = 200 50.0 × 10 -6 T ( ) 39.0 cm 2 ( ) 1 m 2 10 4 cm 2 ( ) cos62.0 ° 1.80 s = 1.02 × 10 5 V = 10.2 μ V 20.18 From = B l v , the required speed is v = B l = IR B

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## This note was uploaded on 09/13/2010 for the course PHYS PHYS 1B taught by Professor Berman during the Summer '09 term at UCSD.

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ch20solnsAM - Problem Solutions 20.1 The magnetic flux...

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